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Book Title: LUMO Mathematics 9
ISBN: 978-93-49697-17-1
Published by Lumo Learning Private Limited
Corporate Office Address:
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S. No.
1.
REAL NUMBERS
UNIT 1: NUMBER SYSTEMS
1. Review of representation of natural numbers, integers, rational numbers on the number line. Representation of terminating/ non-terminating recurring decimals on the number line through successive magnification, Rational numbers as recurring/terminating decimals. Operations on real numbers.
2. Examples of non-recurring/non-terminating decimals. Existence of non-rational numbers (irrational numbers) such as 2 , 3 and their representation on the number line. Explaining that every real number is represented by a unique point on the number line and conversely, viz. every point on the number line represents a unique real number.
3. Definition of nth root of a real number.
4. Rationalization (with precise meaning) of real numbers of the type 1 abx + and 1 xy + (and their combinations), where x and y are natural numbers and a and b are integers.
5. Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing learner to arrive at the general laws.)
• Develops a deeper understanding of numbers, including the set of real numbers and its properties.
• Recognizes and appropriately uses powers and exponents.
• Computes powers and roots and applies them to solve problems.
• Differentiates rational and irrational numbers based on decimal representation.
• Represents rational and irrational numbers on the number line.
• Rationalizes real number expressions such as 1 abx + and 1 xy + , where x, y are natural numbers and a, b are integers.
• Applies laws of exponents.
1. POLYNOMIALS
1. Definition of a polynomial in one variable, with examples and counter examples. Coefficients of a polynomial, terms of a polynomial and zero polynomial.
2. Degree of a polynomial.
3. Constant, linear, quadratic and cubic polynomials. Monomials, binomials, trinomials. Factors and multiples.
4. Zeroes of a polynomial.
5. Motivate and State the Remainder Theorem with examples.
6. Statement and proof of the Factor Theorem. Factorization of ax2 + bx + c,a ≠ 0 where a, b and c are real numbers, and of cubic polynomials using the Factor theorem.
7. Recall of algebraic expressions and identities. Verification of identities:
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
(x ± y)3 = x3 ± y3 ± 3xy(x ± y)
x3 + y3 = (x + y)(x2 xy + y2)
x3 y3 = (xy)(x2 + xy + y2
x3 + y3 + z3 3xyz
= (x + y + z)(x2 + y2 + z2 xyyzzx) and their use in factorization of polynomials.
2. LINEAR EQUATIONS IN TWO VARIABLES
1. Recall of linear equations in one variable.
2. Introduction to the equation in two variables. Focus on linear equations of the type ax + by + c = 0. Explain that a linear equation in two variables has infinitely many solutions and justify their being written as ordered pairs of real numbers, plotting them and showing that they lie on a line.
• Learns the art of factoring polynomials.
• Defines polynomials in one variable.
• Identifies different terms and different types of polynomials.
• Finds zeros of a polynomial
• Proves factor theorem and applies the theorem to factorize polynomials.
• Proves and applies algebraic identities up to degree three.
• Visualizes solutions of a linear equation in two variables as ordered pair of real numbers on its graph.
UNIT III: COORDINATE GEOMETRY
• Describes and plot a linear equation in two variables.
1. Coordinate Geometry:
1. The Cartesian plane, coordinates of a point
2. Names and terms associated with the coordinate plane, notations.
• Specifies locations and describes spatial relationships using coordinate geometry.
• Describes cartesian plane and its associated terms and notations.
1. INTRODUCTION TO EUCLID’S GEOMETRY
1. History - Geometry in India and Euclid's geometry. Euclid's method of formalizing observed phenomenon into rigorous Mathematics with definitions, common/obvious notions, axioms/ postulates and theorems.
2. The five postulates of Euclid. Equivalent versions of the fifth postulate. Showing the relationship between axiom and theorem, for example:
(a) Given two distinct points, there exists one and only one line through them. (Axiom)
(b) (Prove) Two distinct lines cannot have more than one point in common. (Theorem)
2. LINES AND ANGLES
1. (State without proof) If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and the converse.
2. (P rove) If two lines intersect, vertically opposite angles are equal.
3. (State without proof) Lines which are parallel to a given line are parallel.
3. TRIANGLES
1. (State without proof) Two triangles are congruent if any two sides and the included angle of one triangle is equal (respectively) to any two sides and the included angle of the other triangle (SAS Congruence).
2. (Prove) Two triangles are congruent if any two angles and the included side of one triangle is equal (respectively) to any two angles and the included side of the other triangle (ASA Congruence).
UNIT IV: GEOMETRY
• Proves theorems using Euclid’s axioms and postulates—for triangles, quadrilaterals, and circles and applies them to solve geometric problems.
• Understands historical relevance of Indian and Euclidean Geometry.
• Defines axioms, postulates, theorems with reference to Euclidean Geometry.
• derives proofs of mathematical statements particularly related to geometrical concepts, like parallel lines by applying axiomatic approach and solves problems using them.
• Visualizes, explains and applies relations between different pairs of angles on a set of parallel lines and intersecting transversal.
• Solves problems based on parallel lines and intersecting transversal.
• Describe relationships including congruency of two-dimensional geometrical shapes (lines, angle, triangles) to make and test conjectures and solve problems.
• derives proofs of mathematical statements particularly related to geometrical concepts triangles by applying axiomatic approach and solves problems using them.
• Visualizes and explains congruence properties of two triangles.
• Applies congruency criteria to solve problems.
3. (State without proof) Two triangles are congruent if the three sides of one triangle are equal (respectively) to three sides of the other triangle (SSS Congruence).
4. (State without proof) Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle. (RHS Congruence).
5. (Prove) The angles opposite to equal sides of a triangle are equal.
6. (State without proof) The sides opposite to equal angles of a triangle are equal.
4. QUADRILATERALS
1. (Prove) The diagonal divides a parallelogram into two congruent triangles.
2. (State without proof) In a parallelogram opposite sides are equal, and conversely.
3. (State without proof) In a parallelogram opposite angles are equal, and conversely.
4. (State without proof) A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal.
5. (State without proof) In a parallelogram, the diagonals bisect each other and conversely.
6. (State without proof) In a triangle, the line segment joining the mid points of any two sides is parallel to the third side and is half of it and (State without proof) its converse.
5. CIRCLES
1. (Prove) Equal chords of a circle subtend equal angles at the center and (State without proof) its converse.
2. (State without proof) The perpendicular from the center of a circle to a chord bisects the chord and conversely, the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.
• derives proofs of mathematical statements particularly related to geometrical concepts of quadrilaterals by applying axiomatic approach and solves problems using them.
• Visualizes and explains properties of quadrilaterals.
• Solves problems based on properties of quadrilaterals.
• Proves theorems about the geometry of a circle, including its chords and subtended angles
• Visualizes and explains properties of circles.
• Solves problems based on properties of circle.
3. (State without proof) Equal chords of a circle (or of congruent circles) are equidistant from the center (or their respective centers) and conversely.
4. (Prove) The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
5. (State without proof) Angles in the same segment of a circle are equal.
6. (State without proof) If a line segment joining two points subtends equal angle at two other points lying on the same side of the line containing the segment, the four points lie on a circle.
7. (State without proof) The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse.
1. AREAS
1. Area of a triangle using Heron's formula (without proof).
2. SURFACE AREAS AND VOLUMES
1. Surface areas and volumes of spheres (including hemispheres) and right circular cones.
1. STATISTICS
1. Bar graphs
2. Histograms (with varying base lengths)
3. Frequency polygons.
UNIT V: MENSURATION
• Visualizes, represents, and calculates the area of a triangle using Heron’s formula.
• Visualizes and uses mathematical thinking to discover formulas to calculate surface areas and volumes of solid objects (spheres, hemispheres and right circular cones).
UNIT VI: STATISTICS
• Draws and interprets bar graph, histogram and frequency polygon.
• States and applies Heron’s Formula to find area of a triangle.
• Solves problems based on surface areas and volumes of three-dimensional shapes (spheres/hemisphere, right circular cones).
• Represents data using Bar Graph, Histogram and frequency polygon.
Time: 3 Hours
MATHEMATICS QUESTION PAPER DESIGN
CLASS – IX (2025-26)
Remembering: Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers.
Understanding: Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas.
Applying: Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.
Analysing:
Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations.
Evaluating:
Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria.
Creating:
Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions.
Max. Marks: 80
Chapter Walk Through
QR Code: Provides access to the digital content of each chapter.
Chapter Number and Chapter Name
Chapter at a Glance: A short and crisp summary of a chapter with all the important ideas/concepts/relations.
NCERT Zone: Detailed model answers to NCERT questions (Intext and Exercises) given in simple, easy to understand language.
Constructed Response Questions: A set of solved questions (Very Short Answers, Short Answers, Long Answers), that are based on constructed response.
Competency Based Questions: A set of questions (Multiple Choice, Assertion-Reason and Case-based questions), that are based on competencies defined by CBSE.
Additional Practice Questions: A set of unsolved question, covering the entire chapter for practice.
Challenge Yourself: A unique set of higher order questions that are designed to challenge you and your understanding of the chapter. Additional Questions can be found on scanning the QR code.
Answers to Unsolved Problems: Solution to objective/numerical questions from the Additional Practice Questions and Challenge Yourself sections. Complete solutions can be accessed on scanning the QR code.
Complete Exemplar Solutions: Scan the QR code at end of each chapter for complete NCERT Exemplar solutions.
Self-Assessment: Assessment at the end of each chapter to help learners prepare for tests and exams. Complete Solutions will be available on QR code.
1 Number Systems
Chapter at a Glance
1. Natural Numbers: Counting numbers are called natural numbers. The collection of natural numbers is denoted by N and is written as N = {1, 2, 3, 4, 5, 6, …}.
2. Whole Numbers: The collection of all natural numbers along with zero is called the set of whole numbers. It is denoted by W, and written as: W = {0, 1, 2, 3, 4, 5, …}.
3. Integers: The collection of all natural numbers, zero, and the negatives of natural numbers is called the set of integers. It is denoted by Z, and written as: Z = {…, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, ….}.
4. Rational Numbers: The numbers of the form p q , where p and q are integers and q ≠ 0, are known as rational numbers. The collection of rational numbers is denoted by Q
5. Irrational Numbers: A number is called irrational number, if it cannot be written in the form p q , where p and q are integers and q ≠ 0. For example: 2 , 3 , π, etc.
6. Real Numbers: A collection of rational and irrational numbers.
7. Decimal Expansion of Real Numbers
(a) A rational number has a decimal expansion that is either terminating (ends after a finite number of digits) or non-terminating but recurring (infinite decimal with a repeating pattern).
(b) Conversely, any number with a terminating or non-terminating recurring decimal expansion is a rational number.
(c) An irrational number has a decimal expansion that is non-terminating and non-recurring (infinite decimal without any repeating pattern).
(d) Likewise, any number whose decimal expansion is non-terminating and non-recurring is an irrational number.
8. If r is a rational number and s is an irrational number then r + s and rs are irrational numbers and rs and r s are also irrational numbers, r ≠ 0.
9. Laws of Exponents for Real Numbers
If a, b are positive real numbers and m, n are rational numbers. Then,
10. For positive real numbers a and b
(a) abab =
(c) () () ababab +−=−
(e) () () 2 ababab +−=−
()2 2 abaabb +=++
11. If a and b are positive integers, then
(a) Rationalizing factor of 1 a is a
(b) Rationalizing factor of 1 ab ± is ab
(c) Rationalizing factor of 1 ab ± is ab
NCERT Zone
NCERT Exercise 1.1
1. Is zero a rational number? Can you write it in the form p q , where p and q are integers and 0 q ≠ ?
Sol. Yes, 0 is a rational number. It can be written as 0 1 , 00 , 23 , etc., in the form p q , where p and q are integers and 0. q ≠
2. Find six rational numbers between 3 and 4.
Sol. To find six rational numbers between 3 and 4 first we need to make the denominators 617. +=
Therefore, 37214728 3 , 4 7777 ×× ====
So, six rational numbers can be found between 3 and 4 by changing the numerators between 21 and 28.
Hence, the numbers are 222324252627 ,,,,,. 777777
3. Find five rational numbers between 3 5 and 4 5
Sol. Since, we want five rational numbers, we write 3 5 and 4 5 .
So, multiply in numerator and denominator by 5 + 1 = 6 and we get, 36 318 55630 × == × and 46 424 55630 × == × We know that, 19 < 20 < 21 < 22 < 23
⇒ 18192021222324 30 303030303030 <<<<<<
Hence, 5 rational numbers between 3 5 and 4 5 are:
⇒ 1920212223 , , , , 30 30303030
4. State whether the following statements are true or false. Give reasons for your answers.
(a) Every natural number is a whole number.
(b) Every integer is a whole number.
(c) Every rational number is a whole number.
Sol. (a) True, because whole numbers are the collection of all natural numbers in addition to 0. So, every natural number is a whole number.
(b) False, because negative numbers are also integers. Whole numbers cannot be negative. So, every integer is not a whole number.
(c) False, because numbers like 12462 ,,,, 33583 , etc., are rational numbers but not whole numbers.
NCERT Exercise 1.2
1. State whether the following statements are true or false. Justify your answers.
(a) Every irrational number is a real number.
(b) Every point on the number line is of the form m , where m is a natural number.
(c) Every real number is an irrational number.
Sol. (a) True, since all the rational and irrational numbers together make a real number.
(b) False, no negative number can be the square root of any natural number.
(c) False, for example 2 is real but not irrational.
2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Sol. No, because it can be rational also. For example: 42, 93.==
3. Show how 5 can be represented on the number line.
Sol. To show 5 on a number line, first we need to take a length of 2 units starting from 0 on the number line towards the positive numbers. Then we need to draw a line of 1 unit length perpendicular to it. The hypotenuse of the triangle thus formed is of length 5 .
Finally with the help of divider we draw a length equal to the hypotenuse of 5 units on the number line as shown in picture below.
NCERT Exercise 1.3
1. Write the following in decimal form and say what kind of decimal expansion each has:
2. You know that 1 0.142857 7 = . Can you predict what the decimal expansions of 23456 , , ,, 77777 are, without actually doing the long division? If so, how?
Sol. 21 20.285714
3. Express the following in the form p q , where p and q are integers and 0 q ≠ .
(a) 0.6 (b) 0.47 (c) 0.001
Sol. (a) 0.60.6666 x == {as 6 is repeating}
We multiply both the sides by 10
106.6 x == 6.0 0.6 + 106xx =+
106 xx−=
96 x = 62 93 x ==
(b) 0.470.4777 x == {as 7 is repeating}
We multiply both the sides by 10.
104.7 x =
104.30.47 x =+ 104.3xx =+
104.3 xx−= 4.343 990 x ==
(c) 0.0010.001001001 x == {as 001 is repeating}
We multiply both the sides by 1000.
10001.001 x = 10001xx =+ 9991 x = 1 999 x =
4. Express 0.99999 .... in the form p q . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Sol. 0.99999.0.9 x =…=
As only one digit is repeating so we multiply both sides by 10.
109.9 x = 109xx =+
109 xx−= 99 x = 9 1 9 x ==
We got the answer as 1. Yes, many students are surprised that 0.99999… is actually equal to 1 not just close to 1. But it’s mathematically true and accepted.
5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1 17 ? Perform the division to check your answer.
Sol. After dividing 1 by the 17, the maximum number of digits in the repeating block we get is () 16 17. < 1 0.0588235294117647 17 =
As we found there are 16 digits in the repeating block.
6. Look at several examples of rational numbers in the form () 0 pq q ≠ , where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Sol. 2327 0.4, 0.03,1.6875 510016 ===
012240 22332727 , , 510016252525 === ×××
We observe that the denominators of the above rational numbers are in the form of 2a × 5b, where a and b are whole numbers. Hence if q is in the form 2a × 5b then p q is a terminating decimal.
7. Write three numbers whose decimal expansions are non-terminating non-recurring.
Sol. (a) 6.341411411141114…….
(b) 0.101020203030……….. (c) 3.1416
8. Find three different irrational numbers between the rational numbers 59and. 711
Sol. After dividing both the fractions we get: 5 0.714285 7 = and 9 0.81 11 =
There can be infinite number of irrational numbers between both the given rational numbers. We can choose any three among those.
0.714283……..
0.756843……..
0.804246……..
9. Classify the following numbers as rational or irrational:
(a) 23 (b) 225 (c) 0.3796
(d) 7.478478….. (e) 1.101001000100001….
Sol. (a) Irrational (b) Rational
(c) Rational (d) Rational (e) Irrational
NCERT Exercise 1.4
1. Classify the following numbers as rational or irrational:
By using the Identity ()() abab −+ , we get 2 33936 −=−=
(c) ()2 52 +
By using the Identity ()2 22 2 ababab +=++ , we get ()2 5252252 +=++ = 7210 +
(d) () ()5252 −+
By using property ()() 22ababab −+=− , we get () ()() 525252 3 −+=−=
3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d ). That is, c d π= . This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Sol. π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d), that is, π = c d . Hence, we see that π is a rational number as it is expressed in the form of p q . But, we know that π is an irrational number. Let’s resolve the contradiction.
Writing π as 22 7 is only an approximated value and so we can’t conclude that it is in the form of a rational number. In fact, the value of π is calculated as the non-terminating, non-recurring decimal number as π= 3.14159... whereas, if we calculate the value of 22 7 , it gives 3.142857 and hence π is not exactly equal to 22 7 . In conclusion, π is an irrational number.
4. Represent 9.3 on the number line.
Sol. To represent 9.3 , first we need to draw a segment of 9.3 units on the number line. Let A represent 9.3.
Then, extend this segment by 1 unit to reach a new point A', making the total length OA' = 10.3 units.
1. Find:
2. Find:
Next, find the midpoint of OA', which is 10.3 2 = 5.15 units from point O. With this midpoint as the center and a radius of 5.15 units, draw a semicircle passing through points O and A'. At point A (located at 9.3 units), draw a perpendicular line to intersect the semicircle at point B. The length of segment AB is 9.3 units. Finally, with O as the center and AB as the radius, mark this length on the number line. This point represents 9.3 on the number line.
5. Rationalize the denominators of the following: (a) 1
1. Every rational number is (a) a natural number (b) an integer (c) a real number (d) a whole number
2. Decimal representation of a rational number cannot be (a) terminating (b) non-terminating (c) non-terminating repeating (d) non-terminating non-repeating (NCERTExemplar)
3. Between two rational numbers (a) there is no rational number (b) there is exactly one rational number (c) there are infinitely many rational numbers (d) there are only rational numbers and no irrational numbers (NCERTExemplar)
4. The decimal expansion of the number 2 is (a) a finite decimal (b) 1.41421 (c) non-terminating recurring (d) non-terminating non-recurring (NCERTExemplar)
5. Which of the following is a rational number? (a) 2 3 (b) 1 5 (c) 13 5 (d) 2 3 (CBSEQB)
6. Which of the following is irrational? (a) 4 9 (b) 12 3 (c) 7 (d) 81 (NCERTExemplar)
7. The number () ()3232 −+ is (a) a natural number (b) an irrational number (c) a rational number (d) both (a) and (c)
8. A rational number between 2 and 3 is (a) 32 2 (b) 1.52 (c) 1.92 (d) 32 2 ×
9. The product of any two irrational numbers is (a) always an irrational number (b) always a rational number (c) always an integer (d) sometimes rational, sometimes irrational (NCERTExemplar)
10. The p q form of the number 0.9 is (a) 1 9 (b) 9 10 (c) 9 100 (d) 1
11. The value of 233 + is (a) 26 (b) 6 (c) 33 (d) 46 (NCERTExemplar)
7. Find the value of a in the following: 6 323 3223 a =− (NCERTExemplar) Sol. 663223 322332233223 + =× −−+ () ()() () 2 2 6322363223 1812 3223 ++ ==
Therefore, 3223323 a +=− or 2 a =−
8. Insert a rational number and an irrational number between the following:
(a) 6.375289 and 6.375738
(b) 2 and 3 (NCERTExemplar)
Sol. (a) Rational number between 6.375289 and 6.375738 = 6.3753
Irrational number between 6.375289 and 6.375738 = 6.375414114111...
(b) Rational number between 2 and 3 = rational number between 1.41 and 1.732 = 1.5
Irrational number between 2 and 3 = irrational number between 1.41 and 1.732 = 1.58585558...
9. Find the value of x
Long Answer Questions
3. Represent 8.5 on the number line.
Sol. Mark the distance 8.5 units from a fixed point A on a given line to obtain a point B such that AB = 8.5 units.
From B, mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark the point O
Draw a semi-circle with center O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D
Then, BD = 8.5 units.
Let us treat the line BC as the number line, with B as zero, C as 1, and so on. Draw an arc with center B and radius BD = 8.5 units, which intersects the number line in E.
Then, E represents 8.5 on the number line.
4. If 526 a =+ and 1 b a = , then what will be the value of 22ab + ?
Sol. 526 a =+ 111526 526526526 b a ===× ++−
= ()2 2 526526 526 2524 526 ==−
Therefore, 222()2 ababab +=+−
Here, ()() 52652610 ab+=++−= () () 22 5265265(26)25241 ab =+−=−=−=
Therefore, 2221021100298ab+=−×=−=
5. Find the value of: 231 345 412 (256) (216) (243) ++ (NCERTExemplar)
4. On a number line, 3 18 is halfway located between 0 and a . What is the value of a?
(a) 2 (b) 4 (c) 6 (d) 8
5. Which of the following statements is/are correct?
(a) Every integer is a rational number.
(b) Every rational number is an integer.
(c) A real number is either a rational or irrational number.
(d) Every whole number is a natural number.
6. Which of these is equivalent to 1 3 5 4 9 27 × ?
(d) 33 45 3
7. Which of the following rational numbers is equivalent to a decimal that terminates?
(a) 1 3 (b) 8 9
8. Which of these is a way to convert 15 6320 + to an equivalent number whose denominator is a rational number?
(a) 153725 63203725 × +−
(b) 153745 63203745 × +−
(c) () ()() 153745 63203745 +− +×−
(d) () ()() 153725 63203725 +− +×−
9. How many digits are there in the repeating block of digits in the decimal expansion of 17 7 ?
(a) 16 (b) 6 (c) 26 (d) 7
10. The value of 1 16 4 4 256 81 x y
is
Assertion-Reason Based Questions
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): The decimal value of 437 999 is 0.437437…which is rational.
Reason (R): A non-terminating repeating decimal number is a rational number.
2. Assertion (A): If a = 3 and b = 4, then the value of ba is 64.
Reason (R): If a = 4 and b = 5, then the value of 22ab +
Case Study Based Questions
1. Read the following and answer questions from (a) to (d).
Sameer went to a book fair. He was attracted towards the cover page of the book with ‘π’ written on it. He tries to find about this unique number.
(a) What is π?
(b) What is the nature of the product of any two irrational numbers?
(c) Find a rational number between 3 and 5 .
Answers
Multiple Choice Questions
1. (c) 2 let 1.999 x =…
Multiply both sides by 10: 1019.999 x =…
Now subtract the first equation from the second: 1019.9991.999 xx−=…−… 918 x = 18 2 9 x ==
So, the value of 1.999... in the form of p q is: 2.
2. (c) 5 3 let 2.66 x =
Multiply both sides by 10 to shift the decimal: 1026.6 x =
Subtract the two equations: 1026.62.66 xx−=− 924 x = 248 93 x == Let 0.99 y =
(d) Find rationalizing factor for the denominator of the expression 1 7 2 .
2. Read the following and answer questions from (a) to (d).
Seerat and Meera are playing with matchsticks by making different figures from them.
Seerat kept one matchstick horizontally and then two matchsticks vertically as shown in the figure. Meera joined the open ends with a string. They ask their elder brother whether he can find the length of this string. He replied, length of this string can be calculated using Pythagoras theorem and it is equal to 2122 41 5 +=+=
(a) What is 5 ?
(b) Evaluate 355 +
(c) What is the decimal representation of an irrational number? Explain with a reason.
(d) Evaluate () 2 25 .
3. (c) 1 310 First, 21 4 2 33342 == 11 1 5 25 2 33 × = 1 310 = 4. (a) 2 A number located halfway between 0 and a on the number line is 0 2 a + = 1 2 a 13 2 18 a =⇒ 1 2 2 a ×=⇒ 2 2 a = Simplify, 2222 2 2 22 ×== ⇒ 2 a = . Hence, 2 a = String
Multiply both sides by 10: 109.9 y =
Subtract: 109.90.99 yy−=− 99 y = 1 y = Now, subtract 8 2.660.991 3 −=− 835 333 =−=
5. (a) and (c)
(a) Every integer is a rational number. Yes 1 Becauseanyintegerncanbewrittenan s
(b) Every rational number is an integer. No 1 2 . Becauseisrationalbutnotaninteger
(c) A real number is either a rational or irrational number. Yes (That’sthedefinitionofrealnumbers!)
(d) Every whole number is a natural number. No (Zeroisawholenumberbutnotanatural numbertraditionally.)
6. (a) 33 25 3 +
Now multiply:
7. (c) 3 8 A rational number has a terminating decimal if, after simplifying, its denominator (in lowest terms) has only 2 and/or 5 as prime factors. Now check each option:
We want to rationalize the denominator, which means we multiply numerator and denominator by 6320 . Thus, 156320 63206320 × +− () () () 156320 63206320 = +− () () 63206320 +− ()() 22 6320632043=−=−=
Thus, () 156320 43
Please note, 639737 =×= also 204525 =×=
Hence the correct one is 153725 37253725 × +−
9. (b) 6 Simplify, 17 2.428571428571 7 = …
Here, the repeating part is 428571. Clearly, the repeating block is 428571, which has 6 digits.
Apply power on numerator and denominator separately () () 1 4 4 1 16 4 81 256 y x = Thus, 4 3 4 y x =
Assertion-Reason Based Questions
1. (a) Both A and R are true, and R is the correct explanation of A. 437 999 = 0.437437437... (non-terminating but repeating decimal). 437 999 is a rational number (because it is of the form p q , where p and q are integers and q ≠ 0).
So, Assertion is correct.
A non-terminating repeating decimal is always a rational number. So, reason is also correct.
2. (b) Both A and R are true, but R is not the correct explanation of A. Given 4 a = , 5 b = Then, 222245162541ab+=+=+=
But Reason just says “the value of 22ab + ” — it does not state the value clearly (like “41”). Thus, the Reason is correct in concept, but it is not explaining the Assertion.
Assertion is about ba , and Reason is about 22ab + They are different topics.
Assertion is correct. Reason is correct. But Reason is NOT the explanation of Assertion.
Case Study Based Questions
1. (a) π is a non-terminating, non-repeating decimal, and cannot be expressed as p q .
(b) Sometimes the product of two irrational numbers is rational () example: 222 ×= Sometimes the product is irrational () example: 236 ×=
(c) 1.7322.2363.968 1.984 22 + =≈
This is between1.732 and 2.236, but 35 2 + is irrational.
Now, 532.2361.7320.504 0.252 222 ==≈
Way below 3 and 5 35 >
2 lies between 1.732 and 2.236 and 2 is rational.
(d) To rationalize 72 we multiply numerator and denominator by 72 + .
Additional Practice Questions
1. 24x = 162, then x is equal to (a) 2 (b) 3 (c) 1 (d) 4
2. The value of 15 20 7 7 × is (a) 5 7 (b) 5 7 (c) 35 7 (d) 35 7
3. If 322 x =+ then find
(a) 34
4. If 21.4142 = , then 21 is equal to 21 + (a) 2.4142 (b) 5.8282 (c) 0.4142 (d) 0.1718
5. Simplified value of 11 161644 ()×() (a) 0 (b) 1 (c) 4 (d) 16
6. Which of the following is equal to x ? (a) 12 7 7 x
()
7. Represent 7.3 on the number line.
8. Find the value of x, if 575693 757 x =
2. (a) 5 is an irrational number, because it cannot be expressed as a fraction and its decimal expansion is non-terminating and nonrepeating
(b) () 351531545 45 +=+==
(c) Irrational numbers have decimal expansions that:
• do not terminate (i.e., never end)
• do not repeat in a pattern.
(d) 2 2 11 25 625 25 == 111 62525 625 ==
9. Are there two irrational numbers whose sum and product both are rationals? Justify.
10. Find the value of 1218 × .
11. Find the value of a and b in the following: 23 6 3223 ab + =−
12. If x = 7 + 40 , find the value of x + 1 x .
13. If 2121 and 2121 ab+− == −+ then find the value of 22 4.abab +−
14. Rahul recently bought a new smartphone with 128 GB of storage. His phone showed the following usage: System files: 14.25 GB
Apps: 28.5 GB
Photos and videos: 43.75 GB
Miscellaneous: 9.5 GB
He also installed a storage cleaner app that claimed it could save up to 16 GB by removing duplicate files. His friend Anjali said, “That number seems to be a perfect square!” Rahul was confused about whether numbers like 14.25 or 16 are rational or irrational. To make things more interesting, Rahul noticed that the cleaner app’s rating on the store was 4.666..., and the app download size was π MB.
Based on the above information, answer the following questions:
(a) Identify one rational number from the above?
(b) Identify one irrational number from Rahul’s phone data and justify why it is irrational.
(c) Rahul claims the storage cleaner can clean 16 GB. Is this number rational? What is its value?
15. Vasu represents 4.5 on the number line PW. The length of TS = 1 unit. His representation is shown below.
X PQRSTUVW
Challenge Yourself
1. If ()() 2 2 2 33 933 27 1 27 32 n nn m ××− = × , prove that 1. mn−=
2. If 223 53135, xx ×= find the value of x.
Answers
Additional Practice Questions
1. (a) 2
2. (c) 35 7
3. (a) 34
4. (c) 0.4142
5. (b) 1
6. (c) () 2 3 3 x 7. AOC C D
1
8. x = 5
9. Yes
10. 6 6
11. a = 2, b = 5 6
Based on the above information, answer the following questions:
(a) Which letter represents 0 on the number line?
(b) Between which two points does 5.2 lie on this number line?
(c) Write three non-terminating rational numbers that lie between the points T and U.
Scan me
3. Express 3 325 −+ with rational denominator.
4. Show that 111 388776 −+ 11 5 6552 −+=
12. () 2 252 3 + 13. 30
14. (a) 14.25
(b) π is an irrational number because its decimal expansion is non-terminating and nonrepeating, and it cannot be written as a ratio of two integers.
(c) Yes, 164 = which is an integer, hence rational.
15. (a) Letter S
(b) 5.2 lies between point U and V on the number line.
(c) 1.3,1.6and1.9
Challenge Yourself
2. 3 3. 233032 4 +−
Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 60 mins
Multiple Choice Questions
Scan me for Solutions
Max. Marks: 40
[3 × 1 = 3Marks]
1. Which of the following is a rational number? (a) 1 + 3 (b) π (c) 23 (d) 0
2. A rational number between 3 and 3 is (a) 0 (b) 4.3 (c) 3.4 (d) 1.101100110001
3. If 72.646 = then 1 7 = ?
(a) 0.375 (b) 0.378 (c) 0.441 (d) none of these
Assertion-Reason Based Questions
[2 × 1 = 2Marks]
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
4. Assertion (A): Three rational numbers between 2 5 and 3 5 are 91011 , , 202020
Reason (R): A rational number between two rational numbers p and q is () 1 2 pq + .
5. Assertion (A): 3 is an irrational number.
Reason (R): Square root of a positive integer which is not a perfect square is an irrational number.
Very Short Answer Questions
6. Write the sum of 0.3 and 0.4 in the form p q , where p and q are integers and q ≠ 0.
7. If 7x = 1, then find the decimal expansion of x.
Short Answer Questions
8. Insert an irrational number between 2 5 and 1 2 .
9. Simplify: 333 3404320 5
10. Find the product of () ()52325 2. ++
11.
[2 × 2 = 4Marks]
[4 × 3 = 12Marks]
12. Express 1.320.35 + in the form p q , where p and q are integers and q ≠ 0.
13. Simplify: 2454 89 +
14. If 415 x =− , then find the value of 2 1 x x +
Case Study Based Questions
15. Democracy has given people a powerful right-that is to VOTE. In India, every citizen over 18 years of age has the right to vote. Instead of enjoying it as a holiday, one must vote if he/she truly wants to contribute to the nation-building process and bring about a change. A survey was done in a small area in which 922xx+− voters were men and 5 9 2x + voters were women.
(a) Find x, if number of men is equal to number of women.
(b) Calculate the value of () 6 3 2 1 x xx ×−
(c) x 2 . xq × 0.33... equals
2 Polynomial
Chapter at a Glance
● Polynomial: A polynomial p(x) in one variable x is an algebraic expression in x of the form p(x) = a n xn + a n - 1 xn - 1 + a n - 2 xn - 2 ++ a 2 x 2 + a 1 x + a 0,
Where a 0, a 1, a 2, , a n are constants and a n ≠ 0, where n is a positive integer.
● Terms of Polynomial: The parts of a polynomial separated by + or - signs are called terms.
Example: In 9x 6 - 3x 5 + x + 5, the terms are 9x 6 , -3x 5 , x, and 5.
● Degree of Polynomial: The highest power of variable x in a polynomial p(x) is called the degree of the polynomial p(x).
Example: The degree of the polynomial 9x 6 - 3x 5 + x + 5 is 6.
Types of polynomials
1. On the basis of terms
(a) Monomial: A polynomial having one term is called a monomial e.g., 4x, 5y 2, 7x 3, 3x 2y, and 2xyz
(b) Binomial: A polynomial having two terms is called a binomial e.g., x + 4, x 2 - 9, x + y, x 2 - y 2 etc.
(c) Trinomial: A polynomial having three terms is called a trinomial e.g., x 2 + 5xy + y 2 , x + y + z, etc.
2. On the basis of degree:
(a) Constant Polynomial: A polynomial of degree 0 is called a constant polynomial.e.g. 2 7, 3 etc.
(b) Linear Polynomial: A polynomial of degree 1 is called a linear polynomial e.g., 7x + 9, 8y + 2, 6x etc.
(c) Quadratic Polynomial: A polynomial of degree 2 is called a quadratic polynomial e.g., -2x 2 + x - 6, x 2 + 4x + 9, 5x 2 + 7 etc.
(d) Cubic Polynomial: A polynomial of degree 3 is called a cubic polynomial. e.g., 2x 3 + 9x 2 - 7x + 2, 8x 3 + 9x 2 + 72 etc.
(e) Zero Polynomial: 0 is called a zero polynomial. The degree of a zero polynomial is not defined.
(f) Zero of a Polynomial: A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0.
Here ‘a’ is called a root of the equation p(x) = 0.
Example: Let p(x) = x 2 - 5x + 6
p(2) = 2 2 - 5(2) + 6 = 4 - 10 + 6 = 0
p(3) = 3 2 - 5(3) + 6 = 9 - 15 + 6 = 0
Hence, 2 and 3 are zero or roots of p(x).
Factorisation of polynomial
1. Remainder Theorem: If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by the linear polynomial (x - a), then the remainder is p(a), where a is any real number.
2. Factor Theorem: If p(x) is a polynomial of degree greater than or equal to 1 and a is any real number, then (i) (x - a) is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if (x - a) is a factor of p(x).
Some Algebraic Identities
(a) (x + y)2 = x 2 + 2xy + y 2
(c) (x2 - y 2) = (x + y)(x - y)
(e) (x + y + z)2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx
(g) (x - y)3 = x 3 - y 3 - 3xy(x - y) = x 3 - 3x 2 y + 3xy 2 - y 3
(i) x 3 - y 3 = (x - y)(x 2 + xy + y 2)
(b) (x - y)2 = x 2 - 2xy + y 2
(d) (x+a)(x + b) = x 2 + (a + b)x + ab (f) (x + y)3 = x 3 + y 3 + 3xy(x + y) = x 3 + 3x 2
(h) x 3 + y 3 = (x + y)(x 2 - xy + y 2)
+ 3xy 2 + y 3
(j) x 3 + y 3 + z 3 - 3xyz = (x + y + z)(x 2 + y 2 + z 2 - xy - yz - zx)
NCERT Zone
NCERT Exercise 2.1
1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer
(a) 4x 2 - 3x + 7 (b) 2 2 y +
(c) 32 tt + (d) 2 y y +
(e) x 10 + y 3 + t 50
Sol. (a) 4x2 - 3x + 7, yes, it is polynomial in one variable because it has only x as variable.
(b) 2 2 y + , yes, it is polynomial in one variable because it has only y as variable.
(c) 32 tt + , no, it is not polynomial in terms of power of t because t is not a whole number.
(d) 2 y y + , no it is not a polynomial as y in the power of second term is negative.
(e) x 10 + y 3 + t 50, no it is not the polynomial in one variable as it has x, y and t as variables.
2. Write the coefficients of x2 in each of the following:
(a) 2 + x 2 + x (b) 2 - x 2 + x 3
(c) 2 2 xx π + (d) 21 x
Sol. (a) Coefficient of x 2 in 2 + x 2 + x is 1.
(b) Coefficient of x 2 in 2 - x 2 + x 3 is -1.
(c) Coefficient of x 2 in 2 is . 22 xx ππ +
(d) Coefficient of x 2 in 21 x is 0. As x 2 not present in equation.
3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Sol. (a) Binomial of degree 35 is: x 35 + 10.
(b) Monomial of degree 100 is: 5x100
4. Write the degree of each of the following polynomials:
(a) 5x 3 + 4x 2 + 7x (b) 4 - y 2
(c) 57 t (d) 3
Sol. (a) In 5x 3 + 4x 2 + 7x, degree is 3 as x 3 is the maximum power.
(b) In 4 - y 2, degree is 2 as y 2 is the maximum power.
(c) In 57 t , degree is 1, as t is the maximum power.
(d) In 3, degree is 0, as x 0 is the maximum power.
5. Classify the following as linear, quadratic and cubic polynomials.
(a) x 2 + x (b) x - x 3
(c) y + y 2 + 4 (d) 1 + x
(e) 3t (f) r 2
(g) 7x 3
Sol.
NCERT Exercise 2.2
1. Find the value of the polynomial 5x - 4x 2 + 3 at (a) x = 0
(b) x =-1
(c) x = 2
Sol. Let p(x) = 5x - 4x 2 + 3
(a) While x = 0, p(0) = 5(0) - 4(0)2 + 3 = 3
(b) While x =-1, p(-1) = 5(-1) -4(-1)2 + 3 =-5 -4 + 3 =-6
(c) While x = 2, p(2) = 5(2) - 4(2)2 + 3 = 10 - 16 + 3 =-3
2. Find p(0), p(1) and p(2) for each of the following polynomials:
(a) p(y) = y 2 - y + 1 (b) p(t) = 2 + t + 2t 2 - t 3
(c) p(x) = x 3 (d)
Sol. (a) p(y) = y 2 - y + 1
p(0) = (0)2 - 0 + 1 = 1
p(1) = (1)2 - 1 + 1 = 1
p(2) = (2)2 - 2 + 1 = 3
(b) p(t) = 2 + t + 2t 2 - t 3
p(0) = 2 + 0 + 2(0)2 - (0)3 = 2
p(1) = 2 + 1 + 2(1)2 - (1)3 = 4
=
p(2) = 2 + 2 + 2(2)2 - (2)3 = 2 + 2 + 8 - 8 = 4
(c) p(y) = x 3
p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8
(d) p(t) = (x - 1)(x + 1)
p(0) = (0 - 1)(0 + 1) =-1
p(1) = (1 - 1)(1 + 1) = 0 × 2 = 0
p(2) = (2 - 1)(2 + 1) = 1 × 3 = 3
3. Verify whether the following are zeroes of the polynomial, indicated against them.
(a) p(x) = 3x + 1, x = 1 3
(b) p(x) = 5x -π, x = 4 5
(c) p(x) = x 2 - 1, x = 1, -1
(d) p(x) = (x + 1)(x - 2), x =-1, 2
(e) p(x) = x 2 , x = 0
(f) p(x) = lx + m, x = m l
(g) p(x) = 3x 2 - 1, x 12 , 33 =−
(h) p(x) = 2x + 1, 1 2 x =
Sol. (a) p(x) = 3x + 1, x = 1 3
Putting the value of x = 1 3 , we get 11 31110 33 p
3x + 1 = 0
So, it is a zero of polynomial.
(b) p(x) = 5x -π, x = 4 5
Putting the value of x = 4 5 , we get p 4 5
= 5 × 4 5 -π= 4 -π
5 × 4 5 -π≠ 0
So, it is not a zero of polynomial.
(c) p(x) = x 2 - 1, x = 1, -1
Putting the value of x = 1, we get p(1) = (1)2 - 1 = 1 - 1 = 0
Putting the value of x =-1, we get
p(-1) = (-1)2 - 1
= 1 - 1 = 0
So, they are zeros of polynomial.
(d) p(x) = (x + 1)(x - 2), x =-1, 2
Putting the value of x =-1, we get
p(-1) = (-1 + 1)(-1 - 2)
= 0 × (-3) = 0
Putting the value of x = 2, we get p(2) = (2 + 1)(2 - 2)
= 3 × 0 = 0
So, they are zeros of polynomial.
(e) p(x) = x 2 , x = 0
Putting the value of x = 0, we get
p(0) = (0)2 = 0
(0)2 = 0
So, it is a zero of polynomial.
(f) p(x) = lx + m, x = m l
Putting the value of x = m l , we get p m l
= l × m l
+ m =-m + m = 0
So, it is zero of polynomial.
(g) p(x) = 3x 2 - 1, x = 12 , 33 =−
Putting the value of x = 1 3 , we get p 2 111 31310 2 33
So, it is a zero of polynomial.
Putting the value of 2 3 x = , we get
p 2 224 31313 3 33
So, it is not a zero of polynomial.
(h) p(x) = 2x + 1, 1 2 x = (Given)
Putting the value of 1 2 x = , we get
p 11 2120 22
So, it is not zeros of polynomial.
4. Find the zero of the polynomial in each of the following cases:
(a) p(x) = 1 + 5 (b) p(x) = x - 5
(c) p(x) = 2x + 5 (d) p(x) = 3x - 2
(e) p(x) = 3x (f) p(x) = ax, a ≠ 0
(g) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Sol. (a) p(x) = x + 5
x + 5 = 0
x =-5
So, -5 is the zero of x + 5.
(b) (x) = x - 5
x - 5 = 0
x = 5
So, 5 is the zero of x - 5.
(c) p(x) = 2x + 5
2x + 5 = 0
2x =-5, x = 5 2
So, 5 2 is the zero of 2x + 5.
(d) p(x) = 3x - 2
3x - 2 = 0, 3x = 2, x = 2 3
So, 2 3 is the zero of 3x - 2.
(e) (x) = 3x
3x = 0
x = 0
So, 0 is the zero of 3x.
(f) p(x) = ax
ax = 0, a ≠ 0
x = 0
So, 0 is the zero of ax.
(g) p(x) = cx + d, c ≠ 0
cx + d = 0
cx =-d, x = d c
So, d c is the zero of cx + d.
NCERT Exercise 2.3
1. Determine which of the following polynomials has (x + 1) a factor:
(a) x 3 + x 2 + x + 1
(b) x 4 + x 3 + x 2 + x + 1
(c) x 4 + 3x 3 + 3x 2 + x + 1
(d) x 3 - x 2 -()222 x ++
Sol. To have (x + 1) as a factor, substituting x =-1 and we must get p(-1) = 0
Using the identity: (a - b)3 - (a + b)3 =-2b(3a 2 + b 2)
Let a = 5m and b = 3n
=-2(3n)(3(5m)2 + (3n)2)
=-6n(75m 2 + 9n 2)
=-450m 2 n - 54n 3
Short Answer Questions
1. Factorise: x 8 - y 8
Sol. x 8 - y 8 = (x 4)2 - (y 4)2 = (x 4 + y 4)(x 4 - y 4)
= (x 4 + y 4){(x 2)2 - (y 2)2}
= (x 4 + y 4)(x 2 + y 2)(x 2 - y 2)
= (x4 + y4)(x2 + y2)(x + y)(x - y)
2. Factorise 9x 2 + y 2 + z 2 - 6xy + 2yz - 6zx and hence find its value when x = 1, y =-2 and z = 1.
Sol. We observe that -6xy and -6zx are negative terms in the given expression, so x is negative; y and z are positive. Therefore, we write 9x 2 as (-3x)2
Thus, we have,
9x 2 + y 2 + z 2 - 6xy + 2yz - 6zx
Using: a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = (a + b + c)2
9. (a) Check whether p(x) is a multiple of g(x) or not, where p(x) = x 3 - x + 1, g(x) = 2 - 3x
(b) Check whether g(x) is a factor of p(x) or not, where p(x) = 8x 3 - 6x 2 - 4x + 3, g (x) = 1 34 x
Sol. (a) p(x) will be a multiple of g(x) if g(x) divides p(x).
Now, g(x) = 2 - 3x = 0 ⇒ x = 2 3
g(x) will be a factor of p(x) if 2 0 3 p
3 222 1 333 p
Competency Based Questions
Multiple Choice Questions
1. The value of () ()() () ()() 22abbc bccaabca +++() ()() 2 ca abbc + + is
(a) -1 (b) 0 (c) 1 (d) 2
2. If 2 2 2 11 24, then aa aa ++=+=
(a) 12 (b) 13
(c) 14 (d) -14
3. For what value of p is the coefficient of x2 in the product (2x - 1)(x - k)(px + 1) equal to 0 and the constant term equal to 2?
(a) k = 2 5 , p = 2 (b) k = 2, p = 2 5
(c) k = 5, p = 2 (d) k = 2, p = 5
4. Which one of the following is one of the factors of x 2(y - z) + y 2(z - x) - z(xy - yz - zx)?
(a) (x - y)
(b) (x + y - z)
(c) (x - y - z)
(d) (x + y + z)
5. If (x + 2) and (x - 1) are factors of f (x) = x 3 + ax 2 + bx - 6, then the value of a and b is:
(a) a = 4, b = 1 (b) a = 1, b = 4
(c) a = 2, b = 3 (d) a = 3, b = 2
Since remainder ≠ 0
So, p(x) is not a multiple of g(x). (b) Solve g(x) = 0 13 0 344 x x −=⇒= g(x) will be a factor of p(x) if 3 0 4 p
Now, 32793 8643 464164 p
=−−+
21654 33 6416 2727 0 88 =−−+ =−=
Since, 3 0 4 p
So, g(x) is a factor of p(x).
6. The ratio of the remainders when f (x) = x 2 + ax + b is divided by (x - 2) and (x - 1) respectively is 4 : 3.
If (x + 1) is a factor of f (x), then
(a) a = 9, b =-10
(b) a =-9, b = 10
(c) a = 9, b = 10
(d) a =-9, b =-10
7. If 2x + 3 y = 12 and xy = 30, then 8x3 + 3 24 y =
(a) 1008 (b) 168
(c) 106 (d) none of these
8. If 112 yzzxxy += +++ , then what is (x 2 + y 2) equal to?
(a) z2 (b) 2 1 z
(c) 2z2 (d) 2 2 z
9. The value of ()() ()() 33 22 0.0130.007 , is 0.0130.0130.0070.007 + −×+
(a) 0.006 (b) 0.02
(c) 0.0091 (d) 0.00185
10. The remainder when f (x) = x 45 + x 25 + x 14 + x 9 + x is divided by g(x) = x 2 - 1, is:
(a) 4x - 1
(b) 4x + 2
(c) 4x + 1
(d) 4x - 2
Assertion-Reason Based Questions
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): The degree of the polynomial x 5 + 2x 3 - 7x + 4 is 5
Reason (R): The degree of a polynomial is the sum of the exponents of all its terms.
2. Assertion (A): A polynomial of degree 0 is always a constant.
Reason (R): polynomial with only a constant term has no variable, so its highest exponent is considered 0.
3. Assertion (A): If x + 2a is a factor of f (x) = x 5 - 4a 2x 3 + 2x + 2a + 3, then 2a - 3 = 0
Reason (R): If f (x)is divisible by (ax + b) then 0 fb a
Case Study Based Questions
1. Pranjal formed a square using four pieces of origami, as shown in the figure. Based on above information answer the following questions.
(a) (i) Pranjal wants to calculate the total area of the square she made using algebra. Which trinomial correctly represents the area of the square in terms of x.
Answers
Multiple Choice Questions
1.
(ii) If Pranjal changes the arrangement and the new area of the square becomes x 2 - 10x + 25; what is the expression for the new side length of her square?
(b) (i) Pranjal cuts a square out of colored paper, where the side length is (x + 5). She then reduces the side length by 2 cm. What will be the new area of the square in terms of x?
(ii) If Pranjal increases the original side length (x + 5) by 50%, what will be the new expression for the side length, and what is the expression for the new area of the enlarged square?
2. A farmer is planning to set up a rectangular garden on his farm. The area of the garden (in square meters) is represented by the polynomial:
a(x) = x 2 + 6x + 8
where x represents the width of the garden in meters. Based on this, answer the following:
(a) The farmer wants to know the possible dimensions (length and width) of the garden. Help him by factoring the area expression.
(b) If the width of the garden is chosen to be 2 meters, calculate the actual area of the garden.
(c) The farmer is trying to understand at which values of width the area of the garden becomes zero.
(d) What is the sum of the zeroes of the polynomial?
2. (c) 14
We have, 2 1 24 a a ++=
⇒++=±
⇒+=+=
1 22 11 0 or, 4 a a aa aa
Now, 2 1 010,aa a +=⇒+= which is impossible.
6. (d) a =-9, b =-10
It is given that (2)4 and (1)0 (1)3 ff f =−=
424 and 10 13 abab ab ++ ⇒=−+= ++
⇒ 2a - b + 8 = 0 and -a + b + 1 = 0
⇒ a =-9, b =-10
7. (a) 1008
⇒++=⇒+=
Therefore, 1 0 a a +≠ 2 22 22 11 4 16 11 216 14 aa aa aa aa
3. (b) k = 2, 2 5 p = (2x - 1)(x - k)(px + 1)
= (2x - 1)(px 2 + x - kpx - k)
= 2px 3 + 2x 2 - 2kpx2 - 2kx - px 2 - x + kpx + k
= 2px 3 + x 2[2 - 2kp - p] - x[2k + 1 - kp] + k
Here constant term k = 2.
Coefficient of x 2 = 2 - 2kp - p = 2 - 4p - p = 2 - 5p
20. Find the value of 1 27 r 3 - s 3 + 125t 3 + 5rst, when 5 3 r st =+
21. If xy = 6 and 3x + 2y = 12, compute the value of 9x 2 + 4y 2 .
22. If the polynomials az 3 + 4z 2 + 3z - 4 and z 3 - 4z + a leave the same remainder when divided by z - 3, find the value of a
23. Calculate the perimeter of a rectangle whose area is 25x 2 - 35x + 12.
24. Without actual division, prove that 2x 4 + x 3 - 14x 219x - 6 is exactly divisible by x 2 + 3x + 2.
25. Show that 3 2 and -3 are zeroes of the polynomial
6x 3 + 23x 2 + 9x - 18. Also, find the third zero of the polynomial.
Challenge Yourself
1. Let R 1 and R 2 be the remainders when polynomial f (x) = 4x 3 + 3x 2 + 12ax - 5 and g(x) = 2x 3 + ax 2 - 6x - 2 are divided by (x - 1) and (x - 2) respectively. If 3R 1 + R 2 - 28 = 0 find the value of a.
2. Find the product of:
Answers
Additional Practice Questions
1. (a)
2. (a) -2 3. (c) 12
4. (d) -5 5. (a) (x - 5)(x - 4)
6. (b) 9 7. (a) -3
8. (a) 0 9. (b) x2 - 1
10. (d) A polynomial can have only non-negative integer exponents.
11. 4 x π =
12. No, 1 is not a zero of the polynomial.
13. a 3 + b 3 + c 3 - 3abc = 70
14. 93
3. Simplify:
4. If x 2 - 1 is a factor of ax 4 + bx 3 + cx 2 + dx + e, show that a + c + e = b + d = 0.
15. k = 6 16. 40 17. -540 18. m = 1 19. (a) (x - 2)(2x - 5)(x + 3) (b) () 3 23ab + 20. 0 21. 72 22. a =-1 23. P = 20x - 14 25. 2 3 is the third zero of the given polynomial.
Challenge Yourself 1. 1 2 a = 2. 8 8 1 x x 3. (x 2 + xy + y 2)(y 2 + yz + z 2)(z 2 + zx + x 2)
me for Exemplar Solutions
SELF-ASSESSMENT
Time: 60 mins
Multiple Choice Questions
1. The polynomial which has degree 1 is known as
(a) cubic polynomial
(c) quadratic polynomial
2. (x + 8)(x - 10) in the expanded form is
(a) (x + 8)(x - 10)
(c) x2 + 2x + 80
Scan me for Solutions
Max. Marks: 40
[3×1=3Marks]
(b) Linear polynomial
(d) bi-quadratic polynomial
(b) x 2 - 2x - 80
(d) x 2 - 2x + 80
3. According to Factor Theorem, if p(a) = 0, which of the following is a factor of p(x)?
(a) (x + a)
(c) (x - 1)
Assertion-Reason Based Questions
(b) (x - a)
(d) Can’t be decided
[2×1=2Marks]
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
4. Assertion (A): (x + 2) is a factor of x 3 + 3x 2 + 5x + 6
Reason (R): If p(x) is a polynomial of degreen ≥ 1and a is any real number, then(x - a) is a factor of p(x) if p(a) = 0.
5. Assertion (A): (x - 2y)3 + (2y -
Reason (R): If a + b + c = 0, then a 3 + b 3 + c 3 = 3abc
Very Short Answer Questions
6. Find the value of 10152 - 10142.
7. If 49x2 - b = 11 77, 22 xx +−
then find the value of b.
Short Answer Questions
).
[2 × 2=4Marks]
[4 × 3=12Marks]
8. Find the remainder when p(x) = x 3 - 2x 2 - 4x - 1 is divided by g(x) = x + 1.
9. For what value of m is x 3 - 2mx 2 + 16 divisible by x - 2?
10. Factorise 9m 2 - 66mn + 121n 2 .
11. If a, b, c are all non-zeros and a + b + c = 0, prove that
3. abc
++=
Long Answer Questions
12. If (x + a) is a factor of the polynomials x 2 + px + q and x 2 + mx + n, prove that . anq mp =
13. Using factor theorem, factorise the polynomial x 3 - 2x 2 - x + 2.
14. If 2 2 1 18 x x += , find the value of 3 3 1 . x x
Case Study Based Questions
15. Seemant is a delivery boy. He rides scooty on all working days. The distance covered on scooty is given by the polynomial expression p(x) = x 2 + 3x - 10.
Assuming that he rides at the uniform speed and takes the time given by g(x) = x - 2, x > 2.
Use the information stated above and answer the questions given below:
(a) What is the speed of the scooty?
(b) Find the sum of the degrees of the polynomial p(x) and g(x).
(c) If p (x) is replaced by r (x) = 5x 2 - kx - 18 and g (x) is a factor of r(x), then what will be the value of k?
(d) If p (x) is replaced by 4x 2 + 4x - 3, then what are the possible expressions for time and speed?
3 Coordinate Geometry
Chapter at a Glance
Coordinate Plane and Axes: To locate the position of a point on a plane, two perpendicular lines are used: one horizontal and one vertical. This plane is known as the Cartesian Plane or Coordinate Plane, and the lines are called Coordinate Axes.
x-axis and y-axis: The horizontal line is called x-axis, and the vertical line is called the y-axis.
Origin: The point of intersection of the axes is called the origin.
Abscissa
Ordinate
Coordinates of a point
Abscissa and Ordinate: The distance of a point from the y-axis is called its x-coordinate or Abscissa, and the distance of the point from the x-axis is called its y-coordinate or Ordinate.
Coordinates of a Point: If the abscissa of a point is x and the ordinate is y, then (x, y) are called coordinates of the point.
Points on the y-axis: The x -coordinate of every point on y -axis is zero. So, the coordinates of any point on y -axis are (0, y ).
Points on the x-axis: The y -coordinate of every point on x -axis is zero. So, the coordinates of any point on x -axis are ( x , 0).
Coordinates of the Origin: The coordinates of the origin are (0, 0).
Signs of Coordinates in Quadrants: The signs of the coordinates in various quadrants are as follows:
NCERT Exercise 3.1
1. How will you describe the position of a table lamp on your study table to another person?
Sol. The table lamp is 2 feet from the bottom side of the desk and 1 feet from is right side. Taking point ‘O’ as origin, we can write position of a lamp as (2, 1).
2. (Street Plan): A city has two main roads which cross each other at the centre of the City. These two roads are along the North-South direction and East-West direction. All the other street of the city run parallel to these roads and are 200 m apart. There are about 5 streets in each direction. Using 1 cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross streets in your model. A particular cross street is made by two streets, one running in north south direction and another in the East-West direction. which cross street is refer to in the following manner: If the 2nd Street running in the north south direction and 5th in the east west direction meet at some crossing, then we will call this cross street (2, 5). Using this convention, find:
(a) How many cross streets can be referred to as (4, 3).
(b) How many cross streets can be referred to as (3, 4).
Sol. Let us draw two perpendicular lines as the two main roads of the city that cross each other at the center. Let us mark them as North–South and East–West. As given in the question, let us take the scale as 1 cm = 200 m. Draw five streets that are parallel to both the main roads (which intersect), to get the given below figure. Street plan is as shown in the figure:
NCERT Exercise 3.2
1. Write the answer of each of the following questions:
(a) What is the name of the horizontal and vertical lines drawn to determine the position of any point in the Cartesian plane?
(b) What is the name of each part of the plane formed by these two lines?
(c) Write the name of the point where these two lines intersect.
Sol. (a) The horizontal line that is drawn to determine the position of any point in the Cartesian plane is called as x-axis. The vertical line that is drawn to determine the position of any point in the Cartesian plane is called as y-axis.
(b) The name of each part of the plane that is formed by x-axis and y-axis is called as quadrant.
(c) The point, where the x-axis and the y-axis intersect is called as origin.
2. See figure and write the following:
(a) There is only one cross street, which can be referred as (4, 3).
(b) There is only one cross street, which can be referred as (3, 4).
(a) The coordinates of B.
(b) The coordinates of C
(c) The point identified by the coordinates (-3, -5).
(d) The point identified by the coordinates (2, -4).
(e) The abscissa of the point D
(f ) The ordinate of the point H
(g) The coordinates of the point L.
(h) The coordinates of the point M.
Sol. (a) (-5, 2) (b) (5, -5)
(c) E (d) G
(e) 6 (f ) -3
(g) (0, 5) (h) (-3, 0)
Multiple Choice Questions
1. If the coordinates of the two points are P(-5, 3) and Q(8, -9), then (abscissa of Q) - (abscissa of P) is
(a) 4
(b) -12
(c) 13
(d) -13
2. The point whose ordinate is 3 and which lies on the y-axis is
(a) (0, 3)
(b) (0, -3)
(c) (3, 0)
(d) (-3, 0)
3. The coordinates of a point whose ordinate is 3 4 and abscissa is 5 are
(a) 3 ,5 4
(b) 3 5, 4
(c) 3 ,5 4
(d) 3 5, 4
4. The abscissa of a point is the (a) x-coordinate
(b) y-coordinate
(c) Sum of x and y coordinates
(d) Product of x and y coordinates
5. A point with coordinates (0, -6) lies on the (a) x-axis
(b) y-axis
(c) First quadrant
(d) Fourth quadrant
6. If a point lies in the third quadrant, then (a) Both coordinates are positive
(b) x is negative, y is positive
(c) x is positive, y is negative
(d) Both coordinates are negative
7. Which of the following points lies in the fourth quadrant?
(a) (-2, 4)
(b) (3, -5)
(c) (-6, -3)
(d) (2, 3)
8. A point on the x-axis has its y-coordinate always
(a) 0
(b) Positive
(c) Negative
(d) Any real number
9. On plotting the points O(0, 0), A(5, 0), B(5, 3), C(0, 3) and joining OA, AB, BC and CO, which of the following figures is obtained?
(a) rhombus
(b) square
(c) trapezium
(d) rectangle
10. If a point lies in the second quadrant, then its coordinates must satisfy
(a) x > 0, y > 0 (b) x < 0, y > 0
(c) x > 0, y < 0 (d) x < 0, y < 0
11. What is the coordinate of the point P shown on the coordinate grid?
(a) (-4, 5) (b) (4, 5)
(c) (5, 4) (d) (5, -4)
12. Abscissa of all the points on the x-axis is
(a) 0
(b) 1
(c) 2
(d) any number (NCERTExemplar)
13. The point A(k, k - 2) lies in the first quadrant and the point does not lie on any of the axis. Another point P(m, 2m - 5) is such that m is equal to the least possible integer value of k. Which of these statements is true?
(a) Point P lies in the first quadrant.
(b) Point P lies in the second quadrant.
(c) Point P lies in the third quadrant.
(d) Point P lies in the fourth quadrant.
14. The reflection of the point P(-4, 5) in y-axis has the coordinates
(a) (-4, -5)
(b) (4, 5)
(c) (4, -5)
(d) (5, -4)
15. The point P(5, -1) on reflection in x-axis is mapped as Q and the point Q on reflection in y-axis is mapped as R, the coordinates of R are
(a) (-5, -1)
(b) (-5, 1)
(c) (5, 1)
(d) (1, 5)
16. The distance between the points A(-5, 12) and (7, 12) is
(a) 5 units
(b) 7 units
(c) 12 units
(d) 17 units
17. If the point P(4, 2) is translated parallel to x-axis through 8 units, then the coordinates of new position of P are
(a) (4, 10)
(b) (4, 6)
(c) (12, 2)
(d) (4, -6)
18. Point (-3, 5) lies in the (a) first quadrant (b) second quadrant
(c) third quadrant
(d) fourth quadrant (NCERTExemplar)
19. In given figure coordinate of P are
(a) (-4, 2)
(b) (-2, 4)
(c) (4, -2)
(d) (2, -4)
20. The point R(2a + 5, 2b + 1) lies in the third quadrant, where a and b are integers and a ≠ b. Which of these could be the point Q with coordinates (a, b)?
Answers
1. (c) 13
3. (b) 3 5, 4
5. (b) y-axis
2. (a) (0, 3)
4. (a) x-coordinate
6. (d) Both coordinates are negative
7. (b) (3, -5)
9. (d) rectangle
11. (b) (4, 5)
8. (a) 0
10. (b) x < 0, y > 0
12. (d) any number
13. (a) Point P lies in the first quadrant.
14. (b) (4, 5)
16. (c) 12 units
18. (b) second quadrant
15. (b) (-5, 1)
17. (a) (4, 10)
19. (b) (-2, 4)
Constructed Response Questions
Very Short Answer Questions
1. In which quadrant does a point lie if:
(a) Both coordinates are positive?
(b) x-coordinate is negative and y-coordinate is positive?
(c) x-coordinate is positive and y-coordinate is negative?
(d) Both coordinates are negative?
Sol. (a) First Quadrant (I)
(b) Second Quadrant (II)
(c) Fourth Quadrant (IV) (d) Third Quadrant (III)
2. Find the point which lies on the line y = 5x + 2 having abscissa 4.
Sol. Given equation: y = 5x + 2
Substituting x = 4
y = 5(4) + 2 = 20 + 2 = 22
So, the required point is (4, 22).
3. A point P is at a perpendicular distance of 8 units from the x-axis and the foot of the perpendicular lies on the positive direction of x-axis at 5 units from y-axis. Find the coordinates of P.
Sol. The point is 5 units away from the y-axis, so the x-coordinate = 5.
The point is 8 units away from the x-axis, so the y-coordinate = 8 or -8.
As it is on positive direction, thus coordinates of P is (5, 8).
4. Which axis is parallel to the line on which the two points with coordinates (2, 5) and (2, -3) lie?
Sol. Both points have the same x-coordinate = 2, which means they lie on a vertical line.
A vertical line is parallel to the y-axis.
20. (a)
5. If the coordinates of two points are A(3, -2) and B(-5, 4), then find (abscissa of A) - (abscissa of B).
Sol. Abscissa of A = 3; Abscissa of B =-5
Difference: 3 - (-5) = 3 + 5 = 8
6. Without plotting the points, indicate the quadrant in which the point will lie if:
(a) Ordinate is 7 and abscissa is -4.
(b) Abscissa is -6 and ordinate is -2.
Sol. (a) (-4, 7): x is negative, y is positive
→ Second Quadrant (II).
(b) (-6, -2): x is negative, y is negative
→ Third Quadrant (III).
7. If a point lies in the third quadrant and has an abscissa of -6, what can be the possible values of its ordinate?
Sol. In the third quadrant, both x and y coordinates are negative.
Given x =-6, the y-coordinate must also be negative.
Possible values: any negative number (e.g., -1, -2, -3, -4, -5, etc.)
8. If two points A(5, 2) and B(5, -4) lie on a vertical line, what does this tell you about their relationship to the axes?
Sol. Since both points have the same x-coordinate (5), they lie on a vertical line.
A vertical line is always parallel to the y-axis and perpendicular to the x-axis.
The line passing through A and B is parallel to the y-axis and perpendicular to the x-axis.
Short Answer Questions
1. From figure, write the following:
(a) Coordinates of B, C, and E
(b) The point identified by the coordinates (0, -2)
(c) The abscissa of the point H
(d) The ordinate of the point D
Sol. (a) Coordinates:
B = (-5, 2)
C = (-2, -3)
E = (3, -1)
(b) The point identified by (0, -2) is F.
(c) The abscissa of H is 1.
(d) The ordinate of D is 0.
2. Write the coordinates of each of the points P, Q, R, S, T, and O. Y
So, the point (-2, 3) lies in Quadrant II.
In the point (5, 4), both abscissa and ordinate are positive. So, the point (5, 4) lies in Quadrant I. In the point (4, -2), the abscissa is positive and ordinate is negative.
So, the point (4, -2) lies in Quadrant IV. In the point (-2, -2), both abscissa and ordinate are negative. So, the point (-2, -2) lies in Quadrant III.
4. Write the coordinates of the vertices of a rectangle whose length and breadth are 7 and 4 units respectively, one vertex at the origin, the Ionger side lies on the x-axis and one of the vertices lies in the third quadrant.
Sol. (0, 0), (-7, 0), (-7, -4), (0, -4)
Sol. The coordinates of the points P, Q, R, S, T, and O are as follows:
P = (1, 1)
Q = (-3, 0)
R = (-2, -3)
S = (2, 1)
T = (4, -2)
O = (0, 0)
3. State the quadrants in which the following points lie: (-2, 3), (5, 4), (4, -2), (-2, -2)
Sol. In the point (-2, 3), the abscissa is negative and ordinate is positive.
( 7, 0)
( 7, 4)
(0, 4)
5. In given figure, LM is a line parallel to the y-axis at a distance of 2 units.
(a) What are the coordinates of the points P, R and Q ?
(b) What is the difference between the abscissa of the point L and M?
Sol. (a) Coordinates of the points P, Q and R are:
P = (2, 2), Q = (2, -1), R = (2, -3)
(b) 2 - 2 = 0
Long Answer Questions
1. Points A(5, 3), B(-2, 3), and D(5, -4) are three vertices of a square ABCD. Plot these points on a graph paper and hence find the coordinates of the vertex C.
Sol. Y
(5, 3)
(5, –4) C(x, y) B(-2, 3)
From the graph, we get that, The coordinates of C = (-2, -4).
2. Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, one vertex at the origin, the longer side lies on the x-axis and one of the vertices lies in the third quadrant.
Sol. From the graph, we get that, Y
(-5, 0)
(0, -3)
(-5, -3)
The coordinates of the points of the rectangle are (0, 0), (-5, 0), (-5, -3) and (0, -3).
3. Write the quadrant in which each of the following points lie:
(a) (-3, -5) (b) (2, -5) (c) (-3, 5) Also, verify by locating them on the cartesian plane. (NCERTExemplar)
Sol. (a) (-3, -5) lies in III quadrant, as x < 0 and y < 0.
(b) (2, -5) lies in IV quadrant, as x > 0 and y < 0.
(c) (-3, 5) lies in II quadrant, as x < 0 and y > 0.
Verification: The points (-3, -5), (2, -5) and (-3, 5) are plotted ast shown in figure.
(I)
Result is verified:
(a) (-3,-5) lies in III quadrant.
(b) (2, -5) lies in IV quadrant.
(c) (-3, 5) lies in II quadrant.
4. See figure and write the following:
(a) The co-ordinate of B.
(b) The point identified by the coordinates (-3, -2).
(c) The abscissa of the point D.
(d) The ordinate of the point C
Sol. (a) The co-ordinates of B = (-2, 3)
(b) E is the point which is identified by the co-ordinates (-3, -2)
(c) The co-ordinates of the point D is (6, 2)
∴ Abscissa is 6.
(d) The co-ordinates of the point C is (3, -1)
∴ Ordinate is -1.
Competency Based Questions
Multiple Choice Questions
1. The perpendicular distance of the point P(m, 2n) from the x-axis is 6 units. Given that m < 0 and n > 0, which of these represents the point Q with coordinates (n + 1, m)? (a)
2. Amit’s school is 5 km to the west and 3 km north of his house. He represented his house and his school on a coordinate grid, with his house located at the origin, and the positive x-axis represent the direction that is east of his house. If 1 unit on the coordinated grid represents 1 km, what will be the coordinate of his school? (a) (5, 3) (b) (-5, 3) (c) (3, 5) (d) (-3, 5)
3. If the coordinates of a point P(x, y) satisfy the relation xy > 0, then P may lie (a) I or II quadrant (b) II or III quadrant (c) I or III quadrant (d) I or IV quadrant
4. The area of the triangle the coordinates of whose vertices are O(0, 0), A(6, 0) and B(0, 8) is (a) 48 sq. units (b) 24 sq. units (c) 14 sq. units (d) 12 sq. units
5. The area of the figure formed by joining points A(-1, 5) and B(-2, 4) and their reflections in y-axis is (a) 3 sq. units (b) 6 sq. units (c) 4 sq. units (d) 12 sq. units
6. The ordinate of a point P is 20% more than its abscissa. If the sum of its coordinates is 33, find the coordinates of point P
7. A point P lies in the first quadrant. Its abscissa is p % of its ordinate. Also, the ratio of its abscissa to its ordinate is m : n. Find the relation between p, m, and n to satisfy both conditions.
(a) 100 m p n =× (b) 100 n p m =×
(c) 100 m p mn =× + (d) 100 mn p n =×
8. A point P(h, 3h - 5) lies on the y-axis. Another point Q is located 8 units horizontally away from the reflection of point P in the x-axis. The coordinates of Q is
(a) (8, 5) (b) (-8, -5) (c) (0, 13) (d) (0, -13)
9. A boy starts from his home and walks 5 km north, then turns right and walks 3 km. He then walks 2 km south and finally turns east and walks 4 km. If his home is at the origin and directions follow the standard coordinate grid (positive x-axis towards east, positive y-axis towards north), what are the coordinates of his final position?
(a) (7, 3) (b) (5, 3)
(c) (4, 5) (d) (7, 5)
10. The coordinates of a point P are in the ratio 5 : 2. If the difference of the coordinates is 21, what is the value of its ordinate?
(a) 14 (b) 15 (c) 12 (d) 13
Assertion-Reason Based Questions
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): The point (0, -2) lies on y-axis.
Reason (R): Ordinate of every point on x-axis is zero.
2. Assertion (A): The perpendicular distance of the point P(4, -7) from x-axis is 4.
Reason (R): The perpendicular distance of a point from x-axis is the absolute value of its y-coordinate.
3. Assertion (A): Point (-3, -3) lies on the angle bisectors of first and third quadrant angles.
Reason (R): The numeric value of ordinate and abscissa of every point on the bisectors of first and third quadrant angles are equal.
Case Study Based Questions
1. For a Math’s integrated project, Sonia created a symmetrical design on Cartesian plain. She drew a fish in a rectangle ABCD in the 2nd quadrant as shown in figure.
Based on the above information, answer the following questions:
(a) Find the sum of abscissa of points A and B.
(b) Find the area of rectangle ABCD
(c) What will be the new coordinates of A, B, C and D to draw the reflection of fish in the 3rd quadrant across x-axis.
(d) What will be the new coordinates of A, B, C and D to draw the fish by shifting each vertex of the rectangle 5 units to the right.
2. On Environment day class 9 students got five plants of mango, silver oak, Orange, banyan and Amla from soil department. Students planted the plants and noted their locations as (x, y).
Observing the graph given below, Answer the following questions.
Amla Mango
Answers
Multiple Choice Questions
1. (c)
The perpendicular distance of point P(m, 2n) from the x-axis is given its y-coordinate, which is 2n
2n = 6
Since, n > 0, 2n = 6, so n = 3.
Now, find the coordinates of point Q.
Q has coordinates (n + 1, m).
Substitute n = 3:
x-coordinate of Q = 3 + 1 = 4. y-coordinate of Q = m (which remains unknown but m < 0).
So, point Q has coordinates (4, m).
Since x > 0 and m < 0, the point Q lies in the fourth quadrant. Point Q has coordinates (4, m) and lies in the fourth quadrant. Thus, (c) is correct.
(a) Find in which Quadrant does the Banyan plant lie?
(b) Which plant lie on x-axis?
(c) Find the co-ordinates of location point of Orange tree. Also, write that in which quadrant do these coordinates lie?
(d) What is the distance of the Silver Oak plant from the x-axis?
2. (b) (-5, 3)
Amit’s house is at the origin (0, 0).
His school is 5 km to the west, so the x-coordinate becomes 0 - 5 =-5.
His school is also 3 km to the north, so the y-coordinate becomes 0 + 3 = 3.
The coordinates of his school are (-5, 3). The correct answer is (b) (-5, 3).
3. (c) I or III quadrant
xy > 0
⇒ (x > 0 and y > 0) or (x < 0 and y < 0)
⇒ P lies either in I or in II quadrant.
4. (b) 24 sq. units
By plotting the points O, A and B, we observe that these points are vertices of a right triangle having two legs OA = 6 units and OB = 8 units.
A(6, 0)
∴ Area of ∆OAB () 1 2 OAOB = × () 1 24 sq. units
268 == ×
5. (a) 3 sq. units
Reflections of points A(-1, 5) and B(-2, 4) in y-axis are points D(1, 5) and C(2, 4) respectively. Clearly, ABCD is a trapezium with AD = 2, BC = 4 and distance between parallel sides AD and BC is 1 unit. Therefore,
Area of trapezium ABCD = 1 2 ( AD + BC ) × 1 sq. units = 1 2 (2 + 4) = 3 sq. units.
6. (a) (15, 18)
Let the abscissa (x-coordinate) of point P be x.
Then, the ordinate (y-coordinate) is 20% more than x
⇒ y = x + 20% of x
⇒ y = x + 0.2x = 1.2x
The sum of the coordinates is 33:
x + y = 33
x + 1.2x = 33
2.2x = 33 33 15
2.2 x ==
y = 1.2x = 1.2 × 15 = 18
P(15, 18)
7. (a) 100 m p n =×
Abscissa is p % of the ordinate:
100 p xy =×
Also, xm yn =
Thus, m xy n =×
On equating above equations we get:
100 pmyy n ×=×
100 pm n = 100 m p n =×
8. (a) (8, 5)
Since P lies on the y-axis, its x-coordinate is 0.
So, h = 0.
Substituting h = 0 into P’s coordinates:
P(0, 3 × 0 – 5) → P(0, -5)
Reflect over x-axis → (0, 5)
Q is located 8 units horizontally away:
Q can be either 8 units to the right → x = 0 + 8 = 8 OR
8 units to the left → x = 0 - 8 =-8 (8, 5) → 8 units to the right
9. (a) (7, 3)
Initial position of the boy is at the origin (0, 0).
Step 1: He walks 5 km north. His new position becomes (0, 5).
Step 2: He turns right. Facing north, a right turn means he now moves east. He walks 3 km east. His new position becomes (3, 5).
Step 3: He walks 2 km south. His new position becomes (3, 5 - 2) = (3, 3).
Step 4: He turns east again and walks 4 km. His final position becomes (3 + 4, 3) = (7, 3).
The final coordinates of the boy are (7, 3).
10. (a) 14
Let the abscissa (x-coordinate) be 5k.
Let the ordinate (y-coordinate) be 2k, where k is a positive real number.
The difference between the coordinates is 21.
5k - 2k = 21
3k = 21 21
7 3 k ==
Thus ordinate, y = 2 × 7 = 14
Assertion-Reason Based Questions
1. (b) Both A and R are true, but R is not the correct explanation of A.
The abscissa of P is zero.
So, it lies on y-axis.
So, A is true. R is also true but not the correct explanation of A.
Hence, option (b) is correct.
2. (d) A is false, but R is true.
The R is correct.
Using this, the perpendicular distance of the point P(4, -7) from x-axis is 7.
So, A is not true.
Hence, option (d) is correct.
3. (a) Both A and R are true, and R is the correct explanation of A.
R is true, because every point on the bisectors of first and third quadrant angles is equidistant from the coordinate axes.
As the ordinate and abscissa of P are equal in magnitude.
So, point (-3, -3) lies on the bisector of third quadrant angle.
Case Study Based Questions
1. (a) The x-coordinate of point A is -2 and the x-coordinate of point B is -5.
The sum of these values is -2 + (-5) =-7.
Additional Practice Questions
1. The perpendicular distance of the point P(3, 4) from x-axis is
(a) 3
(b) 4
(c) 5
(d) None of these
2. If points P(3, 0) and Q(a, 0) are equidistant from the origin, then a =
(a) 3
(b) -3
(c) 6
(d) 9
(b) The length is the difference in x-coordinates of points A and B, which is |-2 - (-5)|= 3 units. The breadth is the difference in y-coordinates of points A and C, which is |3 - 1|= 2 units.
Area is 3 × 2 = 6 square units.
(c) For reflection across the x-axis, change the sign of the y-coordinates while keeping the x-coordinates unchanged.
Point A(-2, 3) becomes (-2, -3).
Point B(-5, 3) becomes (-5, -3).
Point C(-5, 1) becomes (-5, -1).
Point D(-2, 1) becomes (-2, -1).
(d) For shifting each vertex 5 units to the right, add 5 to the x-coordinate of each point while keeping the y-coordinate unchanged.
Point A(-2, 3) becomes (3, 3).
Point B(-5, 3) becomes (0, 3).
Point C(-5, 1) becomes (0, 1).
Point D(-2, 1) becomes (3, 1).
2. (a) Co-ordinates of Banyan plant are (-3, 4) so, it lies in II quadrant.
(b) Coordinates of Amla are (-2, 0) and that of mango are (2, 0) so these both plants lie on x-axis.
(c) Co-ordinates of Orange tree are (3, 4).
(d) The Silver Oak plant has coordinates (0, 7). Its distance from the x-axis is given by the absolute value of its y-coordinate, which is |7|= 7 units.
3. Two points having same abscissa but different ordinates lie on
(a) x-axis
(b) y-axis
(c) a line parallel to y-axis
(d) a line parallel to x-axis
4. The distance between the images of points P(-7, 4) and Q(7, 4) in x-axis is
(a) 7 units
(b) 8 units
(c) 11 units
(d) 14 units
5. If the perpendicular distance of a point P from the y-axis is 4 units and the foot of the perpendicular lies on the negative direction of y-axis, then what is the y-coordinate of P ?
6. Write the co-ordinates of each point P, Q, R, S, T and O from the figure given below. (NCERTExemplar)
11. In given figure, line l is parallel to the x-axis at a distance of 2 units.
7. Without plotting the points, indicate the quadrant in which they will lie, if:
(a) Ordinate is 5 and abscissa is -3
(b) Abscissa is -5 and ordinate is -3
(c) Abscissa is -5 and ordinate is 3
(d) Ordinate is 5 and abscissa is 3
8. In the adjoining figure, PQR is an equilateral triangle in which coordinates of Q and R are (0, 4) and (0, -4), respectively. Find the coordinates of the vertex P.
(a) What are the coordinates of the point A, B and C ?
(b) What is the difference between the ordinates of the point A and C ?
(c) Find (abscissa of B) - (abscissa of A).
12. From the given figure, write the points whose (a) abscissa = 0
(b) ordinate = 0
(c) abșcissa =-3
(d) ordinate = 4
9. A point lies on the x-axis at a distance of 7 units from the y-axis. What are its coordinates? What will be the coordinates if it lies on y-axis at a distance of -7 units from x-axis?
10. Without plotting the points, indicate the quadrant in which the following points will lie.
(a) Point whose ordinate is -7 and abscissa is -1.
(b) Point whose abscissa =-4 and ordinate =-4.
(c) Points whose abscissa is 2 and ordinate is 5.
13. If the coordinates of a point A are (-2, 9) which can also be expressed as (1 + x, y 2) and y > 0, then find in which quadrant do the following points lie: P(y, x), Q(2, x), R(x 2 , y - 1), S(2x, -3y)
14. (a) Find values of a and b, if two ordered pairs (a - 3, -6) and (4, a + b) are equal.
(b) Find distances of point (a, b) obtained from x-axis and y-axis.
(c) Find in which quadrant they lie.
15. In the given figure, ABCD is a rhombus with diagonals AC = 16 cm and BD = 8 cm.
Find the coordinates of A, B, C and D.
Challenge Yourself
1. In the given figure, ∆ABC and ∆ABD are equilateral triangles. Find the coordinates of points C and D. [Hint: Altitude of an equilateral triangle, OC = OD
= 3 2 × Side]
2. If ,4 3 Pa
is the mid-point of the line segment
joining the points Q(-6, 5) and R(-2, 3), then find the value of a?
Answers
Additional Practice Questions
1. (b) 4
2. (b) -3
3. (c) a line parallel to y-axis
4. (d) 14 units
5. Cant say
6. Co-ordinates of P =(1, 1)
Co-ordinates of Q = (-3, 0)
Co-ordinates of R =(-2, -3)
Co-ordinates of S = (2, 1)
Co-ordinates of T = (4, -2)
Co-ordinates of O = (0, 0)
16. In which quadrant or on which axis do each of the points (-2, 4), (3, -1), (-1, 0), (1, 2), and (-3, -5) lie?
17. If the coordinates of a point A are (-4, 16), which can also be expressed as (x + 2, y 2) and y > 0, then find in which quadrant do the following points lie:
(a) P(y, x)
(b) Q(3, x)
(c) R(x 2 , y - 4)
(d) S(2x, -2y)
3. Shown below are 2 identical rectangles such that their breadth is half their length. What are the coordinates of point A?
7. (a) II quadrant
(b) III quadrant
(c) II quadrant
(d) I quadrant
8. () 43,0
9. (7, 0), (0, -7)
10. (a) III quadrant
(b) III quadrant
(c) I quadrant
11. (a) A(-2, 2), B(1, 2), C(4, 2)
(b) 0
(c) 3
(4, -3)
12. (a) P and T
(b) Q and N
(c) R and Q
(d) L and S
13. P in IV quadrant, Q in IV quadrant, R in I quadrant and S in III quadrant
14. (a) a = 7 and b =-13
(b) 13 units from x-axis and 7 units from y-axis.
(c) IV quadrant
15. A = (-8, 0); B = (0, 4); C = (8, 0); D = (0, -4)
16. (a) (-2, 4): The abscissa is negative and ordinate is positive. So, the point (-2, 4) lies in Quadrant II.
(b) (3, -1): The abscissa is positive and ordinate is negative. So, the point (3, -1) lies in Quadrant IV.
(c) (-1, 0): The ordinate is zero. So, the point (-1, 0) lies on the x-axis IV.
(d) (1, 2): Both abscissa and ordinate are positive. So, the point (1, 2) lies in Quadrant I.
(e) (-3, -5): Both abscissa and ordinate are negative. So, the point (-3, -5) lies in Quadrant III.
17. (a) IV Quadrant
(b) IV Quadrant
(c) Lies on the x-axis
(d) III Quadrant
Challenge Yourself
1. ()() 0,3 and 0,3 aa
2. -12
3. (-2, -7)
Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 60 mins
Multiple Choice Questions
Scan me for Solutions
Max. Marks: 40
[3×1=3Marks]
1. The point (-4, 5) lies in the (a) First quadrant (b) Second quadrant (c) Third quadrant (d) Fourth quadrant
2. The point at which the x-axis and y-axis intersect is called the (a) Abscissa (b) Ordinate (c) Origin (d) Quadrant
3. If the y-coordinate of a point is zero, then this point always lies (a) In I quadrant (b) In II quadrant (c) On x-axis (d) On y-axis
Assertion-Reason Based Questions
[2×1=2Marks]
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
4. Assertion (A): The point (0, 4) lies on the y-axis.
Reason (R): A point on the y-axis always has an x-coordinate of 0.
5. Assertion (A): A point in the third quadrant has both x and y coordinates negative.
Reason (R): The third quadrant consists of points where x-coordinates are positive and y-coordinates are negative.
Very Short Answer Questions
6. (a) If the coordinates of a point are (4, 3), in which quadrant does it lie? (b) Find the distance of the point(-2, -5) from the y-axis.
7. (a) A point lies on the x-axis at a distance of 8 units from the origin. What will be its coordinates? (b) Write ordinates of the following points: (3, 4), (4, 0), (0, 4), (5, -3)
Short Answer Questions
[2 × 2=4Marks]
[4 × 3=12Marks]
8. (a) The coordinates of point P are (-6, 2). Find its mirror image with respect to the x-axis. (b) If the coordinates of the two points are A(-2, 5) and B(-7, 8), then find (abscissa of B) - (abscissa of A).
9. (a) The point (-3, b) lies on the line with equationy = 2x + 1. Find the value of b. (b) The distance of a point from the x-axis is 3 units and from the y-axis is 5 units. If the point lies in the third quadrant, find the coordinates of the point.
10. (a) If the point P(-7, k) lies in the third quadrant, what can be the possible values of k ? (b) The distance of a point from the x-axis is 4 units and from the y-axis is 7 units. If the point lies in the second quadrant, find its coordinates.
11. ABCD is a rectangle in the coordinate plane such that O is the midpoint of side AB. The length of the rectangle is 6 units and the breadth is 3 units. The rectangle is placed so that side AB lies along the x-axis and the midpoint O is at the origin (0, 0). One of the rectangle’s vertices lies in the first quadrant. Find the coordinates of points A, B, C, and D.
Long Answer Questions
12. If the coordinates of a point A are (-2, 9), which can also be expressed as (1 + x, y 2) and y > 0, then find in which quadrant do the following points lie:
P(y, x), Q(2, x), R(x 2 , y - 1), S(2x, -3y)
13. If a point P(x, y) is given as (2, -5), then find the coordinates of the points obtained by the following transformations:
(a) Reflection in the x-axis
(c) Reflection in the origin
(b) Reflection in the y-axis
(d) Shifting 3 units right and 4 units up
14. If two ordered pairs (a + 2, -7) and (6, a + b) are equal, then find:
(a) The values of a and b
(b) The distances of the point (a, b) from the x-axis and y-axis.
(c) The quadrant in which the point (a, b) lies.
Case Study Based Questions
15. Treasure Hunt on a Coordinate Grid
A treasure is hidden at a point in a Cartesian plane. A pirate starts at point A(2, 3) and moves 6 units left and 5 units down to reach the treasure.
(a) Find the coordinates of the treasure.
(b) In which quadrant does the treasure lie?
(c) What is the distance of the treasure from the origin?
(d) If the pirate instead moved 6 units right and 5 units up from A, where would he land?
4 Linear Equations in Two Variables
Chapter at a Glance
Linear Equations in Two Variables
A linear equation in two variables is an equation the form ax + by + c = 0, where a, b and c are real numbers, and a ≠ 0, b ≠ 0.
Example: 236, xy+= 4 xy−=
Solution of a Linear Equation: The solution of a linear equation in two variables is a pair of values (x, y) that satisfies the equation. There are infinitely many solutions to such an equation.
Example: For 25 xy+= , () 2, 1 is a solution because ()+=⇒= 221555
Graph of a Linear Equation: The graph of a linear equation in two variables is always a straight line and each point on the line represents a solution to the equation.
● Graph of x = a is a straight line parallel to y-axis.
-axis
● Graph of y = a is a straight line parallel to x-axis.
-axis
● Graph of y = mx is a straight line passing through the origin.
-axis
-axis
-axis
NCERT Zone
NCERT Exercise 4.1
1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
( Take the cost of a notebook to be x and that of a pen to be y).
Sol. Cost of a notebook = x
Cost of a pen = y
Then according to given statement
x = 2y
⇒ x 2y = 0
2. Express the following linear equations in the form
ax + by + c = 0 and indicate the values of a, b and c in each case:
(a) 239.35 xy+= (b) 100 5 xy−−=
(c) –236 xy+= (d) 3 xy =
(e) 25xy =−
(f) 320 x +=
(g) 20 y −= (h) 52x =
Sol. (a) 239.35 xy+=
23 9.35 0 xy ⇒+−=
So, ===− 2, 3, 9.35 abc
(b) 100 5 xy−−=
So, ==−=− 1 1, , 10 5 abc
(c) 236 xy −+=
2360 xy ⇒+−=
So, ===− 2, 3, 6 abc
(d) 3 xy =
300xy ⇒−+=
So, ==−= 1, 3, 0 abc
(e) 25xy =−
2500 xy ⇒++=
So, === 2, 5, 0 abc
(f) 320 x +=
3020 xy ⇒++=
So, a === 3, 0, 2 bc
(g) 20 y −=
⇒+−=0120 xy
So, ===− 0, 1, 2 abc
(h) 52x =
250 x ⇒−=
() 2050 xy ⇒+−=
So, ===− 2, 0, 5 abc
NCERT Exercise 4.2
1. Which one of the following options is true, and why?
y = 3x + 5 has
(a) a unique solution
(b) only two solutions
(c) infinitely many solutions
Sol. (c) infinitely many solutions. It is because a linear equation in two variables has infinitely many solutions. Hence, we can change the value of x and solve the linear equation for the corresponding value of y
2. Write four solutions for each of the following equations:
(a) 27 xy+= (b) 9 xy π+= (c) 4 xy =
Sol. (a) 27 xy+=
Let x = 1
Then, 2 × 1 + y = 7
⇒ y = 7 2 = 5.
Hence, (1, 5) is a solution.
Let x = 2
Then, 2 × 2 + y = 7
⇒ y = 7 4 = 3
Hence, (2, 3) is another solution.
Let x = 3
Then, 2 × 3 + y = 7
⇒ y = 7 6 = 1
⇒(3,1) is another solution
Let x = 4
Then, 2 × 4 + y = 7
⇒ y = 8 + 7 = 1
()⇒− 4, 1 is another solution.
Hence (1, 5), (2, 3), (3, 1) and (4, 1) are all solutions of 2x + y = 7.
(b) 9 xy π+=
Let 1 x π = ,
Then, 1 9 y π π ×+=
y = 9 1 = 8
So, π
1 , 8 is a solution.
Let 2 x π = ,
Then, 2 9 y π π ×+=
y = 9 2 = 7
2 , 7 is a solution.
So, π
Let 3 x π =
Then, 3 9 y π π ×+=
y = 9 3 = 6
3 , 6 is a solution.
So, π
Let 4 x π =
Then, 4 9 y π π ×+= y = 5
So, π 4 , 5 is a solution.
Therefore, ππππ
1234 , 8, , 7, , 6, , 5 are all solutions of the equation π x + y = 9.
(c) x = 4y
Let x = 8
Then, 8 = 4y
⇒ y = 2
So,(8, 2) is a solution.
Let x = 12
Then, 12 = 4y
⇒ y = 3
So, (12, 3) is a solution.
Let x = 16
Then, 16 = 4y,
⇒ y = 4
So, (16, 4) is a solution.
Let x = 20
Then, 20 = 4y,
⇒ y = 5
So, (20, 5) is a solution.
Therefore, (8, 2), (12, 3), (16, 4) and (20, 5) are all solutions of equation x = 4y
3. Check which of the following are solutions of the equation x 2y = 4 and which are not:
(a) (0, 2) (b) (2, 0)
(c) (4, 0) (d) ( 2,42 )
(e) (1, 1)
Sol. 24xy−=
(a) When 0, 2 xy== , then () 02244 −=−≠ .
This implies R.H.S ≠ L.H.S. Therefore, it is not a solution.
(b) When x = 4, y = 0
4 2(0) = 4 = 4.
This implies R.H.S ≠ L.H.S. Therefore, it is not a solution
(c) When 4,0xy== () 42044 −==
This implies R.H.S = L.H.S. Therefore, it is a solution.
(d) When 2 x = , y = 42 () 2242282724 −=−=−≠
This implies R.H.S ≠ L.H.S. Therefore, it is not a solution
(e) When x = 1, y = 1
1 2(1) = 1 ≠ 4
This implies R.H.S ≠ L.H.S.
Therefore, it is not a solution.
4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Sol. x = 2, y = 1 is a solution of 2x + 3y = k.
Thus, 2 × 2 + 3 × 1 = k
⇒ 4 + 3 = k
k = 7 is the solution.
Multiple Choice Questions
1. The linear equation 3xy = x − 1 has (a) A unique solution
(b) Two solutions
(c) Infinitely many solutions
(d) No solution
2. A linear equation in two variables is of the form ax + by + c = 0, where
(a) a ≠ 0, b ≠ 0 (b) a = 0, b ≠ 0 (c) a ≠ 0, b = 0 (d) a = 0, c = 0
3. Any point on the y-axis is of the form (a) (x, 0) (b) (x, y) (c) (0, y) (d) (y, y)
4. If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is (a) 4 (b) 6 (c) 5 (d) 2 (NCERTExemplar)
5. Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form (a)
9 , 2 n (c)
9 0, 2 (d) ( 9, 0)
(NCERTExemplar)
6. If C = S + 3, t then the value of t when 3 5 C = and 1 10 S = is (a) 1 2 (b) 1 3 (c) 1 6 (d) 1 5
7. The point of the form , 2 a a always lies on the line
(a) 2x = y (b) x = 2y
(c) x = y + 2 (d) x + 2 = y
8. If a linear equation has solutions ( 2, 2), (0, 0) and (2, 2), then it is of the form
(a) yx = 0 (b) xy = 0
(c) 2x + y = 0 (d) x + y = 0
9. The positive solutions of the equation ax + by + c = 0 always lie in the
10. If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation
(a) changes (b) remains the same (c) changes in case of multiplication only (d) changes in case of division only (NCERTExemplar)
11. If 235 xy=+ is written in standard form ax + by + c = 0, then the value of a, b and c are
(a) a = 2 , b = 3 , c = 5
(b) a = 2 , b = 3 , c = 5
(c) a = 2 , b = 3 , c = 5
(d) a = 2 , b = 3 , c = 5
12. The equation x = 7, in two variables, can be written as (a)1 × x + 1 × y = 7 (b) 1 × x + 0 × y = 7 (c) 0 × x + 1 × y =7 (d) 0 × x + 1 × y =7
(NCERTExemplar)
13. The solution of the equation 5 23 xx += is
(a) 5 (b) 6 (c) 4 (d) 7
14. Which option shows 5y 8x = 7(x + y) 9 expressed in the form of ax + by + c = 0?
(a) 15x + 2y 9 = 0 (b) 15x 4y 9 = 0
(c) x + 6y 9 = 0 (d) x + 12y 9 = 0
15. In the equation (k + 1)x + ky 5k = 1 2ky, the equation when expressed in the form ax + by + c = 0 gives c = 6. What are the values of a and b?
(a) =−= 221 , 55 ab (b) ==− 714 , 55 ab
(c) ==− 221 , 55 ab (d) ==− 521 , 22 ab
16. The equation 2x + y = 4 has a unique solution if x and y are
(a) rational numbers (b) real numbers
(c) natural numbers (d) positive real numbers
17. Which of these equations has (1.5, 4) as one of the solutions?
(a) 20x + 5y = 50 (b) 20x + 5y = 87.5
(c) 20x + 5y = 270 (d) 20x + 5y = 520
18. Any solution of the linear equation 0 × x + 2 × y + 9 = 0 in two variables is of the form
9 , 2 n (c)
9 , 2 m (b)
9 0, 2 (d) () 9, 0
Answers
1. (c) Infinitely many solutions
2. (a) a ≠ 0, b ≠ 0
3. (c) (0, y)
4. (a) 4
19. (a, 0) is the solution of the equation 3x + 2y = 6 for a is equal to (a) 2 (b) 2 (c) 3 (d) 3
20. If y = x + 4t, then the value of t when x = 4 and y = 12 is (a) 2 (b) 4 (c) 8 (d) 12
11. (c) a = 2 , b = 3 , c = 5
12. (b) 1 × x + 0 × y = 7
13. (b) 6
14. (a) 15290 xy+−=
5. (a)
6. (c) 1 6
9 , 2 m
7. (b) x = 2y
8. (d) x + y = 0
9. (a) 1st quadrant
10. (b) remains the same
Constructed Response Questions
Very Short Answer Questions
1. If a line represented by the equation 3x +αy = 8 passes through (1, 1), then find the value of α
Sol. We have 3x +αy = 8 (i)
On substituting x = 1 and y = 1 in (i), we get
3 × 1 +α× 1 = 8
⇒ 3 +α= 8 ⇒α= 8 3
⇒α= 5
2. Draw the graph of the equation represented by the straight line which is parallel to the x-axis and is 4 units above it.
(NCERTExemplar)
15. (c) ==− 221 , 55 ab
16. (c) natural numbers
17. (a) 20550 xy+=
18. (b) 9 , 2 n
19. (b) 2
20. (a) 2
Sol. Any straight line parallel to x-axis is given by y = k, where k is the distance of the line from the x-axis.
Here k = 4. Therefore, the equation of the line is y = 4.
To draw the graph of this equation, plot the points (1, 4) and (2, 4) and join them.
3. Find the points where the graph of the equation 3x + 4y = 12 cuts the x-axis and the y-axis.
(NCERTExemplar)
Sol. The graph of the linear equation 3x + 4y = 12 cuts the x-axis at the point where y = 0.
On substituting y = 0 in the linear equation, we have 3x = 12, which gives x = 4.
Thus, the required point is (4, 0).
The graph of the linear equation 3x + 4y = 12 cuts the y-axis at the point where x = 0.
On substituting x = 0 in the given equation, we have 4y = 12, which gives y = 3.
Thus, the required point is (0, 3).
Hence, the graph cuts the x-axis at (4, 0) and the y-axis at (0, 3)
4. The weight of a man is four times the weight of a child. Write an expression in two variable for the given situation. If the child weighs 15 kg, find the weight of the man.
Sol. Let the weight of the man be x kg, Let the weight of the child be y kg.
Then the required equation is 4 xy = or 40xy−=
Given: Weight of the child = 15 kg
Therefore, weight of the man: 41560 x =×= kg
5. For what value of c, the linear equation 2x + cy = 8 has equal values of x and y as its solution?
Sol. Given 2x + cy = 8.
For equal values of x and y, put y = x
∴ 2x + cx = 8
⇒ cx = 8 − 2x
⇒ 8 2 , x c x = x ≠ 0
(NCERTExemplar)
6. If the point (3, 4) lies on the graph of 3x = ay + 7, then find the value of a
Sol. (3, 4) lies on the graph of 3x = ay + 7
Substitute x = 3 and y = 4
∴ 3(3) = a × 4 + 7
⇒ 9 = 4a + 7
⇒ 4a = 2
⇒ 1 2 a =
7. Give the equations of three lines passing through the point (4, 2). How many more such lines are possible, and why? Also, draw the graphs of any three such lines.
Sol. Let the equation of a line passing through point () 4,2 be 0 axbyc++= . Since line passes through () 4,2 ,
∴ 4a + 2b + c = 0
For each pair of a and b, there exists a value of c which can satisfy this equation. Hence, infinitely many lines can pass through () 4,2 .
Equations of the three such lines passing through (4, 2) are as follows:
(a) a = 1, b = 1
∴ 4 ×1 + 1 × 2 + c = 0 ⇒ c = 2
Equation of line: 2 0 xy+−= or 2 yx =−
(b) a = 1, b = 1
∴ 4 × 1 + 1 × 2 + c = 0 ⇒ c = 6
Equation of line: xy 6 = 0 or y = x 6
(c) a = 2, b = 1
∴ 4 × 2 + 1 × 2 + c = 0 ⇒ c = 10
Equation of line: 2xy 10 = 0 or y = 2x 10
8. Find whether () 2, 32 is a solution of x 3y = 9 or not.
Sol. To check whether () 2, 32 is a solution of the given equation.
Substitute
2 x = and 32 y = in 39xy−= LHS = x 3y =() 2332
= 292829RHS −=−≠=
Since LHS ≠ RHS
Hence, () 2, 32 is not a solution of the given equation.
9. Give the geometric interpretations of 5x + 3 = 3x 7 as an equation
(a) in one variable (b) in two variables.
Sol. Given, 5x + 3 = 3x 7
⇒ 5x 3x = 7 3
⇒ 2x = 10
x = 5
(a) The given equation represents point x = 5 on the number line when treated as an equation in one variable.
(b) The equation x = 5 can be written as 1 × x + 0 × y + 5 = 0
which is a linear equation in two variables x and y. Geometrically, it represents a vertical straight line parallel to the y-axis, passing through x = 5 on the Cartesian plane.
10. From a bus stand in Delhi, if we buy 2 tickets to Agra and 3 tickets to Mathura, the total cost is `440. Express this situation in linear equation. If the cost of one ticket to Agra is `120, find the cost of one ticket to Mathura.
Sol. Let: x = cost of one ticket to Agra (in `)
y = cost of one ticket to Mathura (in `)
Linear equation of given situation is 2x + 3y = 440.
Given: cost of one ticket to Agra is `120.
Substitute in the equation: 2 × 120 + 3y = 440
⇒ 240 + 3y = 440
⇒ 3y = 200
⇒=≈ 200 66.67 3 y `
11. The cost of a ball pen is `5 less than half of the cost of a fountain pen. Write this statement as a linear equation in two variables.
Sol. Let cost of a ball pen = `x and cost of a fountain pen = `y
According to the question,
Cost of a ball pen = Half of the cost of a fountain pen 5
2 xy
⇒ 2x = y 10 ⇒ 2xy +10 = 0
The required linear equation is 2xy + 5 = 0.
12. Find any two solutions of the equation 4x + 3y = 12.
Sol. Given equation 4x + 3y = 12 (i)
On substituting x = 0 in Eq. (i), we get
⇒ 4(0) + 3y = 12
⇒ 3y = 12
12 4 3 y ⇒==
So, (0, 4) is a solution of the given equation.
On substituting y = 0 in Eq. (i), we get
4x + 3(0) = 12
⇒ 4x = 12 12 3 4 x ⇒==
So, (3, 0) is a solution of the given equation.
13. Find the co-ordinates of points where the line 2xy = 4 intersects x-axis and y-axis.
Sol. Given equation: 2xy = 4
The line intersects x-axis at y = 0.
∴ 2x 0 = 4
⇒ 2x = 4
⇒ x = 2
Hence, the co-ordinate where the equation 24 xy−= intersects x-axis is (2, 0).
Similarly, the line intersects y-axis at 0 x = .
∴ 2(0) y = 4
⇒ y = 4
⇒ y = 4
Hence, the co-ordinate where the equation 2xy = 4 intersects y-axis is () 0,4 .
14. At what point does the graph of the linear equation x + y = 5 meet a line which is parallel to the y-axis, at a distance 2 units from the origin and in the positive direction of x-axis. (NCERTExemplar)
Sol. The line parallel to the y-axis and at a distance of 2 units from the origin in the positive x-direction has the equation x = 2.
The coordinates of any point on this line are of the form () 2, . a
Substituting x = 2,y = a in the equation x + y = 5, we get 2 + a = 5
⇒ a = 5 − 2
⇒ a = 3
Thus, the required point is (2, 3).
The graph of the equation x + y = 5 meets the required line at the point (2, 3).
15. Check by substituting whether 6 and 3 xy=−=− is a solution of the equation () 2151 xy−−= or not. Find one more solution. How many more solutions can you find?
Sol. Given equation: () 2151 xy−−=
Substituting x = 6, y = 3
LHS = 2( 6 − 1) − 5( 3)
= 2( 7) + 15
= 14 + 15 = 1 = RHS
So yes, it is a solution.
Let’s find one more solution:
Let x = 0
2(0 − 1) − 5y = 1 ⇒ 2 − 5y = 1
⇒ 5y = 3 3 5 y ⇒=−
So, another solution is
3 0, 5 .
A linear equation in two variables has infinitely many solutions, as each value of x leads to a corresponding value of y satisfying the equation.
16. Solve for x: 7 31118 99 x x ++=+ . What will be the graph of the equation?
Sol. Given: 7 31118 99 x x ++=+
LHS: 27 28 311 1111 9999 xxxx x ++=++=+
RHS: 77162155 18 9999 −+=−+=
LHS = RHS 28155 11 99 x ⇒+= 281559956 9999 x =−= 28x = 56
17. The work done by a body under a constant force is the product of the constant force and the distance travelled in the direction of the force.
(a) If the constant force is 3 units, and the distance travelled is x, form the linear equation in two variables relating work done y and distance x.
(b) Also, calculate the work done when the distance travelled is 5 units
Sol. (a) Given: Work Done = Force × Distance
Given: Constant force = 3 units, Distance travelled = x units, Work done = y units
y = 3 × x ⇒ y = 3x
(b) Work done when x = 5.
y = 3 × 5 = 15
When the distance travelled is 5 units, the work done is 15 units.
18. Draw the graph of: xy = 2
Sol. Given: xy = 2 ⇒ x = 2 + y
(a) When y = 0, x = 2 + 0 = 2
(b) When y = 1, x = 2 + 1 = 3
(c) When y = 2, x = 2 + 2 = 4 x 2 3 4 y 0 1 2 y -axis x-axis (2, 0) (3, 1) (4, 2) 8
Short Answer Questions
1. When three times the larger of two numbers is divided by the smaller, then the quotient and remainder are 2 and 7, respectively. Form a linear equation in two variables for above and give its two solutions.
Sol. Let the larger and smaller numbers be x and y, respectively.
According to the question,
3x = 2y + 7
⇒ 3x 2y − 7 = 0
which is the required linear equation in two variables.
Now, substituting x = 0, we get
3(0) 2y 7 = 0 7
27 2 yy ⇒−=⇒=−
So, 7 0, 2 is a solution of 3x 2y 7 = 0
Now, substituting x = 1, we get
3 × 1 − 2y − 7 = 0
⇒ 3 − 2y − 7 = 0
⇒−2y = 4 ⇒ y = 2
So, (1, 2) is another solution of 3x 2y 7 = 0. Hence, two solutions are 7 0, 2 and () 1,2 .
2. Find the value of k for which the point (1, 2) lies on the graph of linear equation 2y + k = 0. Hence, find two more solutions of the equation.
Sol. Since (1, 2) lies on the graph of linear equation x 2y + k = 0 (i)
By substituting 1 x = , 2 y =− in (i), we have () 1220 k −−+=
140 k ⇒++=
505 kk ⇒+=⇒=−
Hence the equation (i) reduces to 250xy−−= (ii)
Let 3 x = and substitute in (ii)
3250 y −−=
221 yy ⇒−=⇒=−
One solution is () 3,1
Let 5 x =− and substitute in (ii)
52501020 yy −−−=⇒−−= 2105 yy ⇒−=⇒=−
Another solution is () 5,5. Hence, (3, −1) and (−5, −5) are two more solutions of the equation.
3. Determine the point on the graph of the equation 2519 xy+= , whose ordinate is 3 2 times its abscissa.
Sol. Let the point be () , xy , where: 3 2 yx =
Substitute 3 2 yx = into the given equation: 3 2519 2 xx += 15 219 2 xx+= 41519 1919 222 xxx +=⇒= 2 x = () 3 23 2 y ⇒==
∴ The required point is (2, 3).
4. Write four solutions for the following equation: 216xy π+= .
Sol. (a) Let us choose y = 0, then with this value of y, the given equation reduces to 16
016 xxπ π +=⇒=
So, π 16 , 0 is a solution of 216xy π+= .
(b) Take 2 x = , then the equation becomes 162 2216 8 2 yy π ππ +=⇒==−
So, () π 2,8 is a solution of 216xy π+=
(c) Take 0 x = , the equation becomes 02168 yy +=⇒=
So, () 0, 8 is a solution of 216xy π+= .
(d) Take 2 x =− , the equation becomes 162 2216 8 2 yy π ππ + −+=⇒==+
So, () −+π 2, 8 is a solution of 216xy π+= Thus, four of the infinitely many solutions of the equation 216xy π+= are:
()()() ππ π −−+ 16 , 0, 2, 8, 0, 8, 2, 8
5. Solve for ()()()() ++−=++ : 5138512.xxxxx
Sol. Given, ()()()() ++−=++ 5138512 xxxx
LHS:
()()++− 5138 xx
()() =+++− 53138 xxx
2 51538 xxx =+++− 2 5165 xx =+−
RHS:
()() 512 xx++
()() 22 522532 xxxxx =+++=++
2 51510 xx =++
LHS = RHS: 22 516551510 xxxx +−=++
1651510 xx−=+
510 x −= 15 x =
6. If the length of a rectangle is decreased by 3 units and breadth increased by 4 units, then the area will increase by 9 sq. units. Represent this situation as a linear equation in two variables. Also, find the breadth when the length is 10 units.
Sol. Let the length be x and breadth be y units.
∴ Area of the rectangle = xy New length =() 3 x New breadth =() 4 y +
∴ New area =()()349xyxy−+=+
()()349xyxy−+=+
43129xyxyxy ⇒+−−=+
43129 xy ⇒−−=
43210 xy ⇒−−=
So, the required linear equation is 43210 xy−−=
Given length 10 x =
Substitute in equation we get:
() 4103210 y −−=
403210 y ⇒−−= 19 193 6.33 units 3 yy ⇒=⇒=≈
Breadth is 6.33 units.
7. If the point (2k 3, k + 2) lies on the graph of the equation 2x + 3y + 15 = 0, find the value of k. Also, find the corresponding coordinates of the point.
Sol. As (2k 3, k + 2) lies on the line 2x + 3y + 15 = 0,
So, substituting x = 2k 3 and y = k + 2 in the equation, we get
()() 22332150 kk−+++=
4636150 kk−+++=
7150 k +=
715 k =−
15 7 k =
Now, calculate the coordinates using the value of 15 7 k =− 15 2323 7 xk
302151 777 =−−=− 15 22 7 yk=+=−+ 15141 777 =−+=−
So, the point is
511 , 77 .
8. The cost of a toy horse is same as that of cost of 3 balls. Express this statement as a linear equation in two variables. Also draw its graph.
Sol. Let the cost of toy horse be `x and cost of one ball be `y
∴ Cost of three balls = 3y According to the given condition, we have (x = 3y) (i)
For graph,
(a) Taking y = 1, in equation (i), we get x = 3(1) = 3
(b) Taking y = 2, in equation (i), we get x = 3(2) = 6
(c) Taking y = 3, in equation (i), we get x = 3(3) = 9
Now draw a graph taking P(3, 1), Q(6, 2) and R(9, 3). x 3 6 9 y 1 2 3
3)
(3, 1)
9. Solve for x: +=≠≠≠− −+ 314 , where 0, 1, 1 11 xxx xxx
Sol. Given ()() ()() 3111 4 11 xx xxx ++− = −+ 2 3314 1 xx xx ++− ⇒=
2 424 1 x xx + ⇒= ()() 2 4241xxx ⇒+=−
22 4244 xxx ⇒+=−
24 x ⇒=−
2 x ⇒=−
10. Find the solutions of the form , 0 xay== and 0, xyb == for the following equations: 2510 xy+= and 236 xy+= . Is there any common solution?
Sol. Given equations: 2510 xy+= and 236 xy+=
To find the solution of form , 0 xay== for 2510 xy+= , substituting 0 y = 2105 xx ∴=⇒=
Solution is () 5, 0 . For 236 xy+= , substituting 0 y = 263 xx=⇒=
Solution is () 3, 0
To find the solution of form 0, xyb == for 2510 xy+= , substituting 0 x =
5102 yy=⇒=
Solution is () 0, 2 . For 236 xy+= , substituting 0 x = 362 yy=⇒=
Solution is () 0, 2
Thus, the common solution is (0, 2).
11. Solve for x:
()()41 322 21 753 xx x ++ +=+
Sol. Given ()()41 322 21 753 xx x ++ +=+ 324442 753 xxx +++ += ()() 532744 42 35 3 xxx ++++ = 1510282842 35 3 xxx ++++ ⇒=
433842 353 xx++ ⇒= ()() 343383542 xx ⇒+=+ 12911414070 xx ⇒+=+ 14012911470 xx ⇒−+=−+ 14012944 xx ⇒−+=− 1144 x ⇒−=−
4 x ⇒=
12. Determine the point on the graph of the equation 2520 xy+= whose x-coordinate is 5 2 times its ordinate. Also, verify the point.
Sol. As the x-coordinate of the point is 5 2 times its ordinate, Therefore, 5 2 xy =
Now substituting 5 2 xy = in 2 5 20 xy+= , we get:
5 2520 2 yy
5520 yy ⇒+= 1020 y ⇒= 2 y ⇒=
Now, 5 25 2 x =×= Therefore, 5. x = Thus, the required point is () 5, 2 .
Verifying: ()() 2552101020 +=+= Correct
13. Write the equation 3 1 25 xy+= in 0 axbyc++= form and also find its two solutions.
15. Write 3218 xy+= in the form of ymxc =+ . Find the value of m and c. Is (4, 3) lies on this linear equation?
Sol. Given: 3218 xy+=
2183yx⇒=−
183 2 x y ⇒=
3 9 2 yx ⇒=−+
Here, 3 2 m =− and the 9 c = .
Verifying if the point (4, 3) lies on the line substituting 4 x = into equation:
3
49 2 y =−×+
69 y =−+
3 y =
Since the calculated y-value matches the given y-coordinate of the point, the point () 4, 3 lies on the line.
16. The sum of the ages of two friends is 40 years. If one of the friends is 17 years old, find the age of the other friend. What will be the sum of their ages after 10 years?
Sol. Let the age of one friend be x years and the age of the other friend be y years.
The sum of their ages: x + y = 40 (i)
If one friend is 17 years old.
Thus, substitute x = 17 into equation (i): 17 + y = 40
401723 y ⇒=−=
So, the other friend is 23 years old.
After 10 years:
Age of first friend 171027 years =+=
Age of second friend 231033 years =+=
Sum of their ages after 10 years: 273360 years +=
The sum of their ages after 10 years will be 60 years.
Long Answer Questions
1. Draw the graph of the linear equation 346 xy+= . At what points does the graph cut the x-axis and the y-axis?
y=6 X Y Y'
Given equation: 346 xy+= 463yx⇒=−
63 4 x y ⇒=
To draw graph, we need 3 points.
Substituting 2 x =− ()632 6612 3 444 y + ====
Point 1: (−2, 3)
Substituting 0 x = : ()630 6 1.5
44 y ===
Point 2: (0, 1.5)
Substituting 2 x = : ()632 660 0 444 y ====
x
Sol.
Point 3: (2,0)
Thus, the table of values is: x 2 0 2 y 3 1.5 0
The line passes through the points (–2, 3), (0, 1.5), and (2, 0).
Clearly, the graph line cuts x-axis and y-axis at (2, 0) and (0, 1.5) respectively.
2. The sum of a two-digit number and the number obtained by reversing the order of its digits is 88. Express this information in linear equation. Find the original number if the difference between the digits is 2.
Sol. Let x as the ten’s digit and y as the unit’s digit.
Then the two-digit number will be 10xy +
Reversed number = 10 yx +
Given: ()() 101088 xyyx+++=
111188 xy ⇒+=
8 xy ⇒+=
Linear equation: 8 xy+= (i)
Also given that the difference between the digits is 2:
(a) when xy >
2 xy−= 2 xy⇒=+
Substitute in equation (i) we get:
28 yy ++= 26 y ⇒= 3 y ⇒=
Using equations (i) again we get:
38 x += 835 x ⇒=−=
So, the original number is: 105353 ×+= (b) when xy <
2 yx−= 2 xy ⇒=−
Substitute in equation (i) we get: 28yy−+= 210 y ⇒= 5 y ⇒=
Using equations (i) again we get:
58 x += 853 x ⇒=−=
So, the original number is: 103535 ×+=
The original number can be 35 or 53.
3. The Auto-rikshaw fare in a city is charged `10 for the first kilometre and @ `4 per kilometre for subsequent distance covered. Write the linear equation to express the above statement. Draw the graph of the linear equation.
(NCERTExemplar)
Sol. Let the total distance covered be x km and the fare charged be y `.
Then for the first km, fare charged is `10 and for remaining () 1 x km fare charged is ` ()41 x ×− .
Therefore, ()1041yx=+− 104446yxx ⇒=+−=+
The required equation is 46yx=+ .
Now, when 0, 6 xy== and when –1, 2 xy== x 0 1 2 y 6 10 14
The graph is given in figure.
20 15 10 (0, 6) (1, 10) (2, 14) y = 4x + 6
4. The work done by a body on application of a constant force is the product of the constant force and the distance travelled by the body in the direction of force.
(a) If the body experiences a constant force of 3 units, express this relationship in the form of a linear equation in two variables
(b) Draw the graph of the linear equation.
(c) Find the work done when the distance travelled is 5 units.
(d) Determine the value of distance when the work done is 27 units.
Sol. (a) From the given information: Work Done = Constant Force × Distance If the constant force is 3 units, then:
Work Done = 3 × Distance
Let the work done be y units and the distance travelled be x units.
So, the linear equation becomes: y = 3x
(b) x 0 1 2 y 0 3 6 Y 6 C (2, 6) B (1, 3) 5 4 3 2 1 A O 1 2 3 4 5 X
(c) Given distance travelled is 5 units. 3 yx = () 3515 y ==
Work done is 15 units.
(d) Given work done is 27 units.
3 yx =
273 9 xx =⇒=
Distance travelled is 9 units.
5. For what value of p; 2, 3 xy== is a solution of ()()12310pxpy+−+−= ?
(a) Write the equation.
(b) How many solutions of this equation are possible?
(c) Is this line passing through the point (−2, 3)? Give justification.
Sol. (a) Given ()()12310pxpy+−+−= (i)
Substitute 2 x = and 3 y = into equation (i): ()()()()1223310pp+−+−= 226910 pp+−−−= 48048 pp −−=⇒−= 2 p ⇒=−
Substitute 2 p =− back into equation (i): ()()2122310 xy −+−−+−=
()() 4310110xyxy −−−+−=⇒−−−−= 10 xy −+−= 10 xy ⇒−+= (ii)
(b) Since equation (ii) is a linear equation in two variables, it has infinitely many solutions.
(c) if the point () 2, 3 lies on the line 10 xy−+= substituting 2 and 3 xy=−=
L.H.S = 2 − 3 + 1 = 4
R.H.S = 0
Since, LHS ≠ RHS. Therefore, the line 10 xy−+= does not pass through (−2, 3).
6. (a) If the point (4, 3) lies on the linear equation 3xay = 6, find whether ( 2, 6) also lies on the same line?
(b) Find the coordinates of the point on the above line when:
(i) abscissa is zero (ii) ordinate is zero
Sol. (a) If point () 4, 3 lies on 36 xay−= , then
3436 a ×−×=
1236 a ⇒−=
36126 a ⇒−=−=−
362 aa ⇒=⇒=
So, linear equation becomes 326 xy−= (i)
Substitute x = 2 and y = 6 in L.H.S. of (i), we get ()() L.H.S.3226 =×−−×− 6126R.H.S.=−+==
Hence, () 2,6 lies on the line 326 xy−=
(b) (i) When abscissa is zero, it means 0 x = .
From (i), we get
3026 y ×−×=
26 y ⇒−=
3 y ⇒=−
Required point is () 0,3
(ii) When ordinate is zero, i.e. 0 y = From (i), we get
3206 x −×=
2 x ⇒=
Required point is (2, 0).
Competency Based Questions
Multiple Choice Questions
1. A number say z is exactly the four times the sum of its digits and twice the product of the digits. Find the numbers.
(a) 36 (B) 37 (c) 38 (d) 39
2. If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of ∆AOB is of length
(a) 4 units (b) 3 units (c) 5 units (d) none of these
3. The larger of two supplementary angles exceeds thrice the smaller by 20°. The supplementary angles are
(a) 40° and 140° . (b) 140° and 30° .
(c) 50° and 130° . (d) 60° and 120° .
4. A boat travels 44 km upstream and 56 km downstream in 10 hours. If the speed of boat in still water is x km/h and the speed of stream is y km/h. Find the value of xy. (a) 2 (b) 3 (c) 5 (d) 6
5. If a linear equation has solutions ( 2, 2), (0, 0) and (2, 2), then it is of the form
(a) 0 yx−= (b) 0 xy+=
(c) 20 xy −+= (d) 20xy −+=
6. A diver rowing at the rate of 5 km/h in still water takes double the time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream (a) 2 5 (b) 3 5 (c) 5 2 (d) 5 3
7. Ravi and Raju can complete a job together in 6 days. If Ravi alone can complete the job in x days and Raju alone can complete it in y days, then express the given information as a linear equation in two variables. If Ravi can complete the work in 9 days, find how many days Raju alone will take to complete the same work.
(a) 16 days (b) 17 days
(c) 18 days (d) 19 days
8. Apeksha pays `1000 as fixed monthly hostel charges and `50 per day for meals. If x is the number of days she avails meals and y is the total monthly charge,
what is the correct linear equation and the amount she pays for 21 days?
(a) 501000yx=+ ; amount for 21 days is `2050
(b) 100050yx=+ ; amount for 21 days is `2050
(c) 1000 yx=+ ; amount for 21 days is `1021
(d) 1050 yx = ; amount for 21 days is `21150
9. Find the four angles of a cyclic Quadrilateral ABCD in which ∠A = (2x 1)° , ∠B = (y + 5)° , ∠C = (2y + 15)°, and ∠D = (4x 7)° .
(a) 65°, 55°, 115°, 125° (b) 55°, 55°, 125°, 125°
(c) 65°, 45°, 125°, 125° (d) 55°, 55°, 105°, 145°
10. A and B are friends. A is elder to B by 5 years. B’s sister C is half the age of B while A’s father D is 8 years older than twice the age of B. If the present age of D is 48 years, what is the present age of A?
(a) 15 years (b) 18 years
(c) 19 years (d) 25 years
Assertion-Reason Based Questions
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): The graph of the linear equation 29180 xy−+= meets x-axis at ( 9, 0).
Reason (R): Coordinates of points on y-axis are of the form () 0, a , where a is a variable.
2. Assertion (A): The graph of the linear equation ymxc =+ passes through the origin.
Reason (R): The linear equation ax + by = 0 represents a straight line passing through the origin.
3. Assertion (A): If ( 3 , 2 k ) is a solution of the linear equation ax + b = 0, then k = 1
Reason (R): The linear equation ax + b = 0 can be expressed as a linear equation in two variables as ax + y + b = 0.
1. Our excessive dependence on motorized road transport imposes significant economic costs on society. These include congestion, road casualties, physical inactivity and ill health caused by it e.g.; obesity. Cycling could substantially reduce these risks, while strengthening local economies in both urban and rural areas, supporting local businesses and boosting the economic productivity.
A bicycle is being lent at fixed charges for the first three days of the week and an additional charge of each day thereafter. If fixed charges are `x and per day charges are `y, then based on the above information, answer the following questions:
(a) Form the linear equation, if Sam paid `27 for a bicycle kept for 7 days.
(b) Form the linear equation, if David paid `30 for a bicycle kept for 6 days.
(c) If fixed charge is `15, then find the additional charge paid by Sam for each extra day.
Answers
Multiple Choice Questions
1. (a) 36
Let x is the tens digit, and y is the units digit.
From equation I we get 5622 yyyyy +=⇒=() 2 0606 yyyy ⇒=−⇒=−
⇒ y = 0 or y = 6 (Exclude 0 as that will make our number 00)
Substitute: 6 y = into equation I, we get 623 xx =⇒=
Required number, () 103636 z =+=
(d) If per day charge for lending the bicycle is `3, find the value of fixed charge paid by David.
2. Prime Minister’s National Relief Fund (PMNRF) in India is the fund raised to provide support for people affected by natural and man-made disasters. It was first consolidated during the time of the first Prime Minister of India, Jawaharlal Nehru. The fund is fully collected from the public. Kamya and Shweta, together contributed `500 towards the Prime Minister’s National Relief Fund.
Based on above information answer the following questions.
(a) Write the linear equation for the above situation.
(b) If Shweta contributed `280, then what is the amount contributed by Kamya?
(c) If both of them contribute an equal amount, then what is the amount contributed by each of them?
(d) If amount paid by Kamya is four times the amount paid by Shweta, then how much does each contribute?
2. (c) 5 units
x-intercept: =⇒=⇒=⇒() 041233, 0 yxxA
y-intercept: =⇒=⇒=⇒() 031240, 4 xyyB
Using Pythagoras theorem, 22 Hypotenuse34916255 =+=+==
3. (a) 40° and 140°
The two angles are supplementary, so their sum is: 180 xy+= (i)
Let’s assume:
Smaller angle = x
Larger angle = y
The larger angle exceeds thrice the smaller by 20°, So equation: 320yx=+ (ii)
From equation (i) and (ii), ()320180420180xxx ++=⇒+= 416040 xx ⇒=⇒= () 32034020140yx=+=+=
Smaller angle = 40; Larger angle = 140
4. (b) 3
Let x = boat speed (km/h), y = stream speed (km/h).
Upstream speed = xy, Downstream speed = xy
Using formula, Distance = Speed × Time
() =−×⇒−=4410 4.4 xyxy (i)
() =+×⇒+=5610 5.6 xyxy (ii)
On adding (i) and (ii),
210 x5 x =⇒=
Putting 5 x = , in equation (i)
54.4 0.6 yy −=⇒=
Therefore, 50.63 xy =×=
5. (b) x + y = 0
Tr y 0 xy+=
( 2 + 2 = 0)
(0 + 0 = 0)
(2 + (–2) = 0)
All satisfied.
6. (d) 5 3
Upstream speed = 5 x
Downstream speed = 5 x + Distance
Time of travel Speed =
So, downstream travel time = 40 5 x + upstream travel time = 4040 2 55xx =× −+
4080 55xx = −+
()()405805xx+=−
2004040080 xx⇒+=−
4080400200 xx+=− 2005 120200 1203 xx ⇒=⇒==
The speed of the stream is 5 km/h 3 .
7. (c) 18 days
Work done by Ravi in 1 day = 1 x
Work done by Raju in 1 day = 1 y
Together in 1 day: 111 6 xy +=
Multiply throughout by 6xy to eliminate denominators:
66660 yxxyxyxy +=⇒−−=
Given 9 x = , () 9696095460 yyyy −−=⇒−−=
35418 yy ⇒=⇒=
So, Raju alone can do the job in 18 days.
8. (a) 501000yx=+ ; amount for 21 days is `2050
Fixed = `1000
Meal cost = `50/day
Let x = days of meals
Let y = total charge
501000yx=+
For 21 days: () 50211000105010002050 y =+=+=
9. (a) 65°, 55°,115°, 125°
We know that sum of opposite angles of a cyclic quadrilateral is 180o
∴∠A +∠C = 180o and ∠B +∠C = 180o
∠A +∠C = 180o ⇒()()++=° 2 – 1 2 15180 xy
⇒+= 83 xy (i)
∠B +∠C = 180o ⇒()()++=° 5 4 – 7180 yx
⇒ 4 182 xy+= (ii)
Subtracting (i) from (ii) we get =⇒= 3182–8333 xx
Substituting in (i), we get = 50 y
∴∠=×=° 2 33 – 1 65 A
∠B = 50 + 5 = 55o
∠C = 2 × 50 + 15 = 115o
∠D = 4 × 33 7 = 125o.
10. (d) 25 years
Let the present age of the sister ‘C’ be x years.
It is given that B’s sister C is half the age of B.
∴ Age of C = 1 2 × Age of B
⇒ Age of B = 2 × Age of C
⇒ Age of B = 2x
Also, it is given that A is elder to B by 5 years.
∴ Age of A = Age of B + 5 years
⇒ Age of A = (2x + 5) years
Further, it is given that A’s father D is 8 years older than twice the age of B.
∴ Age of D = 2 × Age of B + 8 years
⇒ Age of D = (2 × 2x + 8) years
⇒ Age of D = (4x + 8) years
⇒ 48 = 4x + 8
[∴ Present age of D = 48 years (given)]
⇒ 4x = 48 8 ⇒ 4x = 40
⇒ x = 40 ÷ 4 = 10
Hence, the present ages of A, B, and C are: (2 × 10 + 5) years, (2 × 10) years, and 10 years, respectively, i.e., 25 years, 20 years, and 10 years respectively.
Assertion-Reason Based Questions
1. (b) Both A and R are true, but R is not the correct explanation of A.
To find the x-intercept, put y = 0 in the equation: () 29018021809 xxx −+=⇒+=⇒=−
So, the point is () 9, 0 . Assertion is correct.
Reason (R): Yes, any point on the y-axis has the form () 0,a : true
But this does not explain how the graph meets the x-axis.
2. (d) A is false, but R is true.
The line ymxc =+ passes through the origin only if 0 c =
So, this assertion is not always true- False
The equation 0 axby+= always passes through the origin, because when x = 0, y = 0 and vice versa: True
3. (c) A is true, but R is false. Put 3 2 x =− into 230 xy+= :
3 2303301 2 kkk
A is correct.
Reason:
The transformation 00axbaxyb +=⇒++= is mathematically wrong.
Correct way to express it as two-variable equation is 00axyb++= .
Reason is false.
Case Study Based Questions
1. (a) Form the linear equation, if Sam paid `27 for a bicycle kept for 7 days.
Let fixed charges for first 3 days = `x
Additional charges for extra days = `y per day
Number of extra days = 7 3 4 −=
So, total cost = `27
Linear equation: 427xy+= (i)
(b) Form the linear equation, if David paid `30 for a bicycle kept for 6 days.
Number of extra days = 6 3 3 −=
So, total cost = `30
Linear equation: 330xy+= (ii)
(c) If fixed charge is `15, then
Putting 15 x = in the equation from (i): 15427412 yy +=⇒=
3 y ⇒=
Additional charge per extra day = `3
(d) If per day charge for lending the bicycle is `3, find the value of fixed charge paid by Mr. David.
Putting y = 3 in the equation from (ii): () 3330930xx+=⇒+=
⇒ x = 21
Fixed charge = `21
2. (a) Let the amount contributed by Kamya and Shweta be `x and `y
The situation can be represented as a linear equation in two variables as:
500 xy+=
(b) Here, amount contributed by Shweta is `280.
∴ Amount contributed by Kamya will be:
500 yx =−
500280220 y ⇒=−=```
(c) If both of them contribute equal amount (Let x), 5002500xxx+=⇒=
250 x ⇒=
(d) Given, 4 xy = +=⇒=4500 5500yyy
100 y ⇒=
400 x ⇒=
Therefore, Kamya contributed `400 and Shweta contributed `100.
Additional Practice Questions
1. Solve the system of equations: 2x + 3y = 7 and 5x 4y = 3 What is the value of x + y ?
7. The graph of line x + y = 0 passes through (a) (2, 3) (b) (3, 4) (c) (5, 6) (d) (0, 0)
8. Which of the following equations represents a line parallel to the y-axis?
(a) 25yx = (b) 25 y = (c) 25 x = (d) 235 xy+=
9. Equation of line parallel to y-axis and passing through point ( 3, 7) is (a) 3 x =− (b) 3 y =− (c) 7 y = (d) 7 x =
10. The graph of 6 y = is a line
(a) parallel to x-axis at a distance 6 units from the origin
(b) parallel to y-axis at a distance 6 units from the origin
(c) making an intercept 6 on the x-axis
(d) making an intercept 6 on both the axis
11. Write the equation which is represented by the graph given in the figure.
12. Solve for x and y if 237 xy+= and 3 xyx +=
13. A rabbit covers y metres distance by walking 10 metres in slow motion and the remaining by x jumps, each jump contains 2 metres. Express this information in linear equation.
14. Age of x is more than the age of y by 10 years. Express this statement in linear equation.
15. When a number is divided by another number, the quotient and remainder obtained are 9 and 1 respectively. Express this information in linear equation.
16. If x years represents the present age of the father and y years represents the present age of the son, then find the equation of the statement: “Present age of the father is 5 more than 6 times age of the son”.
17. If () + , 21mm is the solution of the equation 5369 xy+= , find the value of m
18. A sum of `90,000 is distributed to three persons are in the ratio 3 : 2 : 5. Determine the share of each of them.
19. Which of the following point(s) lie on the linear equation ()() 523144 xy−++= . (a) A(5, 0) (b) B(0, 17) (c) C(4, 4) (d) D(5, 1)
20. Side of an equilateral triangle is x. If the perimeteris 30 cm, find the value of x
21. Find the coordinates of the point where the line 236 xy−= intersects the y-axis.
22. Find whether , 3 a a lies on the line 3 yx = or not?
23. The perimeter of a rectangle is 22 m. Write the given information in the form of a linear equation in two variables. Also, find two more solutions for the equation.
24. Find the value of k if () 1,1 is a solution of the equation −= 45 xky . Also, find the coordinates of another point lying on its graph.
25. Find three solutions of the equation += 2 35.xy
26. Find the value of x and y in the following system of linear equations:
2x 3y = 5 4x + y = 9
27. Find the equation of a line passing through the points (2, 5) and (4, 7).
28. A part of monthly expenses of a family on milk is fixed is `500 and the remaining varies with the quantity of milk, taken extra at the rate of `20 per kg. Taking the quantity of milk required extra as x kg and the total expenditure on milk `y, write a linear equation for this information.
29. ABC is a right-angled triangle given in the figure. Write equation of AB and find its length. Also, find area of ∆ABC
30. For what value of c, the linear equation 28 xcy+= has equal values of x and y for its solution?
31. Find the solutions of the form x = a,y = 0 and x = 0, y = b for the following equations: 2510 xy+= and 236 xy+= . Is there any common solution?
32. Express the following linear equation in the form ax + by + c = 0 and indicate the value of a, b, and c:
50 2 x y +−=
Find any one solution of the equation.
33. Solve for x: 7 31118 99 x x ++=+ . What will be the graph of the equation?
34. Check by substituting whether 6 x =− and 3 y =− is a solution of the equation () 2151 xy−−= or not. Find one more solution. How many more solutions can you find?
35. Find three solutions of the linear equation: 3412 xy+=
36. How many solution(s) of the equation 5325 xx−=+ are there on the:
(a) Number line?
(b) Cartesian plane?
37. In a one-day International Cricket match, played between India and England in Kanpur, two Indian batsmen, Yuvraj and Abhishek scored 200 runs in a partnership including 5 extra runs. Express this information in the form of an equation.
38. The sum of a two-digit number and the number obtained by reversing the order of its digits is 88. Express this information in linear equation.
Challenge
Yourself
1. The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation:
(a) If the temperature is 86°F, what is the temperature in Celsius?
(b) If the temperature is 35°C, what is the temperature in Fahrenheit?
(c) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(d) What is the numerical value of the temperature which is same in both the scales?
2. A fruit vendor sells 2 apples and 3 oranges for `39 and makes a profit of `6 on this transaction. In another sale, he sells 3 apples and 2 oranges for `38 but incurs a loss of `4. Determine the actual cost price of one apple and one orange.
3. The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as a linear equation of two variables and draw the graph of the same by
Answers
Additional Practice Questions
1. (a) 60 23 xy+=
2. (a) 3 4
3. (a) 6
4. (c) 26xy+=
5. (a) 07xy+=
6. (b) 2 axbyc +=
7. (d) (0, 0)
8. (c) 25 x =
9. (a) 3 x =−
10. (a) parallel to x-axis at a distance 6 units from the origin
11. 240 xy−+=
12. 77 , 84xy==
13. 102 yx =+
14. 10 xy=+
15. 9 1 yx=+
16. 65xy=+
17. 6 m =
18. First person = 27,000 ` ; second person = 18,000` ; third person 45,000 = `
19. Only point B(0, 17) lies on the line.
20. 10 x = cm
21. The point is () 0,2
22. 0 a =
23. 11; xy+= Two solutions: (5, 6) and (7, 4)
taking the constant mass equal to 6 kg. Read from the graph, the force required when the acceleration produced is:
(a) 5 m/sec² (b) 6 m/sec²
4. In a regular polygon with n sides, each interior angle is 360 180 n degrees. How many sides does a polygon have if each interior angle is 156°?
24. 1; k = Another point: () 0,5
25. ()()() 1,1,4,1,2,3
26. 16 7 x = , 1 7 y =−
27. The equation is y = x + 3
28. 20 500 yx=+
29. 4312 yx=+ , 5 units, 9 sq units
30. 82 , x c x = x ≠ 0
31. () 0, 2 is common solution of given equations.
32. One solution is (0, 5).
33. 2 x =
34. Yes, it satisfies the equation. Another solution: 3 0, 5
35. Three solutions: (0, 3); (4, 0); (2, 1.5)
36. (a) On number line: 1 (b) On Cartesian plane: Infinitely many
37. 1950 xy+−=
38. 8 xy+=
Challenge Yourself
1. (a) 30°C (b) 95° F
(c) 0° C = 32° F, 0 17.78 FC°≈− (d) −40
2. Apple: `12 and oranges: `3
3. (a) 30 N (b) 36 N
4. 15 sides
Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 60 mins
Multiple Choice Questions
1. If the equation 3x + 4y = 12 is given, which of the following is a solution?
(a) (2, 3)
(c) (4, 1)
2. The equation x = 4 represents a:
(a) Horizontal line
(c) Diagonal line
Scan me for Solutions
Max. Marks: 40
[3 × 1 = 3Marks]
(b) (1, 3)
(d) (2, 1.5)
(b) Vertical line
(d) None of these
3. The width of a rectangle is one-third of its length. If the perimeter is 96 cm, what is the width of the rectangle?
(a) 12 cm
(c) 24 cm
Assertion-Reason Based Questions
(b) 16 cm
(d) 18 cm
[2 × 1 = 2Marks]
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
4. Assertion (A): The equation 4270 xy−+= does not represents a quadratic equation.
Reason (R): A quadratic equation always has a term with 2 x or 2 y .
5. Assertion (A): The equation 10 xy+= has only whole number solution.
Reason (R): A linear equation in two variables has infinitely many solutions.
Very Short Answer Questions
6. Find the values of x and y for the equation 347 xy+= and 23 xy−= .
[2 × 2 = 4Marks]
7. If a straight line 310xy+= intersects another line 2 y = , find x-coordinate of the point of intersection of lines.
Short Answer Questions
8. For what value of m the point () 2, m lies on the line 35xy+= ?
9. If =+ 3 Cst , then find the value of t when 3 5 C = and 1 10 s = .
[4 × 3 = 12Marks]
10. At what point does the graph of the linear equation += 239 xy meet a line which is parallel to the y-axis, at a distance of 4 units from the origin and on the right of the y-axis?
11. Write two solutions of the given equation 4515 xy−= .
12. If (2, 3) and (4, 0) lie on the graph of the equation 1. axby+= Find the value of a and b, and plot the graph of the obtained equation.
13. Draw the graph of the linear equation 4 x = and 5 y = . Find the area formed by the two graphs and the axes.
14. Two girls have `76 between them. If the first girl gave the second girl `7, they would each have the same amount of money. How much does each girl have?
Case Study Based Questions
15. In a stationery shop, notebooks and pens are sold. Deepanshu purchased 3 notebooks and 6 pens, whereas Suraj purchased 5 notebooks and 9 pens. Deepanshu paid a total of `210 and Suraj paid a total of `330.
(a) Form the pair of linear equations in two variables.
(b) Find the cost of one notebook and one pen.
(c) Find the total cost if 9 notebooks and 10 pens are purchased.
5 Introduction to Euclid Geometry
Chapter at a Glance
Axioms
Self-evident true statements used throughout mathematics and not specific to geometry are called axioms.
Euclid’s Axioms
1. Things which are equal to the same thing are equal to one another.
2. If equals are added to equals, the whole are equal.
3. If equals are subtracted from equals, the remainders are equal.
4. Things which coincide with one another are equal to one another.
5. The whole is greater than the part.
6. Things which are double of the same thing are equal to one another.
7. Things which are halves of the same thing are equal to one another.
Postulates
Statements that are assumed to be true based on basic geometric principles are called postulates.
Euclid’s Postulates
1. A straight line may be drawn from any one point to any other point.
2. A terminated line can be produced indefinitely.
3. A circle can be drawn with any centre and any radius.
4. All right angles are equal to one another.
5. If a straight line falling on two straight lines makes the interior angles on the same side of it, taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.
Though Euclid defined a point, a line and a plane, the definitions are not accepted by mathematicians. Therefore, these terms are now taken as undefined.
Non-Euclidean Geometry
All the attempts to prove Euclid’s fifth postulate using the first 4 postulates failed. But they led to the discovery of several other geometries called non-Euclidean geometries.
1. To distinct lines cannot have more than one point in common.
2. Two lines which are both parallel to same line are parallel each other.
Key Definitions
Statement: A sentence which can be judged to be true or false is called a statement. Theorem: A statement that requires a proof is called a theorem and establishing the truth of a theorem is known as proving the theorem.
Corollary: A statement whose truth can easily be deduced from a theorem, is called its corollary.
Important Points in Geometry
1. A line contains infinitely many points.
2. Through a given point, infinitely many lines can be drawn.
3. Two lines having a common point are called intersecting lines.
4. Three or more lines intersecting at the same point are said to be concurrent lines.
5. Three or more than three points are said to be collinear, if there is a line containing all of them.
6. Two lines in a plane are said to be parallel if they have no point in common. The distance between two parallel lines always remains the same.
NCERT Zone
NCERT Exercise
1. Which of the following statements are true and which are false? Give reasons for your answers.
(a) Only one line can pass through a single point.
(b) There are an infinite number of lines which pass through two distinct points.
(c) A terminated line can be produced indefinitely on both the sides.
(d) If two circles are equal, then their radii are equal.
(e) In the figure, if ABPQ = and PQ = XY, then AB = XY
Sol. (a) False. Infinitely many lines can pass through a point in different directions.
(b) False. As only one line can pass through two distinct points.
(c) True. A terminated line or line segment can be produced indefinitely on both sides to give a line.
(d) True. Two circles with equal area (i.e., equal circles) will have the same radius from the relative area 2 r=π .
(e) True. From the axiom that if two things are separately equal to a third thing, then, they are equal to each other.
2. Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?
(a) Parallel lines (b) Perpendicular lines
(c) Line segment (d) Radius of a circle
(e) Square
Sol. (a) Parallel lines are lines in a plane which do not meet; that is, two lines in a plane that do not intersect or touch each other at any point. ‘Point’ and ‘straight line’ are defined as: A point is an exact position or location on a plane surface. A line is breadth less length and a straight line is a line which lies evenly with the points on itself.
(b) Perpendicular lines: If one among two parallel lines is turned by 90°, the two lines become perpendicular to each other.
(c) Line segment: A line bounded by two distinct end points is a line segment.
(d) Radius of a circle: A straight line from the centre to the circumference of a circle or sphere. ‘Centre’ is defined as a point inside the circle which is at the equally distant from all points on the circle.
(e) Square: A plane figure with four equal straight sides and four right angles.
3. Consider two ‘postulates’ given below:
(a) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(b) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.
Sol. In postulate (a) ‘in between A and B’ remains as an undefined term which appeals to our geometric assumption.
The postulates are consistent. They do not contradict each other. Both of these postulates do not follow from Euclid’s postulates. However, they follow from the axiom given below.
Given two distinct points, there is a unique line that passes through them.
(a) Let AB be a straight line. There are an infinite number of points composing this line. Choose any except the two end-points A and B. This point lies between A and B
(b) Two points can only be connected by a straight line. Therefore, there have to be at least three points for one of them not to fall on the straight line between the other two.
4. If a point C lies between two points A and B such that ACBC = , then prove that 1 2 ACAB = . Explain by drawing the figure.
Sol. AB C
ACCB =
Add AC to both sides of equation: So, . ACACBCAC +=+
BCAC + coincides with AB
⇒ 2 ACAB = ⇒ 1 2 ACAB = Hence, proved.
5. In Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.
Multiple Choice Questions
1. Euclid’s second axiom (as per the order given in the Class IX textbook) is:
(a) The things which are equal to the same thing are equal to one another.
(b) If equals be added to equals, the wholes are equal.
(c) If equals be subtracted from equals, the remainders are equal.
(d) Things which coincide with one another are equal to one another.
Sol. Let there be two mid points C and D on line AB. Then from above theorem 1 2 ACAB = . And 1 2 ADAB = This proves, ACAD = But this can be only possible if D coincides with C. Therefore, we can say that C is the one and only mid-point on AB.
6. In figure, if ACBD = , then prove that ABCD = .
Sol. Given: ACBD = ACABBC =+ BDBCCD =+
As ACBD = (given) ABBCBCCD +=+
ABCD = Hence, proved.
7. Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate.)
Sol. Axiom 5: ‘Whole is always greater than its part.’ Part is always included in the whole and therefore can never be greater than the whole in magnitude. Hence, This is a ‘universal truth’.
2. Euclid’s fifth postulate is:
(a) The whole is greater than the part.
(b) A circle may be described with any center and any radius.
(c) All right angles are equal to one another.
(d) If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.
3. The things which are double of the same thing are: (a) equal (b) unequal (c) halves of the same thing (d) double of the same thing
4. Axioms are assumed as:
(a) universal truths in all branches of mathematics
(b) universal truths specific to geometry (c) theorems
(d) definitions
5. John is of the same age as Mohan. Ram is also of the same age as Mohan. State the Euclid’s axiom that illustrates the relative ages of John and Ram. (a) First Axiom (b) Second Axiom (c) Third Axiom (d) Fourth Axiom
6. The side faces of a pyramid are: (a) Triangles (b) Squares (c) Polygons (d) Trapeziums
7. In Indus Valley Civilisation, the bricks used for construction had dimensions in the ratio: (a) 1 : 3 : 4 (b) 4 : 2 : 1 (c) 4 : 4 : 1 (d) 4 : 3 : 2
8. Boundaries of solids are: (a) surfaces (b) cur ves (c) lines (d) points
9. A pyramid is a solid figure whose base is: (a) only a triangle (b) only a square (c) only a rectangle (d) any polygon
10. Boundaries of surfaces are: (a) surfaces (b) cur ves (c) lines (d) points
Answers
1. (b) If equals be added to equals, the wholes are equal.
2. (d) If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.
3. (a) equal
4. (a) universal truths in all branches of mathematics
11. The total number of propositions in Euclid’s Elements are:
(a) 465 (b) 460
(c) 13 (d) 55
12. The number of dimensions a point has:
(a) 0 (b) 1 (c) 2 (d) 3
13. Euclid divided his famous treatise “The Elements” into:
(a) 13 chapters (b) 12 chapters
(c) 11 chapters (d) 9 chapters
14. The number of dimensions a surface has:
(a) 1 (b) 2 (c) 3 (d) 0
15. The steps from solids to points are:
(a) Solids → Surfaces → Lines → Points
(b) Solids → Lines → Surfaces → Points
(c) Lines → Points → Surfaces → Solids
(d) Lines → Surfaces → Points → Solids
16. The number of dimensions a solid has:
(a) 1 (b) 2 (c) 3 (d) 0
17. The number of interwoven isosceles triangles in Sriyantra (in the Atharva Veda) is:
(a) Seven (b) Eight (c) Nine (d) Eleven
18. Euclid stated that all right angles are equal to each other in the form of:
(a) an axiom (b) a definition
(c) a postulate (d) a proof
19. Which of the following needs a proof?
(a) Theorem (b) Axiom
(c) Definition (d) Postulate
5. (a) First Axiom 6. (a) Triangles
7. (b) 4 : 2 : 1 8. (a) surfaces
9. (d) any polygon 10. (b) cur ves
11. (a) 465
13. (a) 13 chapters
12. (a) 0
14. (b) 2
15. (a) Solids → Surfaces → Lines → Points
16. (c) 3
18. (c) a postulate
17. (c) Nine
19. (a) Theorem
Constructed Response Questions
Very Short Answer Questions
1. If P,Q and R are three points on a line and Q is between P and R, then prove that PRQR = PQ.
Sol.
PQR
In the above figure, PQ coincides with PRQR. So, according to the axiom, “things” which coincide with one another are equal to ‘one another’. We have,
PRQR = PQ
2. Solve the equation u 5 = 15 and state the axiom that you use here.
Sol. u 5 =15
On adding 5 to both sides, we have
u 5 + 5 = 15 + 5
Euclid’s second axiom, when equals are added to equals, the wholes are equal. or u = 20
3. In given figure, AC = XD, C is the mid-point of AB and D is the mid-point of XY. Using Euclid’s axiom, show that AB = XY
Sol. Given:
AC = XD
AB = 2AC (C is the mid-point of AB)
XY = 2XD (D is the mid-point of XY)
Thus, AB = 2 × AC = 2 × XD = XY
∴ AB = XY, because things which are double of the same things are equal to one another.
4. In the given figure, if ∠1 =∠3, ∠2 =∠4 and ∠3 =∠4, write the relation between ∠1 and ∠2, using Euclid’s axiom. 1 2 3 4
Sol. Here, ∠3 =∠4, ∠1 =∠3 and ∠2 =∠4
According to Euclid’s first axiom, the things which are equal to equal things are equal to one another.
∴∠1 =∠2
5. Look at the figure, and show that Length AH > sum of lengths of AB + BC + CD.
ABCDEFGH
Sol. In the given figure, we can see that the lengths AB, BC and CD are parts of line segment AH.
Now since we have ⇒ ABBCCDAD ++= (1)
Also, AD is a part of line AH
Therefore, by Euclid’s fifth axiom, we have “the whole is greater than the part.”
Hence, we can write, AHAD >
That gives, AHABBCCD >++
Therefore, Length AH > sum of lengths of ABBCCD ++
6. Solve the equation, x 10 = 25 and state which axiom you use here.
Sol. Given:
x − 10 = 25
On adding 10 on both sides, we have x 10 + 10 = 25 + 10 ⇒ x = 35
Here, we use Euclid axiom, “If equal be added to the equal, the whole are equal.”
7. Prove that every line segment has one and only one mid-point on it.
Sol. Let R is the mid-point of PQ, such that PR = QR.
Let T be the mid-point of PQ other than R, then 1 2 PRPQ = 1 2 PTPQ⇒= Therefore 1 2 PRPTPQ == .
By Euclid’s First Axiom (Things which are equal to the same thing are equal to one another), This is possible only if R and T coincide, i.e., they represent the same point.
Every line segment has one and only one midpoint. Hence, proved.
8. Show that if p + q = 6 and p = r. Show that r + q = 6.
Sol. Given: p + q = 6 and p = r
So, from Euclid’s axiom, that if equals are added to equals, the wholes are equal.
Therefore, we get
p + q = r + q = 6
⇒ r + q = 6
Hence, proved.
Short Answer Questions
1. Which of the following statements are true and which are false? Justify your answer.
(a) If two circles are equal then their radii are unequal.
(b) Given two distinct points, there is a unique line that passes through them.
Sol. (a) False. If two circles superimpose exactly on one another then they coincide. So their centres and regions bounded by their boundaries must also coincide. Therefore, their radii will also coincide. Hence equal circles have equal radii.
(b) True: Only one line can be passed through two distinct points.
2. In the given figure, if AB = BC and AP = CQ, then prove that BP = BQ.
Sol. Given:
= CQ
To Prove: BP = BQ
From the figure,
AB = AP + BP and BC = BQ + CQ
According to Euclid’s axiom, if equals are subtracted from equals, the remainders are equal.
Therefore, by subtracting (ii) from (i), we get
ABAP = BCCQ (Given AP = CQ)
ABCQ = BCCQ ⇒ BP = BQ
Hence, proved.
3. In the given figure PR = QS, then show that PR = QS. Name the mathematician whose postulate/axiom is used for the same.
PQRS
Sol. Given: PR = QS
Subtracting QR on both sides, we have
⇒ PRQR = QSQR
⇒ PQ = RS
Hence, proved.
We used here Euclid’s axiom to prove the result which states that if equals are subtracted from equals, then remainders are equal.
4. In the following figure, we have AC = DC, CB = CE
Show that AB = DE
Sol. In the above figure, we are given that
AC = DC and CB = CE
Now from Euclid’s second axiom, we know that “if equals are added to equals, the wholes are equal.”
Therefore, after adding the above two equations, we have
AC + CB = DC + CE
From the figure,
AC + CB = AB and DC + CE = DE
That gives us, AB = DE
Hence, proved.
5. Study the following statement:
“Two intersecting lines cannot be perpendicular to the same line”. Check whether it is an equivalent version to Euclid’s fifth postulate.
[Hint: Identify the two intersecting lines l and m and the line n in the above statement.]
Sol. Let the two intersecting lines be l and m, and let line n be the line to which they are both claimed to be perpendicular.
If both l and m are perpendicular to the same line n, then l and m must be parallel.
However, this contradicts the given condition that lines l and m are intersecting lines.
Therefore, it is impossible for two intersecting lines to be perpendicular to the same line.
This reasoning does not represent an equivalent version of Euclid’s Fifth Postulate, which discusses the conditions under which two lines meet or remain parallel based on angle sums, not perpendicularity.
Hence, the given statement is not an equivalent version of Euclid’s fifth postulate.
6. In the given figure, we have ∠ABC =∠ACB, ∠4 =∠3.
Show that ∠1 =∠2.
BC 4 1 3 2
Sol. Given that,
∠ABC =∠ACB And ∠4 =∠3
Now using Euclid’s third axiom, we have “if equals are subtracted from equals, the remainders are equal.”
Therefore, subtracting these equations gives us,
∠ABC − ∠4 =∠ACB − ∠3
Hence, we have ∠1 =∠2 Hence, proved.
Long Answer Questions
1. Read the following statement:
“A square is a polygon made up of four linesegments, out of which, the length of three linesegments are equal to the length of the fourth one and all its angles are right angles”.
Define the terms used in this definition which you feel necessary. Are there any undefined terms in this? Can you justify that all angles and sides of a square are equal?
Sol. The terms used in this definition which needs to be defined are:
Polygon: A polygon is a two-dimensional simple closed figure, made up of three or more linesegments.
Line segments: A line segment is a part of a line with finite length, having two endpoints.
Angle: An angle is the formation obtained after two rays are initiated from a common point.
Right Angle: The measure of an angle is 90° then it is known as a right angle.
Ray: Ray is a part of a line with only one endpoint and moves endlessly in the other direction.
The undefined terms used here are a line and a point.
Here, we are given that all the angles of a square are right angles.
Now, according to Euclid’s fourth postulate which illustrates that, “all right angles are equal to one another,”.
We can say that all the angles of a square are equal to one another.
Also, we are given that the length of three linesegments is equal to the length of the fourth one.
Now, according to Euclid’s first axiom which illustrates that, “things which are equal to the same thing are equal to one another,”.
We can say that all sides of a square are equal to one another.
Hence, all angles and all sides of a square are equal.
2. In the given figure:
(a) AB = BC, M is the mid-point of AB and N is the mid-point of BC. Show that AM = NC.
(b) BM = BN, M is the mid-point of AB and N is the mid-point of BC. Show that AB = BC.
Sol. (a) To show: AMNC = In the above given figure we have, AB = BC
M and N are the mid-points of AB and BC respectively.
Now, using Euclid’s seventh axiom, we have “Things which are halves of the same thing are equal.”
Therefore, we have 11 22 ABBC =
From 1 and 2.
We have, AM = NC Hence, proved.
(b) To show: AB = BC
In the above given figure we have, BN = BM
M and N are the mid-points of AB and BC respectively.
Now since BN = BM, therefore multiplying both sides by 2 we have,
2BM = 2BN
Now by using Euclid’s sixth axiom, we have “Things which are double of the same thing are equal to one another.”
From 3 and 4. 11 22 ABBC =
Therefore, we have ABBC = Hence, proved.
3. Read the following statement:
An equilateral triangle is a polygon made up of three line segments out of which two line segments are equal to the third one and all its angles are 60° each.
Define the terms used in this definition which you feel necessary. Are there any undefined terms in this? Can you justify that all sides and all angles are equal in an equilateral triangle?
Sol. There are some of the terms used in the above definition which can be defined separately. They are:
Triangle: A triangle is a three sided closed figure made up of line segments. It is a three sided polygon.
Polygon: A polygon is a closed figure made up of at least three or more line segments. Example: a square, hexagon.
Line segment: A line segment is a finite part of a line with two endpoints having a finite length.
Angle: When two rays initiate from a common initial point they create an angle.
Ray: A line with one fixed end and extended indefinitely in the other direction.
Acute angle: An angle whose measure is between 0° and 90°.
The undefined terms used here are a line and a point. The justification for an equilateral triangle can be given as: In a triangle, if two line segments are equal to the third line segment then all the three sides of an equilateral triangle are equal.
This is illustrated by Euclid’s first axiom, which states that “things which are equal to the same things are equal to one another.”
If AB = BC and BC = CA, then by the axiom, AB = BC = CA
Also, all of its angles are 60° each. Therefore, all three angles of an equilateral triangle are equal. Therefore, all the sides and angles are equal in an equilateral triangle.
4. Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they and how might you define them?
(a) Parallel lines
(b) Perpendicular lines
(c) Line segment
(d) Radius of a circle
(e) Square
Sol. Yes, we need to have an idea about the terms like point, line, ray, angle, plane, circle and quadrilateral, etc. before defining the required terms.
Definitions of the required terms are given below:
(a) Parallel Lines:
Two lines l and m in the same plane that do not intersect at any point, no matter how far they are extended, are called parallel lines.
We write them as lm l m
(b) Perpendicular Lines:
Two lines p and q lying in the same plane are said to be perpendicular if they intersect and form a right angle (90°).
We write them as pq ⊥ q
(c) Line Segment:
A line segment is a part of line and having a definite length. It has two end-points. In the
Competency Based Questions
Multiple Choice Questions
1. It is known that if x + y = 10 then x + y + z = 10 + z
The Euclid’s axiom that illustrates this statement is (a) First Axiom. (b) Second Axiom. (c) Third Axiom. (d) Fourth Axiom.
(NCERTExemplar)
2. Ramesh drew a line passing through the points A, B and C. Neha drew a line passing through B and C Which statement about the lines they drew is correct?
(a) The lines coincide
(b) The lines are perpendicular.
(c) The lines are parallel.
(d) The lines meet at two points but are not perpendicular.
3. The number of inter woven isosceles triangles in Sriyantra (in the Atharva veda) is
figure, a line segment is shown having end points A and B. It is written as AB or BA . AB
(d) Radius of a Circle:
The distance from the centre to a point on the circle is called the radius of the circle. In the figure,P is centre and Q is a point on the circle, then PQ is the radius.
(e) Square:
A quadrilateral in which all the four angles are right angles and all the four sides are equal is called a square.
Given figure, PQRS is a square. PQ SR
(a) Seven (b) Eight
(c) Nine (d) Eleven
(NCERTExemplar)
4. If a straight line falling on two straight lines makes the interior angles on the same side of it, whose sum is 120°, then the two straight lines, if produced indefinitely, meet on the side on which the sum of the angles is
(a) less than 120°
(b) greater than 120°
(c) is equal to 120°
(d) greater than 180°
5. According to Euclid’s axioms, if AB = CD and AB = EF, which of the following is always true?
(a) CD = EF
(c) CD < EF
(b) CD > EF
(d) CD + EF = AB
6. A transversal cuts two lines such that the sum of the interior angles on the same side is exactly 180°. What can be said about the two lines?
(a) They are parallel
(b) They are intersecting at some point
(c) They coincide
(d) They are skew lines
7. If two distinct lines are both perpendicular to the same line, what is the correct conclusion about them in Euclidean Geometry?
(a) They are parallel and intersecting
(b) They are parallel and non-intersecting
(c) They coincide
(d) They must form a triangle
8. Which of the following directly leads to the proof that a line has infinite length in Euclidean geometry?
(a) Euclid’s First Axiom
(b) Euclid’s Second Postulate
(c) Euclid’s Third Axiom
(d) Euclid’s Second Postulate
9. Which of the following statements contradicts Euclid’s Fifth Postulate?
(a) Two lines never meet, however far they are extended.
(b) Two lines meet at one point only if the sum of interior angles is exactly 180°.
(c) Two lines meet on the side where the sum of interior angles is less than two right angles.
(d) Two lines meet at two points but are not parallel.
Assertion-Reason Based Questions
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): Self-evident true statements are called axioms.
Reason (R): Statements whose truth can be logically established are called theorems.
2. Assertion (A): Through two distinct points there can be only one line that can be drawn.
Reason (R): From the two points A and B, we can draw only one line.
Case Study Based Questions
1. The Indus Valley Civilisation (3000 BCE) demonstrated advanced knowledge of geometry through well-planned cities like Harappa and Mohenjo-Daro, featuring parallel roads, underground drainage systems, and houses with various room designs. Bricks used in construction followed a precise ratio of 4 : 2 : 1 (length : breadth : thickness), indicating practical applications of mensuration.
In ancient India, the Sulbasutras (800 500 BCE) provided guidelines for constructing altars with precise geometric shapes like squares, circles, rectangles, and triangles, especially for Vedic rituals. The Sriyantra from the AtharvaVeda is a notable geometric design consisting of nine interwoven isosceles triangles that form 43 subsidiary triangles. Unlike the Babylonians and Egyptians, who focused on practical geometry, the Greeks emphasized logical reasoning and proofs. Thales gave the first known geometric proof, and Pythagoras made significant contributions to geometric theory. Euclid later organized this knowledge systematically in his famous work, Elements, which influenced mathematical thinking for generations.
Answer below question on based of above text.
(a) What evidence from the Indus Valley Civilisation indicates their understanding of geometry and mensuration?
(b) Explain the significance of the brick ratio (4 : 2 : 1) used in Indus Valley constructions. How does this reflect practical geometry?
(c) What was the purpose of the Sulbasutras, and how did they contribute to the development of geometry in ancient India?
(d) Describe the geometric structure of the Sriyantra mentioned in the AtharvaVeda. How many subsidiary triangles are formed in its design?
Answers
Multiple Choice Questions
1. (b) Second Axiom.
If 10 xy+=
then 10 xyzz ++=+
The Euclid’s second axiom states that if equals are added to equals, the wholes are equal.
2. (a) The lines coincide.
Ramesh drew a line passing through points A, B, and C.
Neha drew a line passing through points B and C
Since both lines pass through points B and C, and a unique line can be drawn through two points (Euclid’s First Postulate), both lines must be the same.
So, the correct statement is “The lines coincide”
3. (c) Nine
4. (a) less than 120
According to Euclid’s Fifth Postulate, if the sum of the interior angles is less than 180°, the lines meet on that side. Since it’s already 120°, the lines will meet on the side where the sum of angles is less than 120° (to reduce it below 180°).
5. (a) CD = EF
By Euclid’s First Axiom: Things which are equal to the same thing are equal to one another.
6. (a) They are parallel
According to Euclid’s Fifth Postulate, if the sum of the interior angles is exactly 180°, the lines are parallel and will never meet.
7. (b) They are parallel and non-intersecting
Since both lines are perpendicular to the same line, they will never meet and are parallel.
Additional Practice Questions
1. Euclid’s second axiom is
(a) The things which are equal to the same thing are equal to one another.
(b) If equals be added to equals, the wholes are equal.
8. (d) Euclid’s Second Postulate
Euclid’s Second Postulate states that a terminated line can be produced indefinitely. This implies lines can be extended infinitely
9. (d) Two lines meet at two points but are not parallel
According to Euclidean geometry, two lines can meet at most at one point or remain parallel. Saying they meet at two points but are not parallel is impossible.
Assertion-Reason Based Questions
1. (b) Both A and R are true, but R is not the correct explanation of A.
2. (a) Both A and R are true, and R is the correct explanation of A.
Case Study Based Questions
1. (a) The cities like Harappa and Mohenjo-Daro were well-planned with parallel roads, underground drainage systems, and houses having rooms of various designs. This shows their knowledge of geometry and mensuration.
(b) The brick ratio of 4 : 2 : 1 (length : breadth : thickness) ensured structural stability and ease of construction, reflecting a practical application of geometric measurements in architecture.
(c) The Sulbasutras provided guidelines for constructing altars used in Vedic rituals. They contributed to the development of geometry by introducing the use of precise geometric shapes like squares, circles, rectangles, and triangles for religious purposes.
(d) The Sriyantra is a complex geometric design consisting of nine interwoven isosceles triangles. These triangles form 43 subsidiary triangles.
(c) If equals be subtracted from equals, the remainders are equals.
(d) Things which coincide with one another are equal to one another.
(NCERTExemplar)
2. The total number of propositions in “The Elements” are (a) 465 (b) 460 (c) 13 (d) 55
3. The number of interwoven isosceles triangles in Sriyantra (in the AtharvaVeda) is: (a) 7 (b) 8 (c) 9 (d) 11
4. Euclid belongs to the country: (a) Babylonia (b) Egypt (c) Greece (d) India
5. Look at the figure. Show that length AG > sum of lengths of AB + BC + CD
6. What are the basic facts which are taken for granted, without proof and which are specific to geometry?
7. (a) Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the 5th postulate.)
(b) How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
8. Refers to the figure given below and answer the following questions,
(a) If AE is subtracted from AB, then AB – AE = _____.
Challenge Yourself
1. “A square is a polygon made up of four line segments, out of which, length of three line segments is equal to the length of fourth one and all its angles are right angles”
Using Euclid’s Axiom/postulates, justify that all angles and sides of a square are equal.
(b) If GE is added to FG, then the result is equal to ______.
(c) If GD is added to AG, the whole is equal to ______.
(d) If AF is subtracted from AC, the remainder is equal to _____.
9. The equation given is x 5 = 25. Solve for x. Mention the Euclid axiom.
10. Consider two ‘postulates’ given below:
(a) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(b) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow with Euclid’s postulates? Explain.
11. A square is a polygon made up of five line segments, out of which, the length of three line segments is equal to the length of a fourth of one and all its angles are right angles” Define the term used in this definition which you feel necessary.
12. Read the following statements which are taken as axioms.
(a) If a transversal intersects two parallel lines, then corresponding angles are not necessarily equal.
(b) If a transversal intersects two parallel lines, then alternate interior angles are equal.
Is this system of axioms consistent? Justify your answer.
2. Read the following statement.
“A rhombus is a polygon made up of four line segments out of which the length of three line segments is equal to the length of the fourth one and its diagonal intersect at right angles.”
Define the term used in this definition which you feel necessary. Are there any undefined terms in this? Can you justify that all sides of a rhombus are equal?
Answers
Additional Practice Questions
1. (b) If equals be added to equals, the wholes are equal.
2. (a) 465
3. (c) 9
4. (c) Greece
6. Postulates
7. (a) Since it is true for things in any part of the universe so, this is a universal truth.
(b) If the sum of the co-interior angles made by a transversal intersecting two straight lines at distinct points is less than 180°, then the lines cannot be parallel.
8. (a) EB
(b) FE
(c) AD (d) CF
9. x = 30. The axiom used here is Euclid’s axiom: “If equals are added to equals, the wholes are equal.”
10. Yes, these postulates contain undefined terms like point and line.
These two statements are consistent as they talk about two different situations meaning different things.
These statements do not follow Euclid’s postulates but one of the axioms about “Given any two points, a unique line that passes through them” is followed.
11. Polygon: A simple closed figure made up of three or more line segments.
Line Segment: Part of a line with two endpoints.
Angle: A figure formed by two rays with a common initial point.
Ray: any of a set of straight lines passing through one point.
Right angle: Angle whose measure is 90°.
12. Statement I is false. Hence it is not an axiom. Statement II is true. Hence it is an axiom.
Challenge Yourself
2. Undefined terms used are: line and point.
Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 60 mins
Multiple Choice Questions
Scan me for Solutions
Max. Marks: 40
[3×1=3Marks]
1. Pythagoras was a student of: (a) Thales (b) Euclid (c) Both A and B (d) Archimedes
2. Greeks emphasized on: (a) Inductive reasoning (b) Deductive reasoning (c) Both A and B (d) Practical use of geometry
3. ‘Lines are parallel if they do not intersect’ is stated in the form of (a) an axiom (b) a definition. (c) a postulate (d) a proof.
Assertion-Reason Based Questions
[2×1=2Marks]
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
4. Assertion (A): According to Euclid’s 1st Axiom” Things which are equal to the same thing are also equal to one another.”
Reason (R): If ABPQ = and PQ = XY, then AB = XY.
5. Assertion (A): If ‘l’ is a line and P is a point not on the line l, there is one and only one line (say m) which passes through P and parallel to l.
Reason (R): If two lines ‘l’ and ‘m’ are both parallel to the same line n, they will also be parallel to each other.
Very Short Answer Questions
6. Explain when a system of axioms is called consistent.
7. Express in variables the things which are double of the same thing are equal
Short Answer Questions
8. P and Q are the centres of two intersecting circles. Prove that PQQRPR == PQ
[2×2=4Marks]
[4×3=12Marks]
9. In a triangle ABC, X and Y are the points on AB and BC such that BX = BY and AB = BC. Show that AX = CY. State the Euclid’s Axiom used.
BC X Y
10. Prove that every line segment has one and only one mid-point.
11. How many planes can be made to pass through:
(a) Three collinear points.
(b) Three non-collinear points.
Long Answer Questions
12. Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.
13. Which of the following statements are true and which are false? Give reasons for your answers.
(a) Only one line can pass through a single point.
(b) There are an infinite number of lines which pass through two distinct points.
(c) A terminated line can be produced indefinitely on both sides.
(d) If two circles are equal, then their radii are equal.
(e) In figure if ABPQ = and PQ = XY, then AB = XY
[3×5=15Marks]
14. Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?
(a) parallel lines (b) perpendicular lines (c) line segment (d) Radius of circle
Case Study Based Questions
15. If a point C be the midpoint of a line segment AB, then the relationship among AB, BC and AC can be explained by:
AB C
(a) Relationship Between BC and AC is
[1×4=4Marks]
(i) 2BC = AB (ii) 1 2 ACAB = (iii) AC = BC (iv) All of these
(b) Which Euclid Axiom’s state the required result.
(i) All right angles are equal to one another.
(ii) Things which are equal to the same thing are equal to one another.
(iii) Things which are halves of the same thing are equal to one another.
(iv) None of the above
(c) Here, A, B and C are
(i) Points of intersection (ii) Collinear Points (iii) Non-collinear Points (iv) None of these
(d) For given line segment , ABC is midpoint of AB , if
(i) C' is an exterior point of AB
(iii) C is not lying on AB
(ii) C is an interior point of AB
(iv) None of the above
6 Lines and Angles
Chapter at a Glance
Basic Terms and Definitions
Point: A point has no size or dimension; it only represents a position.
Line: A line is formed by joining two distinct points. It extends infinitely in both directions and has no endpoints.
Line Segment: It is a part of a line that has two definite endpoints.
Ray: A ray is a part of a line that starts from a point and extends infinitely in one direction.
Collinear and Non-Collinear Points
Collinear Points: Points that lie on the same straight line.
Non-Collinear Points: Points that do not lie on the same straight line.
Intersecting Lines: Lines that cross each other at a point.
Non-Intersecting (Parallel) Lines: Lines that never meet and remain at a constant distance apart.
Concurrent Lines: Three or more lines in a plane are said to be concurrent if they all pass through a single common point. This common point is called the point of concurrency.
Angles
Definition: When two rays originate from the same endpoint, they form an angle.
Complementary Angles: Two angles are called complementary if their sum is exactly 90°.
Example: If one angle is 30°, the other must be 60° to make them complementary.
Supplementary Angles: Two angles are called supplementary if their sum is exactly 180°.
Example: If one angle is 110°, the other must be 70° to make them supplementary.
Relation Between Two Angles
Two angles with the common vertex and a common side.
Two adjacent angles whose noncommon arms form a straight line.
6. It is given that 64 XYZ ∠=° and XY is produced to point P . Draw a figure from the given information. If ray YQ bisects , ZYP∠ find XYQ∠ and reflex QYP∠ QZ 64° PYX
3. In the figure, if || PQST , 110 PQR ∠=° and 130 RST ∠=°, find QRS∠ . P R 110° 130° Q ST
Sol Extend PQ to Y and draw || LMST through R
LRM XY ST PQ
TSXQXS∠=∠ (Alternate angles)
130 QXS ⇒∠=°
180 QXSRXQ ∠+∠=° (Linear pair)
18013050 RXQ ⇒∠=°−°=° (1)
PQRQRM∠=∠ (Alternate angles)
110 QRM ∠=° (2) RXQXRM∠=∠ (Alternate angles)
50 XRM ⇒∠=° (By (1))
QRSQRMXRM ∠=∠−∠ 1105060 QRS ∠=°−°=°
4. In the figure, if || ABCD , 50 APQ ∠=° and 127 PRD ∠=° , find x and y . APB
50° 127° y x
CQRD
Sol. APQPQR∠=∠ (Alternate angles)
50 x ⇒=°
Also, 127 xy+=° (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)
50127 y ⇒°+=°
1275077 y ⇒=°−°=°
Hence, 50 x =° and 77 y =°
5. In the figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD . Prove that ||.ABCD
Sol. At point B , draw BEPQ ⊥ and at point C , draw CFRS ⊥
12∠=∠ (1)
Multiple Choice Questions
1. If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, then the greater of the two angles is
(a) 54° (b) 108°
(c) 120° (d) 136° (NCERTExemplar)
2. Which one of the following is the supplementary angle of 105°?
(a) 65° (b) 75°
(c) 85° (d) 95°
3. Find the angle such that four times the angle is equal to 150° less than twice its complement. (a) 15° (b) 10° (c) 25° (d) 5°
4. Find the value of k in line l || m?
15 3x + 40
(a) 90° (b) 83° (c) 132° (d) 41°
5. What is the measure of an angle whose measure is 32° less than its supplement? (a) 148° (b) 60° (c) 74° (d) 55°
(Angle of incidence is equal to angle of reflection)
34∠=∠ (2)
(Angle of incidence is equal to angle of reflection)
Also, 23∠=∠ (3) (Alternate angles)
14⇒∠=∠ (From (1), (2) and (3))
2124⇒∠=∠
11 44⇒∠+∠=∠+∠
12 34⇒∠+∠=∠+∠ (From (1) and (2))
BCDABC⇒∠=∠
Thus, ||.ABCD (Alternate angles are equal) Hence, proved.
6. In the following figure, lines are l1 || l2 || l3. The value of x is
(a) 40° (b) 170°
(c) 140° (d) 70°
7. In the following figure, AB || CD. The value of x is A x B CD 140° 35°
(a) 40° (b) 70°
(c) 75° (d) 60°
8. In the following figure, AOB is a straight line. The value of angle BOC is AB O C 3x + 35° 2x + 10°
(a) 64° (b) 54° (c) 81° (d) 116°
9. Find the value of y if AB || CD. ABD C Y° 140° 120°
(a) 60° (b) 80°
(c) 132° (d) 280°
10. In the following figure, find the value of y OB 3y 40° 2y A
(a) 60° (b) 80°
(c) 32° (d) 28°
11. If two parallel lines are cut by a transversal, then the pairs of alternate interior angles are:
(a) Acute angles (b) Equal
(c) Complementary (d) Supplementary
12. In the following figure, the value of x is B x 110° 120°
(a) 160° (b) 130°
(c) 132° (d) 128°
13. In the following figure, find the value of angle ABC B x C A E 127° 50°
(a) 53° (b) 50° (c) 77° (d) 127°
14. Find the value of x from the given figure. (3x + 10)° (4x 26)° C
15. Find the value of x, if B lies on AC. Given 3 ABx=+ , 4 – 5 ACx = and 2 BCx = .
(a) 5 (b) 8
(c) 3 (d) 2
16. When two lines are parallel, they:
(a) Intersect at one point
(b) Intersect at two points
(c) Intersect at three points
(d) Does not intersect at any point
17. If the ratio of the angles is 2 : 4 : 3. The value of the smallest angle will be:
(a) 40° (b) 80°
(c) 60° (d) 20°
18. If AOB is a line then the measure of ∠BOC, ∠COD and ∠DOA respectively in the given figure, are:
2y 3y 5y
(a) 36°, 54°, 90°
(b) 90°, 54°, 36°
(c) 90°, 36°, 54°
(d) 36°, 90°, 54°
19. An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is
(a) 37 1 2 ° (b) 1 52 2 °
(c) 72 1 2 ° (d) 75º (NCERTExemplar)
20. If one of the interior angles of a triangle measures 130°, what is the angle between the internal bisectors of the other two angles?
(a) 50° (b) 65°
(b) 145° (c) 155°
(a) 96° (b) 86° (c) 76° (d) 28°
Answers
1. (b) 108°
2. (b) 75°
3. (d) 5°
4. (b) 83°
5. (c) 74°
6. (c) 140°
7. (c) 75°
8. (a) 64°
9. (d) 280°
10. (d) 28°
Constructed Response Questions
Very Short Answer Questions
1. In the given figure, AB, CD and EF are three lines concurrent at O. Find the value of y (NCERTExemplar)
Sol. C BF O EA D 5y 2y n 2y
5 AOEBOFy∠=∠= (Vertically opposite angles)
Also, 180 COEAOEAOD ∠+∠+∠=° (Sum of angles on a line)
So, 2 5 2 180 yyy++=°
9 180 y ⇒=°
20 y ⇒=°
2. If an angle is half of its complementary angle, then find its other angle.
Sol. Let the required angle assumed to be x
∴ Its complement = 90– x °
Now, according to given statement, we obtain
() 1 90 –2 xx =°
2 90–xx⇒=°
3 90 x ⇒=°
30 x ⇒=°
Hence, the required angle is 30º.
11. (b) equal
12. (b) 130°
13. (c) 77°
14. (d) 28°
15. (b) 8
16. (d) does not intersect at any point
17. (a) 40°
18. (a) 36°, 54°, 90°
19. (b) 1 52 2 °
20. (d) 155°
3. Two supplementary angles are in ratio 2 : 7. Find the measures of angles.
Sol. Given: 2 7 180 xx+=°
9 180 x ⇒=°
20 x ⇒=°
So, the angles are
2 2 20 40 x =×°=°
7 7 20 140 x =×°=°
So, two angles are 40° and 140°.
4. In the given figure, xy = and ab = . Prove that || ln
Sol. y x a lmn b
xy = (Given)
Therefore, || lm (Corresponding angles) (1)
Also, ab = (Given)
Therefore, || nm (Corresponding angles) (2)
From (1) and (2), l || n (Lines parallel to the same line)
Hence, proved.
5. Two angles measure (30° a) and (125° + 2a).
If each one is the supplement of the other, then find the value of a
Sol. Angles (30° a) and (125° + 2a) are supplementary of each other, then
301252180 aa °−+°+=°
⇒ 155° + a = 180°
⇒ a = 180° 155° = 25°
Thus, the value of a is 25°.
6. In the following figure, find the value of x.
4x 2x x
Sol. ∠AOC +∠BOC +∠BOD +∠DOA = 360° (complete Angle at a point)
4x + 2x + x + 150° 360°
7360150 x =°=°
7210 x =°
30 x =°
7. In the given figure, if lm , then find the value of x
Sol. Since lm ,
∴∠1 = 60° (Corresponding angles)
Now, 401 x ∠+°=∠ (Exterior angle property)
⇒ 60 40 20 x ∠=°−°=°
8. In a ΔABC, ∠A +∠B = 110°, ∠C +∠A = 135°. Find ∠A.
Sol. Given: 110 AB ∠+∠=° and 135 CA ∠+∠=°
On adding, we get 110 135 ABCA ∠+∠+∠+∠=°+°
2∠A +∠B +∠C = 245°
Using angle sum property, ∠A +∠B +∠C = 180°
So, ∠A + 180° = 245°
∠A = 245° 180° = 65°
9. If the difference between two supplementary angles is 40°, then find the angles.
Sol. Let the two supplementary angles be x and 40 x +
40180 xx++=
218040 x =−
2140 x =
70 x =
Also, 407040110 x +=+=
So, the angles are 70° and 110°.
10. In the given figure, if PQ || RS, then find the measure of angle m.
PQ T RS 56° 14° m
Sol. Here, PQ || RS, PS is a transversal.
⇒() 56 Alternate interior angles
PSRSPQ ∠=∠=°
Also, 180 TRSSTRTSR ∠+∠+∠=°
14 56 180 m °++°=°
⇒ 1801456110 m =°−°−°=°
11. In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that () 1 –. 2 ROSQOSPOS ∠=∠∠ S R
PQ O
Sol. Since given OR is perpendicular to PQ
⇒∠ 90 PORROQ=∠=°
∴ 90 POSROS ∠+∠=°
⇒ 90 ROSPOS∠=°−∠
Adding ∠ROS to both sides, we have () 90–ROSROSROSPOS ∠+∠=°+∠∠
⇒ 2 –ROSQOSPOS ∠=∠∠
⇒() 1 –2 ROSQOSPOS ∠=∠∠
Hence, proved.
Short Answer Questions
1. In the figure AG ||CD, EF ⊥ CD and ∠GFC = 130°. Find x, y and z
2. In the figure given below, an exterior angle of a triangle is 130° and the two interior opposite angles are equal. Find each of these angles. BC A D x x
Sol. In a triangle ABC, ACDABCBAC ∠=∠+∠ (exterior angle property)
130 xx+=°
2130 x =°
65 x =°
Therefore, 65 AB ∠=∠=°
180 ACBACD ∠+∠=° (Linear pair)
130 180 ACB ∠+°=°
50 ACB ∠=°
Hence, ∠ABC = 65°; ∠BAC = 65° and ∠ACB = 50°.
3. In the given figure, ∠DOB = 87° and ∠COA = 82°. If ∠BOA = 35°, then find ∠COB and ∠COD.
Sol. Given: ∠COA = 82°
∠COB +∠BOA = 82°
O B A 82° 35°
∠COB + 35° = 82° (∵ ∠BOA = 35°)
∠COB = 82° 35°
∠COB = 47°
Similarly, ∠DOB =∠DOC +∠COB
87° =∠DOC + 47°
∠DOC = 87° 47°
∠DOC = 40°
4. In the given figure, two straight lines PQ and RS intersect each other at O. If ∠POT = 75°, find the values of a, b, c.
Sol. Here, ROS is a straight line.
Therefore, 180 ROPPOTTOS ∠+∠+∠=°
4 75 3 180 bb+°+=°
7105 b =°
b = 15°
Therefore, 460ab==°
Therefore, 2 180 ac+=° (Linear Pair)
602180 c °+=°
2120 c =°
60 c =°
5. Prove that if two lines intersect each other, then the bisectors of vertically opposite angles are in the same line.
Sol. AB and CD are assumed to be two intersecting lines intersecting themselves at O. OP and OQ are bisectors of ∠AOD and ∠BOC.
7. In given figure, AB || CD and EF || DG, find ∠GDH, ∠AED and ∠DEF.
Sol. Since AB || CD and HE is a transversal.
40 AEDCDH ∴∠=∠=° (Corresponding angles)
Now, 180 AEDDEFFEB ∠+∠+∠=° (Linear angles)
40 45 180 DEF °+∠+°=°
∠ 180–454095 DEF =°−=°
Again, given that EF || DG and HE is a transversal.
∴ 95 GDHDEF ∠=∠=° (Corresponding angles)
Hence, 95, 40 GDHAED ∠=°∠=° and 95 DEF ∠=°
8. In figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA respectively. Find ∠APB
Sol.
Here, AP and BP are bisectors of ∠EAB and ∠RBA respectively.
⇒ 12and34 ∠=∠∠=∠
Since DE || QR and transversal n intersects DE and QR at A and B respectively.
⇒∠ 180 EABRBA+∠=° (since co-interior angles are supplementary)
⇒()() 1 2 3 4180 ∠+∠+∠+∠=°
⇒()() 1 1 3 3180 ∠+∠+∠+∠=° (using (i))
⇒() 21 3180 ∠+∠=°
⇒ 1 390 ∠+∠=°
Now, in ΔABP, by angle sum property, we have 180 ABPBAPAPB ∠+∠+∠=°
⇒ 3 1 180 APB ∠+∠+∠=°
⇒ 90 180 APB °+∠=°
⇒∠APB = 90°
9. If xywz +=+ , then prove AOB is a line.
C
B w O x y
A D z
Sol. xywz +=+
We know that the sum of all the angles around a fixed point is 360º.
Therefore, we can easily determine that 360º AOCBOCAODBOD ∠+∠+∠+∠= Or 360º yxzw+++= R P QB DA 1 3 4 2 E n
But, it’s already given that xywz +=+ (According to the Q).
Therefore, () 2360º yx+=
180º yx+=
We are also aware that if a ray stands on a straight line, then the sum of all the angles of linear pair formed by the ray with respect to the line is 180º.
Thus, AOB is a line.
Hence, proved.
10. In given figure, if
AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Sol. Given: AB || DE, ∠BAC = 35° and ∠CDE = 53°
To find: ∠DCE
According to angle sum property of a triangle, sum of the interior angles of a triangle is 180°.
Since, AB || DE and AE is the transversal,
∠DEC =∠BAC (Alternate interior angles)
Thus, ∠DEC = 35°
Now, in ΔCDE
∠CDE +∠DEC +∠DCE = 180° (Angle sum property of a triangle)
53° + 35° +∠DCE = 180°
∠DCE = 180° 88°
Thus, we have ∠DCE = 92°.
11. In the given figure, m and n are two plane mirrors perpendicular to each other. Show that incident rays CA is parallel to reflected ray BD.
(NCERTExemplar)
Sol. Let PA and PB be perpendicular to mirrors n and m, respectively.
As mirrors are perpendicular to each other, therefore, BP || OA and AP || OB.
So, BPPA ⊥ i.e., 90 BPA ∠=°
Therefore, 3 2 90 ∠+∠=° (angle sum property) (i)
Also, 1 2 and 4 3∠=∠∠=∠ (Angle of incidence = Angle of reflection)
Therefore, 1 4 90 ∠+∠=° (from (i)) (ii)
Adding (i) and (ii), we have 1 2 3 4 180 ∠+∠+∠+∠=°
i.e., 180 CABDBA ∠+∠=°
Thus, CA || BD Hence, proved.
12. In given figure, find the values of x and y and then show that AB || CD.
Sol. Line CD is intersecting with line P. Hence the vertically opposite angles so formed are equal.
Thus, y = 130°.
Similarly, line AB is intersecting with line P forming a linear pair. Hence the sum of adjacent angles formed is 180°.
x + 50° = 180°
x = 180° 50°
x = 130°
We know that, if a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel.
Here we can see that the pair of alternate angles formed when lines AB and CD are intersected by transversal P are equal. i.e, x = y = 130°.
So we can say the two lines AB and CD are parallel. Hence, proved.
Long Answer Questions
1. If two parallel lines are intersected by a transversal, prove that the bisectors of two pairs of interior angles form a rectangle.
Sol. Given, AB || CD and transversal EF cut them at P and Q respectively and the bisectors ofpair of interior angles form a quadrilateral PRQS.
Sol. Let ∠B = 2x and ∠C = 2y
Therefore, OB and OC bisect ∠B and ∠C respectively.
11 2 22 OBCBxx ∠=∠=×=
And 11 2 22 OCBCyy ∠==×=
Now, in Triangle BOC, we have 180 BOCOBCOCB ∠+∠+∠=°
⇒ 180 BOCxy ∠++=°
⇒() 180 BOCxy ∠=°−+
Now, in Triangle ABC, we have 2 180 ABC ∠++=°
⇒ 2 2 180 Axy ∠++=°
Since PS, QR, QS and PR are the bisectors of angles
Subtracting (1) from (2), we have ⇒()–90––190–) QRTPSxTPSx ∠∠=°+∠°−∠ –90––90 QRTPSxTPSx ⇒∠∠=°+∠°+∠+
2 2 –TPSQR⇒∠=∠ () 1 –2 TPSQR ⇒∠=∠
Hence, proved.
6. It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Sol. Z XP Q Y 64°
Given that, XP is a straight line.
So, 180 XYZZYP ∠+∠=°
So by substituting the value of ∠XYZ = 64° we will get, 64 180 ZYP °+∠=°
∴ 116 ZYP ∠=°
From the above diagram, we also get to know that ZYPZYQQYP ∠=∠+∠
Since, YQ bisects ∠ZYP, ZYQQYP∠=∠
Or, 2 ZYPZYQ∠=∠ 58 ZYQQYP ∴∠=∠=°
Again, XYQXYZZYQ ∠=∠+∠
So, by substituting the value of ∠XYZ = 64° and ∠ZYQ = 58° we get.
64 58 XYQ ∠=°+°
Or, 122 XYQ ∠=°
Now, reflex angle of 180 QYPXYQ∠=°+∠
Therefore, we calculate that the value of ∠XYQ = 122°.
So,
180 122 QYP ∠=°+°
∴∠QYP = 302°
7. In given figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT. R 40° 95°
Sol. According to the angle sum property of a triangle, sum of the interior angles of a triangle is 360°.
Competency Based Questions
Multiple Choice Questions
1. In given figure, if ,,25ABCDEFPQRSRQD∠= ‖‖‖ and 60 CQP ∠= , then QRS∠ is equal to
In ΔPRT,
∠PTR +∠PRT +∠RPT = 180° (Angle sum property of a triangle)
∠PTR + 40° + 95° = 180°
180135 PTR ∠=°−°
∠PTR = 45°
Now,
∠QTS =∠PTR (Vertically opposite angles)
∠QTS = 45° (1)
In ΔTSQ,
∠QTS +∠TSQ +∠SQT = 180° (Angle sum property of a triangle)
45° + 75° +∠SQT = 180° (From (1))
∠SQT = 180° 120°
∠SQT = 60°
Thus, ∠SQT = 60°
3. In the given figure ACBD ‖ , and AEBF ‖ . The measure of x∠ is A x B
(a) 85
(c) 145
(b) 135
(d) 110
(NCERTExemplar)
2. In the given figure, lines p and q are parallel, then x equals
(a) 130° (b)110° (c) 70° (d) 50°
4. In the given figure, COE∠ and BOD∠ are right angles. If the measure of BOC∠ is four times the measure of COD∠ , what is the measure of AOB∠ ?
(a) 22
(c) 62
(a) 16 (b) 17 (c) 18 (d) 19
5. In the given figure, ray AD is the bisector of CAB∠ and the measure of ACD∠ is y. What must be the measure of ADC∠ in order for line AB to be parallel to line CD
AB
y °
CD
(a) 90 y (b) 90 2 y (c) 180 2 y (d) 90 y +
6. AB is a straight line and O is a point lying on AB A line OC is drawn from O such that 36 COA ∠= . OD is a line within COA∠ such that 1 3 DOA ∠= COA∠ . If OE is a line within the 1 , 4 BOCBOEBOC ∠∠=∠ , then DOE∠ must be (a) 60 (b) 132 (c) 144 (d) 108
7. The straight lines AB and CD intersect at E If EF and EG are bisectors of DEA∠ and AEC∠ respectively and AEFx∠= and AEGy∠= , then (a) 90 xy+> (b) 90 xy+< (c) 90 xy+=° (d) 180 xy+=
8. Given that AB || DE, ∠ABC = 115°, ∠EDC = 140°, then the value of x is: 115° 140° C D x AB E (a) 45 (b) 55 (c) 65 (d) 75
9. In the given figure ||,40,35PQRSRSFPQF∠=∠= and QFPx∠= . What is the values of x? (a) 75 (b) 105 (c) 135
(d) 140 SQ
10. Lines m and n are cut by a transversal, so that 1∠ and 5∠ are corresponding angles. If 1267 x ∠=− and 52017 x ∠=+ . What value of x makes the lines m and n parallel? ∠1 O m n ∠5
(a) 5 (b) 4 (c) 1 4 2 (d) 1 3 4
Assertion-Reason Based Questions
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): If angles ‘a’ and ‘b’ form a linear pair of angles and a = 40° then b = 150°.
Reason (R): Sum of linear pair of angles is always180°.
2. Assertion (A): A pair of angles measuring 120° and 60° are supplementary.
Reason (R): Two angles whose sum is 90° are called supplementary angles.
3. Assertion (A): In given figure, line l is parallel to line m m 82° 97° l n
Reason (R): If a transversal intersects two lines in such a way that a pair of consecutive interior angles are supplementary, then the two lines are parallel.
4. Assertion (A): In given figure, if ,110ABCDEAB∠= ‖ and 30 AEC ∠= , then 140 DCE ∠= .
Reason (R): If two parallel lines are intersected by a transversal, then each pair of alternate angles are equal.
Case Study Based Questions
1. Tejinder Singh recently bought an electric bicycle for his son. Filled with joy, he noticed that several parts of the bicycle resembled geometrical figures. While observing, he began imagining and calculating angles formed by the bicycle frame and its supporting rods.
Based on the bicycle frame structure and using the given diagram, answer the following:
(a) While examining the handle and supporting rods, Tejinder observed that the handlebar AB is parallel to the rod CD. If the angle between rods BC and CD (∠BCD) is 45°, what is the measure of ∠CBF?
(b) If the angle formed by rods AF and FC (∠AFC) is 75°, what is the measure of ∠CFB?
(c) Tejinder notices that the supporting rod EF forms an angle with rod FB. What is the value of ∠EFB?
(d) While analyzing the triangular part of the frame, Tejinder finds the measure of angle ∠AFE. What is its value?
Answers
Multiple Choice Questions
1. (c) 145
Produce PQ and SR to intersect AB and EF respectively at L and M .
Now, ABCD ‖ and transversalQR cuts them at R and S respectively.
2. In game period, the teacher of Indian Public School decided to play the puzzle game. For this game, firstly the teacher draw a geometrical figure on the ground, which is shown as below.
While drawing this figure, the teacher have no scale for measuring this length, but they know the side which is opposite to the smallest angle, is smaller and the side which is opposite to the largest angle, is larger.
In this game, the teacher invite the two students Vicky and Vishal and said them that Vicky stands on point A and Vishal stands on point B , respectively (assume that both have same space of walking). Then, answer the following questions, which are based on above data.
(a) Measure of ABD∠ is (i) 70 (ii) 80 (ii) 90 (iv) 100
(b) Measure of 1∠θ is (i) 70
(c) Measure of 2∠θ is (i) 60 (ii) 70 (ii) 80 (iv) 90
(d) Measure of 3∠θ is (i) 60 (ii) 70 (iii) 80 (iv) 90
125∴∠= (Alternate angles)
Again, ABCD ‖ and transversal QL cuts them at L and Q respectively.
260∴∠=
But, 23180∠+∠= (Linear pair)
3120∴∠=
But, 3 ARS ∠=∠ (Corresponding angles)
120 ARS ∴∠=
But, 1 QRSARS∠=∠+∠ 25120145 QRS ∴∠=+=
2. (c) 62° O S R 140° 1 2 x 158° T U p q
Draw a line parallel to p and q and passes through angle x.
180 PORPOU ∠+∠=° (Linear Pair)
18014040 POR ∠=°−°=°
140 POR ∠=∠=° (Alternate angle) (1)
180 QSTQSR ∠+∠=° (Linear Pair)
18015822 QST ∠=°−°=°
222 QST ∠=∠=° (Alternate angle) (2)
Adding (1) and (2) we get: 12402262 x ∠+∠==°+°=°
3. (b)
110°
ACBD ‖ and AB is a transversal and when two parallel lines a coy a transtersal then consecutive interior angels formed are supplementary ∴∠+∠= +∠= ∠=−=
180 130 180 10013050 CABDBA DBA DBA
DBG is a straight line 180 DBAABFFBG ∴∠+∠+∠= (Angles in a linear pair) 5060180 18011070 ABF ABF +∠+=
AEBF ‖ and ∠ABF and ∠EAB are consecutive interior angles
By angel sum property of a triangle; In 180 FPQFPQPQFQFP ∆⇒∠+∠+∠=
4035180 75180 18075105 x x x ++= += =−= 10. (b) 4 O m ∠
(Corresponding Angles)
Assertion Reason Based Questions
1. (d) A is false, but R is true
In the case of Assertion (A): a and ‘b’ are linear pairs. So, a + b = 180°
But, 40° + 150° = 190°. Therefore, Assertion is false.
In the case of Reason (R): The sum of a linear pair of angles is always 180°. Therefore, Reason is true.
2. (c) A is true, but R is false.
In the case of assertion (A), the Sum of 120° and 60° is 120° + 60° = 180°, which is supplementary. Therefore, Assertion is true.
In the case of reason (R): Two angles whose sum is 90° are called complementary angles. Therefore, the Reason is false.
3. (d) A is false, but R is true.
In the case of assertion (A): From figure, we find that 8297179180 +=≠
i.e. pair of consecutive interior angles are not supplementary. Therefore, line l is not parallel to line m. Therefore, Assertion is false.
In the case of reason (R): Reason is true as per the theorem.
4. (b) Both A and R are true, but R is not the correct explanation of A.
In the case of assertion (A): From E draw EF parallel to AB or CD . Now, EFCD ‖ and CE is the transversal intersecting CD and EF at C and E respectively.
180 CEFDCE∠∠ ∴+=
180 CEFDCE∠∠⇒=− (1)
Again AB || EF and transversal AE cuts them at A and E respectively.
180 BAEAEF∠∠ ∴+= () Co-interior angles are supplementary
180 BAEAECCEF∠∠∠ ⇒++=
11030180 40 CEFCEF∠∠ ⇒++=⇒= (2)
From (1) and (2), we obtain 40180 18040140. DCEDCE∠∠ =−⇒=−=
Therefore, Assertion is true.
In the case of reason (R): Reason is true as per the theorem.
Thus, both the A and R are true but R is not a correct explanation for A.
Case Study Based Questions
1. (a) We have, ABCD ‖
∴∠BCD =∠CBF (Alternate angles)
⇒ 45° =∠CBF
⇒∠CBF = 45°
Additional Practice Questions
1. If two lines intersect at a point, how many angles are formed?
(a) 1 (b) 2 (c) 3 (d) 4
2. If two adjacent angles are supplementary, then they form a:
(a) Linear pair
(b) Right angle pair
(c) Parallel lines
(d) Perpendicular lines
(b) We have, 180 AFCCFB∠∠+= (Linear Pair)
75180 CFB∠ ⇒+= 18075105 CFB∠ ⇒=−=
(c) We have, EFBAFC∠∠ = (Vertically opposite angles)
(d) (iii) Also, HDA is a straight line. Therefore, 32 3 30180 7030180 18010080 ∠θθ
3. From the given diagram below, which are the pairs of adjacent angles?
(a) ∠PQR and ∠PQS (b) ∠PQS and ∠SQR
(c) ∠PQR and ∠SQR (d) none of these
4. Prove that bisectors of a pair of vertically opposite angles are in the same straight line.
5. Find the measure of an angle whose supplement is equal to itself.
6. If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, then Find the greater of the two angles.
7. In the figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA respectively. Find ∠APB.
11. Prove that the bisectors of a pair of vertically opposite angles are in the same straight line.
12. AB, CD and EF are three concurrent lines passing through the point O such that OF bisect ∠BOD. If ∠BOF = 35°, find ∠BOC and ∠AOD
8. In the given figure, two straight lines PQ and RS intersect each other at O. If ∠POT = 75°, Prove that 153 abc++=°.
13. In the given figure, AB || CD and two parallel lines intersected by a transversal EF. Bisectors of interior angles, angle BMN and angle DNM on the same side of the transversal meet at P. Prove that ∠MPN = 90°.
14. In the given figure, || LMPN and the line l is a transversal of LM and PN . Find the value of a .
9. In the given below figure, show that XY is parallel to EF.
10. In the given figure below, if AB || CD, ∠BEG = 55° and ∠EFC = 80°, then find the value of x and y
55° 80° x y
15. In the figure above, , BCPQBP ‖ and CQ intersect at O . If 80 xy+= and 55 zy−= , then find , xy and z .
16. If two straight lines intersect each other, prove that the ray opposite to the bisector of one of the angles thus formed bisects the vertically opposite angle.
17. In given figure, BAED ‖ and BCEF ‖ . Show that ABCDEF∠=∠
(NCERTExemplar)
18. In given figure, AB and CD are straight lines and OP and OQ are respectively the bisectors of angles BOD∠ and AOC∠ . Show that the rays OP and OQ are in the same line.
19. In the figure above (not to scale), , DAEBC ‖ (270)BADx ∠=− , (20)ACBx ∠=− and (220)FAGx ∠=+ . Find EAC∠ F DE G A
20. In the figure above, ABCDEF ‖ and FG are the bisectors of BEG∠ and DGE∠ , respectively. 10 FEGFGE ∠=∠+ . Find FGE∠ . AB E G F CD
Challenge Yourself
1. AB is a straight line and O is a point lying on AB. A line OC is drawn from O such that 36 COA ∠= OD is a line within COA∠ such that 1 3 DOA ∠= COA∠ If OE is a line within the 1 , 4 BOCBOEBOC ∠∠=∠ , then DOE∠ must be
2. In the given figure, . Find . ABCDx C E 130° 6x 4x D B A
3. In the given figure, ,118 , EBCABC∠= ‖ ∠DAB 42, then is equal to: ADE=∠
D 42° 118° C E
Scan me
4. In the figure above, BA is parallel to DC , and PQ is a transversal of BA and DC. If 70 PMA ∠= and 230DNMx ∠=+ , then find the value of x . BM NC A P Q D
5. If, , ABEDAGCB ‖‖ and AFAB ⊥ . 38,45FAGCDE ∠=∠= . Find the value of x F G A 38° 45° B D E xC
Additional Practice Questions
1. (a) Linear pair
2. (d) 4
3. (b) ∠PQS and ∠SQR
5. 90°
6. 108°
7. 90°
10. 25, 55 xy=°=°
12. 110 BOCAOD ∠=∠=°
24
Yourself
SELF-ASSESSMENT
Time: 60 mins
Multiple Choice Questions
1. The sum of a pair complementary angles and supplementary angles are:
(a) Exactly 180° (b Exactly 270°
2. How many endpoints does a ray have?
(a) One (b) Two
Scan me for Solutions
Max. Marks: 40
(3×1=3Marks)
(c) More than 270° (d) Less than 270°
(c) Three (d) Four
3. In the given diagram, which type of pair do angles ∠POR and ∠ROQ form?
(a) Linear Pair
(c) Vertically Opposite Angles
Assertion-Reason Based Questions
(b) Adjacent Angles
(d) Supplementary Angles
(2×1=2Marks)
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
4. Assertion (A): If two lines are parallel and a transversal cuts them, the alternate exterior angles are equal.
Reason (R): Alternate exterior angles formed by transversal cutting parallel lines are congruent by the Alternate Exterior Angle Theorem.
5. Assertion (A): In given figure, if , BAEDBCEF ‖‖ and 180 ABC ∠= , then 100 DEF ∠=
Reason (R): If a transversal intersects two parallel lines, then each pair of corresponding angles are equal and each pair of consecutive interior angles are supplementary.
Very Short Answer Questions
6. If two angles of a triangle are 30º and 45º what is the measure of the third angle?
7. Two angles measure () 55 3a °+ and () 115–2a ° . If each is the supplement of the other, then calculate the value of a
Short Answer Questions
8 In the given figure below, AOB is a straight line, if ∠AOC = (3x + 10)° and ∠BOC = (4x 26)°, find ∠BOC
9. In the following figure ∠ABD is an exterior angle of ΔABC. Find ABC∠ .
10. In given figure, ABCF ‖ and BCED ‖ . Prove that ABCFDE∠=∠ .
11. Prove that two lines perpendicular to the same line are parallel to each other.
Long Answer Questions
12. In given figure, ABCD ‖ and CDEF ‖ . Also, EAAB ⊥ . If 55 BEF ∠= , find the values of , xy and z .
13. In the given figure, AB and CD are two mirrors placed parallel to each other. An incident ray PQ strikes the mirror AB at Q , the reflected ray moves along the path QR and strikes the mirror CD at R and again reflects back along RS .
(a) Prove PQRS
(b) If 30 a = , then find the value of y .
14. In the given figure, the side QR of ∠PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that 1 2 QTRPQR∠=∠ .
Case Study Based Questions
(1×4=4Marks)
15. Two friends, Rahul and Gaurav, are flying kites in an open ground. On a breezy afternoon, their kites reach the same height and their strings cross at point O. The diagram below shows the situation, with the angles their kite strings make with the ground. Rahul’s string makes an angle of 35° with the ground at point A, while Gaurav’s string makes an angle of 130° at point B. Answer the following questions based on this scenario:
(a) Rahul notices that his string makes an angle of 35° with the ground. Can you help him calculate the value of angle θ that his string makes on the other side of the ground?
(b) At point F, Rahul’s kite string makes an angle x with the horizontal kite line GF. Calculate the value of angle x.
(c) Gaurav wonders about the angle y formed at point D by his kite string. Help him calculate it.
(d) Rahul and Gaurav want to know if the angles θ and ∠OBC together form a straight angle. Check and explain your reasoning.
7 Triangles
Chapter at a Glance
Triangle
A closed figure formed by three intersecting lines is called a triangle.
A triangle has:
● three sides: AB, BC and CA
● three angles: , and ABCBCACAB ∠∠∠
● three vertices: A, B and C
Properties of Triangles
1. Angle opposite to equal sides of an isosceles triangle are equal.
2. The sides opposite to equal angles of a triangle are equal.
3. Each angle of an equilateral triangle is of 60°
4. In an isosceles triangle altitude from the vertex bisects the base. Conversely, if the altitude from one vertex of a triangle bisects the opposite side, then the triangle is isosceles.
5. A point equidistant from two given points lies on the perpendicular bisector of the line segment joining the two points.
6. A point equidistant from two intersecting lines lies on the bisectors of the angles formed by the two lines.
Congruent Triangles
Geometrical figures that have same shape and size are called congruent. In other words, two triangles, ∆ABC and ∆EFG, are congruent if and only if their corresponding sides and corresponding angles are equal.
If two triangles ∆ABC and ∆EFG are congruent under the correspondence A → E, B → F, and C → G, then it is expressed symbolically as: . ABCEFG∆≅∆
Criteria for Congruence of Triangles
1. Side–Angle–Side (SAS) Criterion: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are equal, then the two triangles are congruent.
2. Angle–Side–Angle (ASA) Criterion: If two angles and the included side of one triangle are respectively equal to two angles and the included side of another triangle, then the triangles are congruent.
3. Angle–Angle–Side (AAS) Criterion: If two angles and any one corresponding side of one triangle are respectively equal to two angles and the corresponding side of another triangle, then the triangles are congruent.
4. Side–Side–Side (SSS) Criterion: If all three sides of one triangle are respectively equal to the three sides of another triangle, then the triangles are congruent.
5. Right Angle–Hypotenuse–Side (RHS) Criterion (Applicable only for right-angled triangles): If the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and the corresponding side of another right-angled triangle, then the triangles are congruent.
Conditions of Congruence of Figures
1. Circle: Two circles of same radii are congruent.
2. Square: Two squares of same sides are congruent.
3. Equilateral Triangle: Two equilateral triangles of same sides are always similar.
NCERT Exercise 7.1
1. In quadrilateral , ABCDACAD = and AB bisects A∠ (see figure).
Show that . ABCABD∆≅∆ What can you say about BC and BD ?
Sol. In ABC∆ and ABD∆ , we have
ACAD = (Given)
CABDAB∠=∠ (AB bisects A∠ )
ABAB = (Common side)
ABCABD∴∆≅∆ (Using SAS congruence property)
.
Therefore, . BCBD = (Corresponding parts of congruent triangles)
Hence, proved.
2. ABCD is a quadrilateral in which ADBC = and DABCBA∠=∠ (see figure).
Prove that
(a) ABDBAC∆≅∆
(b) BDAC =
(c) ABDBAC∠=∠
Sol. In the figure, ABCD is a quadrilateral in which ADBC = and DABCBA∠=∠ . In ABD∆ and BAC∆ , we have
ADBC = (Given)
DABCBA∠=∠ (Given)
ABAB = (Common Side)
ABDBAC∴∆≅∆ (Using SAS congruence property)
BDAC∴= (Corresponding parts of congruent triangles) and ABDBAC∠=∠ (Corresponding parts of congruent triangles) Hence, proved.
3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB
Sol. In AOD∆ and BOC∆ , we have
AODBOC∠=∠ (Vertically opposite angles)
CBODAO∠=∠ (Each 90) °
and ADBC = (Given)
AODBOC∴∆≅∆ (By AAS congruence)
Also, AOBO = (Corresponding parts of congruent triangles)
Thus, CD bisects AB
Hence, proved.
4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that . ABCCDA∆≅∆
Sol. In the given figure, ABCD is a parallelogram in which AC is a diagonal, i.e., || ABDC and ||.BCAD
In ABC∆ and , CDA∆ we have,
BACDCA∠=∠ (Alternate angles)
BCADAC∠=∠ (Alternate angles)
AC = AC (Common)
∴ ABCCDA∆≅∆ (By ASA congruence) Hence, proved.
5. Line l is the bisector of an angle A and B is any point on l . BP and BQ are perpendicular from B to the arms of A∠ (see figure). Show that:
(a) APBAQB∆≅∆
(b) BPBQ = or B is equidistant from the arms of A∠
Sol. In APB∆ and , AQB∆ we have
PABQAB∠=∠ (l is the bisector of ) A∠
APBAQB∠=∠ (Each angle is of 90) °
ABAB = (Common) l l
APBAQB∴∆≅∆ (By AAS congruence)
Also, BPBQ = (Corresponding parts of congruent triangles)
i.e., B is equidistant from the arms of A∠ Hence, proved.
6. In the figure given below, ACAE = , ABAD = and . BADEAC∠=∠ Show that BCDE = .
Sol. BADEAC∠=∠ (Given)
⇒ BADDACEACDAC ∠+∠=∠+∠ (Adding DAC∠ to both the sides)
⇒ BACEAD∠=∠ (i)
Now, in ABC∆ and , ADE∆ we have ABAD = (Given)
ACAE = (Given)
BACDAE⇒∠=∠ (From equation (i))
ABCADE∴∆≅∆ (By SAS congruence)
⇒ . BCDE = (Corresponding parts of congruent triangles)
Hence, proved.
7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BADABE∠=∠ and . EPADPB∠=∠ Show that:
(a) DAPEBP∆≅∆
(b) ADBE = A ED
Sol. In DAP∆ and , we have EBP∆
APBP = (P is the midpoint of line segment AB)
BADABE∠=∠ (Given)
EPBDPB∠=∠
( ) EPADPBEPADPEDPBDPE ∠=∠⇒∠+∠=∠+∠
DPAEPB∴∆≅∆ (By ASA congruence)
⇒ ADBE = (Corresponding parts of congruent triangles)
Hence, proved.
8. In right triangle ABC , right angled at C, M is the mid-point of hypotenuse AB . C is joined to M and produced to a point D such that DMCM = . Point D is joined to point B (see figure). Show that:
(a) AMCBMD∆≅∆
(b) DBC∠ is a right angle.
(c) DBCACB∆≅∆
(d) 1 2 CMAB =
Sol. In AMC∆ and , BMD∆ we have
(a) DMCM = (Given)
BMAM = (M is the midpoint of AB )
DMBAMC∠=∠ (Vertically opposite angles)
AMCBMD∴∆≅∆ (By SAS )
ACBD∴= (Corresponding parts of congruent triangles)
ACMBDM∠∠ = (Corresponding parts of congruent triangles)
Hence, proved.
(b) ACMBDM∠∠ =
However, ACM∠ and BDM∠ are alternate interior angles.
Since alternate angles are equal, it can be said that DBAC ‖
DBCACB∠∠+= 180° (Co-interior angles)
DBC ∠+ 90° = 180°
DBC∠ ∴= 90° Hence, proved.
(c) In DBC∆ and , ACB∆ we have
DBAC = (Corresponding parts of congruent triangles)
BCBC = (Common)
DBCACB∠=∠ (Each angle is of 90) °
DBCACB∴∆≅∆ (By SAS ) Hence, proved.
(d) As, DBCACB∆≅∆
ABCD = (Corresponding parts of congruent triangles)
2. In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ABC∆ is an isosceles triangle in which ABAC = .
ABCD⇒=
11 22
Thus, 1 2
ABCM = ( 1 ) 2 CMCD = Hence, proved.
NCERT Exercise 7.2
1. In an isosceles triangle , ABCABAC = , the bisectors of B∠ and C∠ intersect each other at O . Join A to O
Show that:
(a) OBOC =
(b) AO bisects A∠
Sol. (a) ABACABCACB =⇒∠=∠ (Angles opposite to equal sides are equal)
11 22
ABCACB∠=∠
⇒ CBOBCO∠=∠ (OB and OC are bisectors of B∠ and C∠ )
⇒ OBOC = (Sides opposite to equal angles are equal)
Hence, proved.
(b) 11 22
ABCACB∠=∠
ABOACO∠=∠ ( OB∴ and OC are bisectors of B∠ and C∠ )
In ABO∆ and , ACO∆ we have ABAC = . (Given)
OBOC = (Proved above)
AOAO = (Common)
ABOACO∴∆≅∆ (SSS congruence)
⇒ BAOCAO∆=∆ (Corresponding parts of congruent triangles)
⇒ AO bisects A∠ Hence, proved.
B A DC
Sol. In the figure, ABD∆ and ACD∆ , we have
ADBADC∠∠ = (Each angle is of 90) °
BDCD = (AD bisects BC )
ADAD = (Common)
ABDACD∴∆≅∠ (SAS)
∴ ABAC = (Corresponding parts of congruent triangles)
Hence, ABC∆ is an isosceles triangle.
Hence, proved.
3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.
Sol. In AEB∆ and , AFC∆ = AEBAFC∠∠ (Each 90°)
AA∠=∠ (Common angle)
ABAC = (Given)
∴ AEBAFC∆≅∆ (By AAS congruence rule)
∴ BECF = (Corresponding parts of congruent triangles)
Hence, proved.
4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that:
(a) ABEACF∆≅∆
(b) ABAC = , i.e., ABC is an isosceles triangle. A FE BC
Sol. (a) In ABE∆ and , ACF we have
BECF = (Given)
BAECAF∠=∠ (Common)
BEACFA∠=∠ (Each 90 =° )
So, ABEACF∆≅∆ (By AAS congruence)
Hence, proved.
(b) Also, ABAC = (Corresponding parts of congruent triangles)
Hence, ABC is an isosceles triangle. Hence, proved.
5. ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that . ABDACD∠=∠
Sol. In isosceles , ABC∆ we have ABAC =
ABCACB∠=∠ (i)
(Angles opposite to equal sides are equal)
Now, in isosceles DCB,∆ we have BDCD =
DBCDCB∠=∠ (ii) (Angles opposite to equal sides are equal)
Adding (i) and (ii), we have
ABCDBCACBDCB ∠+∠=∠+∠
ABDACD⇒∠=∠
Hence, proved.
6. ∆ABC is an isosceles triangle in which . ABAC = Side BA is produced to D such that ADAB = (see figure). Show that BCD∠ is a right angle.
Sol. ABAC = (Given)
ACBABC∠=∠ (i)
(Angles opposite to equal sides are equal)
ABAD = (Given)
ADAC = (ABAC = )
ACDADC∴∠=∠ (ii) (Angles opposite to equal sides are equal)
Adding (i) and (ii), we have
ACBACDABCADC ∠+∠=∠+∠
BCDABCADC ∠∠⇒∠=+ (iii)
Now, in , BCD∆ we have
⇒
180 BCDDBCBDC ∠+∠+∠=° (Angle sum property of a triangle)
180. BCDBCD ∴∠+∠=°
⇒ 2180 BCD ∠=°
⇒ 90 BCD ∠=°
Thus, 90 BCD ∠=°or a right angle.
Hence, proved.
7. ABC is a right-angled triangle in which 90 A ∠=° and ABAC = . Find B∠ and . C∠
Sol. In , ABC∆ we have
Given: 90 A ∠=° and ABAC =
We know that angles opposite to equal side of an isosceles triangle are equal.
So, BC∠=∠
In ∆ABC,
180 ABC ° ∠+∠+∠= (Angle sum property of a triangle)
90 180 BC °° +∠+∠=
90 180 BB °° +∠+∠= ( BC∠=∠ )
2 90 B ° ∠=
45 B ° ∠=
Thus, 45. BC ∠=∠=°
8. Show that the angles of an equilateral triangle are 60° each.
Sol. In ABC∆ is an equilateral.
So, ABBCAC ==
Now, ABAC = ACBABC⇒∠=∠ (i) (Angles opposite to equal sides of a triangle are equal.)
Again, BCAC = BACABC⇒∠=∠ (ii)
Now in , ABC∆
180
ABCACBBAC ∠+∠+∠=° (Angle sum property of a triangle)
180 ABCABCABC ⇒∠+∠+∠=° (From equation (i) and (ii))
3180 ABC ⇒∠=°
180 60 3 ABC ° ⇒∠==°
Also, From (i) and (ii)
60 ACB ∠=° and 60 BAC ∠=°
Hence, proved.
NCERT Exercise 7.3
1. ABC∆ and DBC∆ are two triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P , show that:
(a) ABDACD∆≅∆
(b) ABPACP∆≅∆
(c) AP bisects A∠ as well as D∠
(d) AP is the perpendicular bisector of BC .
Sol. (a) In ABD∆ and ACD∆ , we have
ABAC = (Given)
BDCD = (Given)
ADAD = (Common)
. ABDACD∴∆≅∆ (Using SSS congruence property)
Hence, proved.
, BADCAD∠=∠ (Corresponding parts of congruent triangles)
BAPCAP⇒∠=∠ (i) (Base angles of an isosceles triangle)
(b) In ABP∆ and ACP∆ , we have
ABAC = (Given)
BAPCAP∠=∠ (From equation (i))
APAP = (Common)
ABPACP∴∆≅∆ (Using SAS congruence property)
∴ BPCP = (ii) (Corresponding parts of congruent triangles) Hence, proved.
(c) From Equation (1),
BAPCAP∠=∠
Hence, AP bisects ∠A
In ∆BDP and ∆CDP,
BD = CD (Given)
DP = DP (Common)
BP = CP (From equation (ii))
∴ BDPCDP ∆≅∆() By SSS Congruence rule
∴ BDPCDP∠=∠ (iii) (Corresponding parts of congruent triangles) Hence, AP bisects ∠D. Hence, proved.
(d) Now, BPCP = (From equation (iii))
And BPACPA∠=∠ (Corresponding parts of congruent triangles)
But 180 BPACPA ∠+∠=° (Linear pair)
So, 2180 BPA ∠=°
90 BPA ⇒∠=°
Since BPCP = , therefore AP is perpendicular bisector of BC Hence, proved.
2. AD is an altitude of an isosceles triangle ABC in which ABAC = . Show that:
(a) AD bisects BC
(b) AD bisects A∠ B A D C
Sol. (a) In ABD∆ and , ACD∆ we have
ADBADC∠=∠ (Each = 90°, as AD is an altitude)
ABAC = (Given)
ADAD = (Common)
ABDACD∴∆≅∆ (RHS Congruence)
BDCD∴= (Corresponding parts of congruent triangles) Thus, AD bisects . BC Hence, proved.
(b) BADCAD∠=∠ (Corresponding parts of congruent triangles)
Thus, AD bisects A∠ Hence, proved.
3. Two sides AB and BC and Median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of PQR∆ (see figure). Show that:
(a) ABMPQN∆≅∆
(b) ABCPQR∆≅∆
Sol. (a) In ABM∆ and PQN,∆ we have
BMQN = 11 22 BCQRBCQR =⇒=
ABPQ = (Given)
AMPN = (Given)
PQN ABM ∴∆≅∆ (By SSS congruence) Hence, proved.
ABMPQN⇒∠=∠ (Corresponding parts of congruent triangles)
ABCPQR⇒∠=∠ (1)
(b) Now, in ABC∆ and ∆PQR, we have
ABPQ = (Given)
ABCPQR∠=∠ (From 1)
Multiple Choice Questions
1. Which of the following is not a criterion for congruence of triangles?
(a) SAS (b) ASA (c) SSA (d) SSS (NCERTExemplar)
BCQR = (Given)
PQR ABC ∴∆≅∆ (By SAS congruence) Hence, proved.
4. BE and CF are two equal altitudes of a triangle ABC . Using RHS congruence rule, prove that the triangle ABC is isosceles.
Sol. BE and CF are altitudes of a . ABC∆ 90 BECCFB ∴∠=∠=°
Now, in right triangles BEC and CFB , we have Hypotenuse BC = Hypotenuse BC (Common)
Side BE = Side CF (Given)
CFB BEC ∴∆≅∆ (By RHS congruence)
CBF BCE ∴∠=∠ (Corresponding parts of congruent triangles)
Now, in , ABCBC ∆∠=∠
ABAC∴= (Sides opposite to equal angles are equal) Hence, ABC∆ is an isosceles triangle. Hence, proved.
5. ABC is an isosceles triangle with ABAC = . Draw APBC ⊥ to show that BC∠=∠
Sol. Draw , APBC ⊥
In ABP∆ and , ACP∆ we have ABAC = (Given)
APBAPC∠=∠ (Each angle is of 90) °
APAP = (Common)
ABPACP∴∆≅∆ (BY RHS congruence)
Thus, BC∠=∠ (Corresponding parts of congruent triangles)
Hence, proved.
2. If AB = QR, BC = RP, and CA = PQ, then (a) ∆ABC ≅∆PQR (b) ∆CBA ≅∆PRQ (c) ∆BAC ≅∆RPQ (d) ∆PQR ≅∆BCA (NCERTExemplar)
3. In ∆ABC, AB = AC and ∠B = 50° Then ∠C is equal to (a) 40° (b) 50° (c) 80° (d) 130°
(NCERTExemplar)
4. In ∆PQR, ∠R =∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is
(a) 4 cm (b) 5 cm (c) 2 cm (d) 2.5 cm
(NCERTExemplar)
5. In triangles ABC and PQR, AB = PQ and ∠B =∠Q. The two triangles will be congruent by SAS axiom if:
(a) BC = QR (b) AC = PR
(c) AB = QR (d) none of these
6. Which of the following is congruent to the given figure?
8. In triangles ABC and PQR, AB = AC, ∠C =∠P, and ∠B =∠Q. The two triangles are:
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but not isosceles
(d) neither congruent nor isosceles
9. In a figure, if AC is the bisector of ∠BAD such that 3 AB = cm and 5 AC = cm, then CD = ____
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm
10. In a figure, ABC is an equilateral triangle and BDC is an isosceles right triangle, right-angled at D. ∠ABD equals?
(a) 45°
(b) 60°
(c) 105°
(d) 120°
11. If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:
(a) an isosceles triangle.
(b) an obtuse triangle.
(c) an equilateral triangle.
(d) a right triangle.
12. In a figure, write the correspondence if the triangles are congruent.
7. Consider the following two statements about the rectangles a student draws.
(i) Rectangles have the same perimeter.
(ii) Rectangles have the same area.
Which of the following statements is sufficient to conclude that the rectangles are congruent?
(a) (i) alone is sufficient, but (ii) alone is not sufficient.
(b) (ii) alone is sufficient, but (i) alone is not sufficient.
(c) Both (i) and (ii) together are sufficient, but neither statement alone is sufficient.
(d) Both (i) and (ii) together are not sufficient.
(a) ∆OAD ≅∆OBC (b) ∆OAD ≅∆OCB
(c) ∆AOD ≅∆OBC (d) None of these
13. In the given figure, ∠ABC is
14. Which of the following statements is correct?
(a) A triangle cannot have an obtuse angle and a right angle.
(b) A triangle cannot have two obtuse angles.
(c) A triangle can have all acute angles.
(d) All of these.
15. If the corresponding angles of two triangles are equal, then they are always congruent is (a) true
(b) cannot be determined
(c) false
(d) None of these
16. In figure, ABCDEF ∠∠∠∠∠∠+++++=
(a) 180 °
(b) 360 °
(c) 540 ° (d) 90 °
17. In the given figure, PM is the bisector of ∠P and PM ⊥ QR. Then ∆PQM and ∆PRM are congruent by which criterion?
(a) ASA
(b) SSS
(c) SAS
(d) None of these
18. Consider the parallelogram as shown.
Based on the given information, can it be concluded that DACBCA∆≅∆ ? Why or why not?
(a) Yes, because AB = CD, AD = BC, ∠DAC =∠ACB, ∠DCA =∠CAB, ∠ADC =∠ABC and AC = AC.
(b) Yes, because AB = CD, AD = BC, ∠DAC =∠CAB, ∠DCA =∠ACB, ∠ADC =∠ABC and AC = AC.
(c) No, because angle measures are not known.
(d) No, because side lengths are not known.
19. Which of these pairs of triangles are not congruent? (a) (c) (b) (d)
20. In given figure, xy+=
° y ° x ° (a) 270° (b) 230° (c) 210° (d) 190°
° 40°
AD
21. In the given figure, the measure of ∠EDF is
(a) 40° (b) 50° (c) 60° (d) 70°
22. In the given figure, if and AEDCABAC = , the value of ∠ABD is
(a) 100° (b) 110° (c) 120° (d) 70°
23. In figure, AM and DN are the medians. Then value of DF is
MN CF 4.5 cm 6 cm 4 cm
(a) 4 cm (b) 7 cm
(c) 9 cm (d) None of these
24. In the given figure, if PM = QM, ∠SMP =∠RMQ. S PMQ R
Therefore, ∆PMR ≅∆QMS by which congruence rule?
(a) SSS (b) AAS (c) ASA (d) SAS
Answers
1. (c) SSA
2. (b) ∆CBA ≅∆PRQ
3. (b) 50°
4. (a) 4 cm
5. (a) BC = QR
6. (b) 6 cm 10 cm 11 cm 12 cm 110° 70°
7. (c) Both (i) and (ii) together are sufficient.
8. (a) Isosceles but not congruent
9. (b) 3 cm
10. (c) 105°
11. (d) Right triangle
12. (a) ∆OAD ≅∆OBC
Constructed Response Questions
Very Short Answer Questions
1. In given figure, PQPR = and QR∠=∠ . Prove that ∆PQS ≅∆PRT. (NCERTExemplar)
25. In the figure, ABC is an isosceles triangle in which AB = AC and PQ is parallel to BC. If ∠A = 40°, find ∠PQB
13. (d) 20°
14. (d) All of these.
15. (c) False
16. (b) 360°
17. (a) ASA
18. (a) Yes, because AB = CD, AD = BC, ∠DAC =∠ACB, ∠DCA =∠CAB, ∠ADC =∠ABC, and AC = AC
19. (b)
20. (b) 230°
21. (a) 40°
22. (b) 110°
23. (d) None of these
24. (c) ASA
25. (c) 110°
Sol. In ∆PQS and ∆PRT, PQ = PR (Given) ∠Q =∠R (Given) and ∠QPS =∠RPT (Same angle) Therefore, ∆PQS ≅∆PRT (ASA) Hence, proved.
2. In given figure, two lines AB and CD intersect each other at the point O such that BC || DA and BC = DA. Show that O is the midpoint of both the linesegments AB and CD.
(NCERTExemplar)
Sol. BC || AD (Given)
Therefore, ∠CBO =∠DAO (Alternate interior angles)
∠BCO =∠ADO (Alternate interior angles)
Also, BC = DA (Given)
So, ∆BOC ≅∆AOD (ASA)
Therefore, OB = OA and OC = OD, i.e., O is the midpoint of both AB and CD. Hence, proved.
3. In given figure, PQPR > , QS and RS are the bisectors of ∠Q and ∠R, respectively. Show that SQSR > (NCERTExemplar)
Sol. PQPR > (Given)
Therefore, RQ∠>∠ (Angles opposite the longer side is greater)
So, SRQSQR∠>∠ (Half of each angle)
Therefore, SQSR > (Side opposite the greater angle)
Therefore, SQSR > (Side opposite the greater angle will be longer) Hence, Proved.
4. In triangles ABC and PQR , AQ∠=∠ and BR∠=∠ . Which side of ∆PQR should be equal to side AB of ∆ABC so that the two triangles are congruent? Give reason for your answer. A BC P QR
Sol. In triangles ABC and QRP
AQ∠=∠ (Given)
BR∠=∠ (Given)
If ABQR = , then ABCQRP∆≅∆ (By ASA).
5. In given figure ABCD is a square and P is the midpoint of ADBP and CP are joined. Prove that ∠PCB =∠PBC.
Sol. In triangles PAB and PDC,
PA = PD (Given)
AB = CD (Sides of square)
∠PAB =∠PDC = 90°
So, ∆PAB ≅∆PDC (By SAS congruency)
∴ PB = PC (Corresponding parts of congruent triangles)
PCBPBC⇒∠=∠ (Angles opposite to equal sides) Hence, proved.
6. In figure, BM and DN are both perpendiculars to the segments AC an BMDN = . Prove that AC bisects BD
Sol. In BMR∆ and DNR , we have and BMACDNAC ⊥⊥ (Given)
Thus, BMRDNR∠=∠ (Each equal to 90° )
BRMDRN∠=∠ (Vertically opposite angles)
and, BMDN = (Given)
So, by AAS criterion of congruence, we obtain
BMRDNR∆≅∆
BRDR⇒= (Corresponding parts of congruent triangles)
is the mid-point of . RBD ⇒
Thus, AC bisects BD Hence, proved.
Short Answer Questions
1. In the given figure, if ABAC = and BDDC = , then find ∠ADB
Sol. In ∆ADB and ∆ADC
ABAC = (Given)
BDDC = (Given)
ADAD = (Common)
So, ADBADC∆≅∆ (By SSS criterion)
ADBADC⇒∠=∠
But 180 ADBADC ∠+∠=°
180 ADBADB ⇒∠+∠=° 2180 ADB ⇒∠=°
90 ADB ⇒∠=°
D A BC
2. Bisectors of the angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O
Show that external angle adjacent to ∠ABC is equal to ∠BOC.
(NCERTExemplar)
3. Prove that angles opposite to equal sides of a triangle are equal.
Sol. Given: ∆ABC in which ABAC = .
To prove: BC∠=∠
Construction: Draw AD, the bisector of ∠A, to meet BC at D
Now, In ∆ABD and ∆ACD, we have
ABAC = (Given)
BADCAD∠=∠ (By construction)
ADAD = (Common)
ABDACD∴∆≅∆ (By SAS congruence criterion)
Hence, BC∠=∠ (By Congruent part of congruent traingle)
Hence, proved.
4. In the given figure, ∠ACB is the right angle ACCD = and CDEF is a parallelogram. If 10 FEC ∠=° then find the value of ∠CDA. B
Sol. In triangle ABC, since AB = AC,
Let BCx∠∠== (1)
OBCOBAOCBOCA ∠∠∠∠====
Let angle bisectors of ∠B and ∠C meet at O. Then, ,. 22 xx
Now, in quadrilateral BOAC, angle ∠BOC is the angle between two bisectors:
So, 180 OBCOCBBOC ∠∠∠ ° ++= (The sum of angles is 180 ° )
22 xxBOC ∠ ++= 180°
⇒ x +∠BOC = 180°.
180 BOCx∠ ° =− (2)
Now consider the external angle adjacent to ∠B, i.e., ∠ABE. From exterior angle property, ABECA ∠∠∠ =+
6. In the given figure, PRQR = , PRAQRB∠=∠ and BPRAQR∠=∠ . Prove that BPQA = .
Sol. We have PRAQRB∠=∠ (1) (Given)
Adding ∠ARB to both sides of (1), we get PRAARBQRBARB ∠+∠=∠+∠
PRBQRA⇒∠=∠ (2)
In ∆BPR and ∆AQR, we have PRBQRA∠=∠ (From 2)
PRQR = (Given)
And BPRAQR∠=∠ (Given)
So, by ASA rule of congruence, we have BPRAQR∆≅∆
⇒ BPAQ = (Corresponding parts of congruent triangles)
() BPQAAQQA⇒=∴= Hence, proved.
7. In the given figure D and E are points on side BC of a ∆ABC such that and BDCEADAE == . Show that ABDACE∆≡∆ . A
Sol. Given, ADAE = ADEAED⇒∠=∠ (1) (Angles opposite to equal sides in a triangle are equal)
We have, 180 ADBADE ∠+∠=° (Linear pair)
180 ADBADE⇒∠=°−∠
180 ADBAED⇒∠=°−∠ (Using 1) (2)
180 AEDAEC ⇒∠+∠=°
180 AECAED⇒∠=°−∠ (3)
From (2) and (3), we have ADBAEC∠=∠ (4)
In ∆ABD and ∆ACE, we have
ADAE = (Given)
ADBAEC∠=∠ (using (4))
BDCE = (Given)
So, by SAS criterion of congruence, we have ABDACE∆≅∆
Hence, proved.
8. BE and CF are two equal altitudes of a triangle ABC Using RHS congruence rule, prove that the triangle ABC is isosceles.
Sol. BE and CF are altitudes of a ∆ABC. 90 BECCFB ∠=∠=°
Now, in the right triangles BEC and CFB, we have
BECCFB∠=∠ (Each angle is equal to 90°)
BCBC = (Common side)
BECF = (Given)
BECCFB∴∆≅∆ (By RHS congruence criterion)
⇒ BCECBF∠=∠ (Corresponding parts of congruent triangles)
BCACBA⇒∠=∠
ABAC∴= (Sides opposite to equal angles are equal)
Thus, ABC is an isosceles triangle. Hence, proved.
9. In ∆ABC, if D is a point on AC such that ADCDBD == , then prove that ABC∆ is a rightangled triangle.
Sol D is a point on AC such that ADCDBD ==
In ∆ABD, we have
ADBD = (Given)
⇒ 31∠=∠ (1)
(Angles opposite to equal sides are equal)
Again, in ∆DCB, we have CDBD =
⇒ 42∠=∠ (2) (Angles opposite to equal sides are equal)
Adding (1) and (2) we get 3412 ∠+∠=∠+∠
12 ABC ⇒∠=∠+∠ (3)
In ∆ABC, we have 1 2180 ABC ∠+∠+∠=°
()12180 ABC ⇒∠+∠+∠=°
180 ABCABC ⇒∠+∠=° 2180 ABC ⇒∠=°
180 2 ABC ⇒∠=°÷
90 ABC ⇒∠=°
Thus, ∆ABC is a right-angled triangle. Hence, proved.
10. If diagonal AC of a quadrilateral ABCD bisects ∠A and ∠C, then prove that and ABADCDCB == .
Sol. Since the diagonal AC of a quadrilateral ABCD bisects ∠A and ∠C
BACDAC∴∠=∠ and BCADCA∠=∠
In ∆ABC and ∆ADC, we have BACDAC∠=∠ (Given)
BCADCA∠=∠ (Given) and, ACAC = (Common)
So, by the ASA rule of congruence, we have ABCADC∆≅∆
⇒ ABAD = (Corresponding parts of congruent triangles) and CBCD = (Corresponding parts of congruent triangles)
Hence, proved.
11. P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meet BC at Q, prove that BPQ is an isosceles triangle.
Sol. Since P is a point on the bisector of ∠ABC
ABPQBP∴∠=∠ (1)
As QP is parallel to BA and BP is a transversal.
QPBABP∴∠=∠ (2) (Alternate interior angles)
From (1) and (2), we have QPBQBP =∠
BQQP⇒= (Sides opposite to equal angles in a triangle are equal)
Thus, ∆BPQ is an isosceles triangle.
Hence, proved.
12. CDE is an equilateral triangle formed on a side CD of a square ABCD (see figure). Show that ADEBCE∆≅∆
Given: Since ABCD is a square, all sides are equal (ABBCCDDA === ) and all angles are 90°.
∆CDE is equilateral, so CEDECD == and all its angles are 60°.
Calculate ∠ADE and ∠BCE: 9060150 ADEADCCDE ∠=∠+∠=°+°=° 9060150 BCEBCDDCE ∠=∠+∠=°+°=°
Therefore, ADEBCE∠=∠ .
In ∆ADE and ∆BCE:
ADBC = (sides of the square)
ADEBCE∠=∠ (Proved above)
DECE = (sides of the equilateral triangle)
By the SAS (Side-Angle-Side) congruence rule, ADEBCE∆≅∆ Hence, proved.
Long Answer Questions
1. In right triangle ABC, right-angled at C, M is the midpoint of hypotenuse ABC is joined to M and produced to a point D such that DMCM = . Point D is joined to point B. Show that:
(a) AMCBMD∆≅∆
(b) ∠DBC is a right angle.
(c) DBCACB∆≅∆
(d) 1 2
CMAB = .
Sol. (a) In ∆AMC and ∆BMD
AMBM = (Given)
CMDM = (Given)
AMCDMB∠=∠ (Vertically opposite angles)
So, AMCBMD∆≅∆ (By SAS congruence rule)
⇒ 12∠=∠ and ACBD = (Corresponding parts of congruent triangles)
Hence, proved.
(b) Since 12∠=∠ and they form a pair of alternate interior angles.
BDAC ∴
180 DBCACB ⇒∠+∠=° (Co interior angles)
1809090 DBC ⇒∠=°−°=°
⇒∠DBC is a right angle.
Hence, proved.
(c) In ∆DBC and ∆ACB
90 DBCACB ∠=∠=°
DBAC = (Proved above)
BCBC = (Common)
DBCACB∴∆≅∆ (By SAS congruence rule)
⇒ DCAB = (Corresponding parts of congruent triangles)
Hence, proved.
(d) Since DCAB = 11 22 DCAB⇒= 1 2
CMAB⇒= ( 1 2 CMMDDC == )
Hence, proved.
2. In the figure, two sides AB and BC and median AM of one triangle ABC are equal to sides PQ and QR and median PN of ∆PQR. Show that ABCPQR∆≅∆ .
Sol. In ∆ABC and ∆PQR,
BCQR = (Given)
11
BCQR⇒=
22
⇒ BMQN = (M and N are midpoint of BC and QR respectively)
In triangles ABM and PQN, we have
ABPQ = (Given)
BMQN = (Proved above)
AMPN = (Given)
ABMPQN∴∆≅∆ (By SSS congruence criterion)
⇒ BQ∠=∠ (Corresponding parts of congruent triangles)
Now, in triangles ABC and PQR, we have
ABPQ = (Given)
BQ∠=∠ (Proved above)
BCQR = (Given)
ABCPQR∴∆≅∆ (By SAS congruence criterion)
Hence, proved.
3. ∆ABC is an isosceles triangle in which ABAC = . Side BA is produced to D such that ADAB = . Show that ∠BCD is a right angle.
Sol. In ∆ABC, we have ABAC =
⇒ ACBABC∠=∠ (1) (Angles opposite to equal sides are equal)
Now, ABAD = (Given)
ADAC∴= (Since ABAC = )
Thus, in ∆ADC, we have ADAC =
⇒ ACDADC∠=∠ (2) (Angles opposite to equal sides are equal)
Adding (1) and (2), we get
ACBACDABCADC ∠+∠=∠+∠
BCDABCADC⇒∠=∠+∠ (Since ADCBDC∠=∠ )
⇒ BCDBCDABCBDCBCD ∠+∠=∠+∠+∠ (Adding ∠BCD on both sides)
⇒ 2
180 BCDDBCBDCBCD ∠=∠+∠+∠=°
(Sum of angles of a ∆ is 180°)
90 BCD ⇒∠=°
Thus, ∠BCD is a right angle. Hence, proved.
4. ABC is a right triangle with ABAC = . Bisector of ∠A meets BC at D. Prove that 2 BCAD = .
Sol. In right ∆ABC, we have ABAC =
⇒ CB∠=∠ (1)
(Angles opposite to equal sides are equal)
90 A ∴∠=° (2)
In a triangle, the sum of all three angles is 180°.
180 ABC ∴∠+∠+∠=°
90 180 BB ⇒°+∠+∠=° (using (1) and (2))
21809090 B ⇒∠=°−°=°
90 45 2 B ° ⇒∠==°
45 BC ⇒∠=∠=° (3)
Since AD is the bisector of 90 A ∠=° 11 9045 22 BADCADA ∴∠=∠=×∠=×°=° (4)
From (3) and (4), we have BADBABD∠=∠=∠ and CADCACD∠=∠=∠
BDAD⇒= and CDAD = Now, BCBDDCADAD =+=+
2 BCAD⇒= Hence, proved.
5. BC is a triangle with 2 BC∠=∠ . D is a point on BC such that AD bisects ∠BAC and ADCD = . Prove that 72 BAC ∠=°
Sol. Since ADCD = CDAC⇒∠=∠ . But 2 BC∠=∠ 2 BDAC⇒∠=∠
BAx⇒∠=∠= (say) ( AD is bisector of ∠BAC ).
Now, 180 ABC ∠+∠+∠=° (Angle Sum Property)
180 2 B xx ∠ ++=°
2180 2 x x ⇒+=°
4 180 2 xx + ⇒=°
5 180 2 x ⇒=°
5
180 2 x x ⇒=°×
72 A ⇒∠=°
72 BAC ⇒∠=° Hence, proved.
6. O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that OCD∆ is an isosceles triangle.
Sol. Since ∆AOB is an equilateral triangle, 60 OABOBA ∴∠=∠=° (1)
Also, 90 DABCBA ∠=∠=° (2)
( ABCD is a square)
Subtracting (1) from (2), we get
90 60 DABOABCBAOBA ∠−∠=∠−∠=°−°
i.e., 30 DAOCBO ∠=∠=°
Now, in ∆AOD and ∆BOC, AOBO = (Given)
DAOCBO∠=∠ (Proved above)
ADBC = (ABCD is a square)
AODBOC∴∆≅∆ (By SAS congruence)
⇒DOOC = (Corresponding parts of congruent triangles)
Since, in ∆COD, COOD = COD∴∆ is an isosceles triangle.
Hence, proved.
7. A point O is taken inside an equilateral four sided figure ABCD such that its distances from the angular points D and B are equal. Show that AO and OC are in one and the same straight line.
Sol. In AOD∆ and AOB , we have
ADAB = (Given)
AOAO = (Common side)
ODOB = (Given)
So, by using SSS criterion of congruence, we obtain AODAOB∆≅∆
⇒ 12∠=∠ (Corresponding parts of congruent triangles)
Similarly, it can be proved that DOCBOC∆≅∆
Competency Based Questions
Multiple Choice Questions
1. In a ∆ABC, it is given that ∠A : ∠B : ∠C = 3 : 2 : 1 and ∠ACD = 90°. If BC is produced to E, then ∠ECD =
(a)60° (b) 30° (c) 50° (d) 40°
2. In figure, a + b =
(a) 117° (b) 130° (c) 127° (d) 158°
3. In figure, X is a point in the interior of square ABCD. AXYZ is also a square. If DY = 3 cm and AZ = 2 cm, then BY =
⇒ 34∠=∠ (Corresponding parts of congruent triangles)
But, 12344 ∠+∠+∠+∠= right angles (Sum of the angles at a point is 4 right angles)
22234⇒∠+∠= right angles (Using (i) and (ii))
232⇒∠+∠= right angles 180 ° =
2⇒∠ and 3∠ form a linear pair.
AO ⇒ and OC are in the same straight line
AC ⇒ is a straight line. Hence, proved.
4. In figure, if l₁ || l₂, the value of x is
(a) 22.5 (b) 30
(c) 45 (d) 60
5. In given figure, what is z in terms of x and y?
z°
(a) x + y + 180° (b) x + y 180°
(c) 180° (x + y) (d) x + y + 360°
6. In a ∆ABC, if 45, BC ∠=∠=° which is the longest side?
(a) BC (b) AC
(c) CA (d) None of these
7. In given figure, ABCD is a quadrilateral in which BN and DM are perpendiculars drawn to AC such that BNDM = . If 4 OB = cm, then BD =
(a) 5 cm (b) 6 cm
(c) 7 cm (d) 8 cm
(a) 6 cm (b) 8 cm
(c) 10 cm (d) 12 cm
8. In ∆ABC, if 45, 65 BC ∠=°∠=° , and the bisector of ∠BAC meets BC at P, then the ascending order of sides is:
9. How many equilateral triangles each of 1 cm² are required to fill the given hexagonal rangoli?
(a) 200 (b) 300 (c) 150 (d)250
10. In given figure, if PT is the bisector of QPR∠ in , 50, 30 PQRPQRPRQ ° ∆∠=°∠= and PSQR ⊥ , then x =
P
(a) 40° (b) 20° (c) 30° (d) 10°
Assertion-Reason Based Questions
Directions: Each question consists of two statements: an Assertion (A) and Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): If ∆≅∆ABCPQR then . BCPQ = Reason (R): Corresponding parts of congruent triangles.
2. Assertion (A): In figure, side BC of ABC∆ is produced to D . If 110 ACD∠ ° = , then 40 x ° = .
° x + 10° 2x − 20
Reason (A): If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.
Case Study Based Questions
1. An aluminum ladder manufacturing company manufactures foldable step ladder shown in figure The length XY and XZ are each equal to 110 cm and the vertical angle is 36°.
(a) What is the ratio of ∠YXZ =∠XZY?
(b) What will be the length of the side YZ , if YXZ∠ is 60°?
(c) Which type of triangle is ∆XYZ?
2. Read the following and answer the questions given below:
Locations of houses, 1234 , , , , HHHH and 5 H of 5 friends around their school are shown in the map below: School S is exactly in the middle of houses 1 H and 2 H . Houses 345 , , HHH , and the school S are on the same line l as shown in the map.
Also, line segments joining the houses 14 , HH and 25 , HH are perpendicular to the line l.
Using the above information answer the following questions:
(a) Show that line 14HH is parallel to 25HH
(b) Show that 4132 HHSHHS∆=∆
(c) Is the school exactly halfway between houses 4 H and 5 H ? Justify your answer.
Answers
Multiple Choice Questions
1. (a) 60°
Let the angles be:∠∠∠=== 3,2, AxBxCx
Sum of angles in ∆ABC:
32180618030 xxxxx °°° ++=⇒=⇒=
Thus, 90,60,30ABC ∠∠∠ °°° ===
If 90 ACD∠ ° = , also if CD is perpendicular to AC:
180903060 ECD∠ °°°° =−−=
2. (c) 127°
Since ARB is a straight line, 5190180 22 xx x °°° +−++=
44180417644 xxx °°°° ⇒+=⇒=⇒=
Using exterior angle property in ∆PQR, we obtain:
QRCab∠=+ 51 35 22 xxababx°° ⇒+−=+⇒+=−
34451325127 ab °°°°° ⇒+=×−=−= a + b = 127°
3. (a) 7 cm
In the following figure we are given,
DY = 3 cm
AZ = 3 cm
Where ABCD is a square and AXYZ is also a square.
From the above figure we have XY = YZ = AZ = AX.
Now in the given figure.
DZ = DY + YZ
= 3 + 2
= 5 cm
So, =+22ADDZAZ
25429cm=+=
Now inn triangle ∆AXB 22BXABAX =− 2945cm=−=
So, BY = XY + BX = 2 + 5 =7cm
4. (c) 45
The transversal creates angles x and another angle (e.g., 135°) with l1 and l2.
If x and 135° are consecutive interior angles, they must sum to 180°:
13518045xx °°° +=⇒=
If x is a corresponding or alternate angle to another given angle (e.g., 45°), then: 45 x ° =
5. (c) () 180 – xy°+
x, y, and z are angles in a triangle or formed by intersecting lines.
If x and y are two angles of a triangle, then the third angle z would be: () 180 zxy ° =−+
6. (a) BC
In ABC∆ , 45 BC∠∠ ° ==
Since the sum of angles in a triangle is 180°: 180 180454590 ABC ∠∠∠ °°°°° =−−=−−=
In any triangle, the side opposite the largest angle is the longest.
Here, ∠A = 90° is the largest angle, and its opposite side is BC.
7. (b) 8 cm
In triangles ONB and OMD , we have ONBOMD∠∠ = (Each equal to 90 ° ) BONDOM∠∠ = (Vertically opposite angles)
BNDM =
So, by using AAS congruence criterion, we obtain 4 cm ONBOMDOBODODOB ∆≅∆⇒=⇒== 4 cm4 cm8 cm
BDOBOD=+=+=
8. (c) CP, BP, AP
Given: 45,65BC∠∠ °° ==
The sum of angles in a triangle is 180°: 180 180456570 ABC ∠∠∠ °°°°° =−−=−−=
The angle bisector of ∠BAC divides it into two equal angles:
70 35 2 BAPCAP∠∠ ° ° ===
Using the triangle angle sum property in ∆APC
180 180356580 APCCAPC ∠∠∠ °°°°° =−−=−−=
Similarly, in ∆APB:
∠APB = 180° ∠BAP ∠B = 180° 35° 45° = 100°
In any triangle, the side opposite the larger angle is longer.
In APB∆ , 100 APB∠ ° = (largest angle)
⇒ AB is the longest side.
45 B∠ ° =⇒ AP is opposite ∠B.
35 BAP∠ ° =⇒ BP is opposite ∠BAP.
Order of sides: BPAPAB <<
In APC∆ : 80 APC∠ ° = (largest angle)
⇒ AC is the longest side.
65 C∠ ° =⇒ AP is opposite ∠C
35 CAP∠ ° =⇒ CP is opposite ∠CAP.
Order of sides: CPAPAC <<
From the above analysis:
BPAP < and CPAP < Hence, the correct ascending order is CP < BP < AP
9. (c) 150
The area A of a regular hexagon with side length s is given by: 2 33 2 As =
Substituting s = 5 cm: 333375322 525 cm 222 A =×=×=
The area a of an equilateral triangle with side length 1 cm is: 3322 1 cm 44 a =×=
Number of triangles A a = 753 7534 2 752150 2 33 4 ==×=×=
10. (d) 10°
Using angle sum property in PQR∆ , we obtain 180 PQRPRQRPQ∠∠∠ ° ++= 5030 180 RPQ∠ °°° ⇒++= 100 RPQ∠ ° ⇒=
1 50 2 QPTRPQ∠∠ ° ∴==
() is bisector of PTQPR ∠
Using exterior angle property in PQS∆ , we obtain PSTPQSQPS∠∠∠ =+ 9050 40 QPSQPS∠∠ °°° ⇒=+⇒= 504010 xQPTQPS ∠∠ °°° ∴=−=−=
Assertion-Reason Based Answers
1. (d) A is false, but R is true.
By the definition of congruent triangles, all corresponding sides and angles are equal.
If ∆ABC ≅∆PQR, then:
AB = PQ; BC = QR; CA = RP
So BC = QR, not PQ —so Assertion is actually false if we consider strict ordering.
However, if the correspondence is:
A ↔ P; B ↔ Q; C ↔ R
Then BC ↔ QR, not PQ
So BC = PQ is false, unless BC is stated to correspond to PQ, which is not evident here.
Thus, A is false unless matching order is clearly defined.
Reason (R): Corresponding parts of congruent triangles.
True. This is a well-known property of congruent triangles, referred to as Corresponding Parts of Congruent Triangles.
2. (a) Both A and R are true, and R is the correct explanation of A.
R is the Exterior angle theorem. So, it is true.
Using R, we obtain 22010110312040 xxxx °°°°° −++=⇒=⇒=
So, A is also true. Also, R is a correct explanation for A.
Case Study Based Questions
1. (a) We have, XY = XZ = 110 cm
⇒∠XZY =∠XYZ (1) (Equal sides have equal opposite angles)
Now, in ∆XYZ, ∠YXZXYZXZY+∠+∠= 180°
⇒ 36° + XZYXZY ∠+∠= 180° (from (1))
⇒ 2∠XZY = 180° 36°
144
72 2 XZY ° ⇒∠==°
⇒∠YXZ : ∠XZY = 36° : 72° = 1 : 2
(b) In ∆XYZ, If ∠YXZ = 60° and XY = XZ = 110 cm, then ∠XYZ =∠XZY ( Equal sides have equal opposite angles)
∴∆XYZ is an equilateral triangle.
Thus, side YZ = 110 cm.
(c) Since two sides of triangle are equal and all angles are acute. It is an isosceles acute triangle.
2. (a) 14 HHl ⊥ (Given)
25 HHl ⊥ (Given)
⇒ 1425 HHHH (Lines perpendicular to the same line are parallel to each other)
Additional Practice Questions
1. (a) In an isosceles right-angled triangle ABC, if ∠A = 90°, find the values of ∠B and ∠C.
(b) What is the measure of the vertical angle of an isosceles triangle, each of whose base angles measures 65°?
2. In the figure, l is the bisector of ∠A and O is any point on lOB and OC are perpendicular from O to arms of ∠A. Show that OBOC = .
3. In given figure, it is given that , ABCFEFBD == and AFECBD∠=∠ . Prove that AFE ∆≅∆ CBD.
(b) In 41 sHHS ∆ ′ and 32 HHS , we have 14 HHS ∠= 25HH∠ S = 90° (Given)
4152 HSHHSH∠=∠ (1)
( Vertically opposite angle)
4152 HHSHHS⇒∠=∠ (2)
(Angle sum property of a triangle and if two angles of one triangle are equal to two angles in another triangle then the third pair of angles must also be equal.)
12 HSHS = (3) (Given)
Using (1), (2), and (3) and applying ASA congruence criterion,
4152 HHSHHS∆≅∆
(b) 4132 HHSHHS∆≅∆ (From (ii))
45 HSHS⇒= (Corresponding parts of congruent triangles) S is exactly halfway between the houses 4 H and 5 H
5. In the figure, ABCD is a quadrilateral in which ABBC = and ADDC = . Find the measure of ∠BCD.
4. Prove that in an isosceles triangle, the altitude from the vertex bisects the base.
6. In the figure, , BAACDEDF ⊥⊥ such that BADE = and BFEC = . Show that ABCDEF∆≅∆ .
7. In the given figure, , 48, 18 PQQRQPRSRP=∠=°∠=°, then find the ∠QRS and ∠PQR.
8. In the figure, ABC is a right-angled triangle at B. BD is drawn perpendicular to AC. Show that 12∠=∠ .
11. In the given figure, AB = BC and ∠ABO =∠CBO, then prove that ∠DAB =∠ECB.
B
9. In the figure, if ABDACE∠=∠ , then prove that ABAC =
10. ABCD is a quadrilateral in which AD = BC and ∠DAB =∠CBA. Prove that ABDBAC∆≅∆ .
12. In the given figure, equilateral ∆ABD and ∆ACE are drawn on the sides of a ∆ABC. Prove that CD = BE.
13. In ∆ABC, AB = AC and AD is bisector of ∠BAC. Alia proved that ∆ABD ≅∆ACD as follows: AB = AC (Given)
∠BAD =∠CAD (Given)
Also, ∠B =∠C ( AB = AC )
So, ∆ABD ≅∆ACD (by ASA congruence rule)
His classmate Manisha pointed out to her that this proof is not correct and gave the correct proof. Alia accepted her mistake and thanked Manisha for the same. Write the correct proof. What is the mistake in Alia’s proof?
Challenge Yourself
1. Rajeev and Amit sit together at B. Their work places are at A and C respectively such that A, B and C form a triangle. BD is the bisector of ∠ABC, meets AC at right angle. Various distances are expressed in terms of x and y as shown in the figure. (3y+1)km
They used bicycles instead of bikes to go to their work places.
Find the distance between A and C.
2. The picture below shows a staircase outside a house. Each step of the staircase is congruent and there are 25 steps in the staircase from the floor to the platform and 25 steps from the platform to the roof. The horizontal length from the start to the platform is 8 m, and each step has a rise of 20 cm. What is the length of the staircase railing? Railing
3. In given figure, RS is diameter and PQ chord of a circle with centre O. Prove that
(a) ∠RPO =∠OQR
(b) ∠POQ = 2∠PRO R
4. ABC is a triangle in which ∠B = 2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°.
5. In the given figure, AB = BC and AC = CD. Prove that : 3:1 BADADB ∠∠=
Answers
Additional Practice Questions
1. (a) ∠B =∠C = 45° (b) 50°
5. 105°
7. ∠QRS =∠1 = 30° and ∠PQR = 84°
13. The correct proof uses the AAS congruence rule. The mistake in Alia’s proof was incorrectly using the ASA rule instead of AAS.
Challenging Yourself
1. x = 3, y = 4
2. 18.87 m
Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 60 mins
Multiple Choice Questions
1. In the given figure, AM and DN are the medians. Then value of DF is
Scan me for Solutions
Max. Marks: 40
[3×1=3Marks]
(a) 4 cm (b) 4.5
(c) 6
(d) 9
2. In , ABCABAC∆= and 50 B ° ∠= . Then C∠ is equal to (a) 40° (b) 50° (c) 80° (d) 130° (NCERTExemplar)
3. In triangle ABC, AB = AC and BD bisects ∠ABC. If the measure of ∠ABC is 50°, which angle measures 25°?
(a) ∠ABD (b) ∠ACB (c) ∠BDC (d) ∠BDA
Assertion-Reason
Based Questions
[2×1=2Marks]
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false
(d) A is false, but R is true.
4. Assertion (A): , 30, 80 ABCPQRAB ∆≅∆∠=°∠=° then 70 R ∠=°.
Reason (R): The corresponding angles of congruent triangles are equal.
5. Assertion (A): In , 125 PQRQPR ∆∠=° and PQPR = then 30 PQR ∠=°.
Reason (R): In a triangle, angles opposite to equal sides are equal.
Very Short Answer Questions
[2×2=4Marks]
6. “If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Why?
7. Angles of a triangle are in ratio 1: 2: 3, find the smallest angle of the triangle.
Short Answer Questions
8. P is a point on the bisector of ∆ABC. If the line through P, parallel to BA meet BC at Q, prove that BPQ is an isosceles triangle.
9. Two lines l and m intersect at the point O and P is a point on a line n passing through the point O such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m.
10. In the given figure, triangles PQC and PRC are such that QC = PR and PQ = CR. Prove that ∠PCQ =∠CPR.
QCR
11. CDE is an equilateral triangle formed on a side CD of a square ABCD (See the given figure). Show that ADEBCE∆≅∆
Long Answer Questions
12. ABC is a right triangle with ABAC = . Bisector of CA meets BC at D. Prove that 2 BCAD = .
[3×5=15Marks]
13. In a right triangle, prove that the line segment joining the mid-point of the hypotenuse to the opposite vertex is half of the hypotenuse.
(NCERTExemplar)
14. In the given figure, ABC is an isosceles triangle in which ABAC = . Side BA is produced to D such that ADAB = . Show that BCD is a right angle.
Case Study Based Questions
[1×4=4Marks]
15. The Egyptian pyramids are ancient structures located in Egypt. The pyramid of Khufu is the largest Egyptian pyramid. It is one of the seven wonders of the ancient world still in existence.
A pyramid is a structure whose outer surfaces are triangular and converge to a single step at the top. The base of the pyramid can be a triangle, quadrilateral, or any polygon. Geeta, a mathematics student, visited Egypt and observed the pyramid (shown in the figure).
Based on the above information, answer the following questions:
(a) Name the triangle which is congruent to ∆ABC.
(b) By which property are the triangles congruent?
(c) Which side will BC be equal to?
(d) What is the classification of ∆ABC based on the lengths of its sides?
8 Quadrilaterals
Chapter at a Glance
Quadrilateral: Any closed figure with 4 sides.
● Sum of interior angles is equal to 360°.
Parallelogram: A quadrilateral is a parallelogram if:
● Opposite sides are parallel and equal in length.
● Opposite angles are equal.
● Diagonals bisect each other.
● Each diagonal divides it into two congruent triangles.
● One pair of opposite sides is both equal and parallel. (This condition is sufficient)
Rectangle: A parallelogram is a rectangle if:
● Each interior angle is 90°.
● Diagonals are equal.
● Diagonal bisect each other.
Square: A parallelogram is a square, if
● All sides equal.
● Each interior angle is 90°.
● Diagonals are equal.
● Diagonals bisect opposite angles.
● Diagonals bisect each other at right angles.
Rhombus: A parallelogram is a rhombus, if
● All sides equal.
● Diagonals bisect each other at right angles.
● Diagonals bisect opposite angles.
Kite: A quadrilateral is a kite, if
● Two pairs of adjacent sides are equal.
● Two pairs of opposite sides are unequal.
● One pair of opposite angles are equal.
● Diagonals bisect each other at right angles.
● One of the diagonals bisects the angles through which it passes.
Trapezium: A quadrilateral is a trapezium, if
● One pair of opposite sides is parallel.
● Non-parallel sides are called legs
The trapezium which has legs of equal length. Here, ADBC = The trapezium which neither has equal angles nor has equal sides. The trapezium which has a pair of right angles.
Key Properties and Theorems
● Diagonals of a parallelogram are equal ⇔ It is a rectangle.
● Diagonal bisecting one angle of a parallelogram also bisects the opposite angle ⇔ It is a rhombus.
● Bisectors of consecutive angles of a parallelogram intersect at 90°. In figure, 90 AOB ∠=°, AO and BO are angle bisector of A∠ and B∠ respectively.
● Angle bisectors of a parallelogram form a rectangle. In figure, EFGH is a rectangle.
● The line joining mid-points of two sides of a triangle is parallel to the third side and is half its length (Midpoint Theorem). In figure, 1 2 DEBC = and || DEBC .
● A line through the mid-point of a side parallel to another side bisects the third side. In figure, D is the mid-point of AB and ||.DEBC Thus, E is the mid-point of AC.
NCERT Exercise 8.1
1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Sol. In ABC∆ and BAD∆ ,
BCAD = ( Opposite sides of a parallelogram are equal)
ACBD = ( Given)
ABAB = ( Common)
Thus, ABCBAD∆≅∆ ( SSS Congruency rule)
⇒ ABCBAD∠=∠ (1) ( Congruent parts of Congruent triangles)
But, 180 ABCBAD ∠+∠=° ( Co-interior angles)
Using (1) we get
⇒ 2180 BAD ∠=°⇒ 180 90 2 BAD ° ∠==°
A parallelogram with one of its angles 90° is a rectangle.
Thus, ABCD is a rectangle. Hence, proved.
2. Show that the diagonals of a square are equal and bisect each other at right angles.
Sol. In BAD∆ and ABC∆ ,
ADBC = ( Opposite sides of a square)
BADABC∠=∠ ( Each 90°)
ABAB = ( Common)
Hence, BADABC∆≅∆ ( SAS Congruency rule)
⇒ BDAC = ( Congruent parts of Congruent triangles)
In AOB∆ and COD∆ ,
OABOCD∠=∠ ( Alternate angles)
ABCD = ( Opposite sides of a square)
OBAODC∠=∠ ( Alternate angles)
Hence, AOBCOD∆≅∆ ( ASA Congruency rule)
⇒ , AOOCBOOD == ( Congruent parts of Congruent triangles)
In AOB∆ and AOD∆ ,
OBOD = ( Proved above)
ABAD = ( Sides of a square)
OAOA = ( Common)
Hence, AOBCOD∆≅∆ ( SSS Congruency rule)
⇒ AOBAOD∠=∠ ( Congruent parts of Congruent triangles)
But, 180 AOBAOD ∠+∠=° ( Linear Pair)
⇒ 2180 AOB ∠=° ( AODAOB∠=∠ )
⇒ 180 90 2 AOB ° ∠==°
Thus, the diagonals of a square are equal and bisect each other at right angles. Hence, proved.
3. Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that
AB DC
(a) it bisects ∠C also, (b) ABCD is a rhombus.
Sol. (a) ∠DAC =∠BAC ( Given) (1)
∠DAC =∠BCA ( Alternate angles) (2)
∠BAC =∠DCA ( Alternate angles) (3)
From the equations (1), (2) and (3), we have
∠ACD =∠BCA (4)
Thus, diagonal AC bisects angle C also. Hence, proved.
(b) From the equation (2) and (4), we have
∠ACD =∠DAC
In ΔADC, ∠ACD =∠DAC ( Proved above)
AD = DC ( In a triangle, the sides opposite to equal angle are equal)
However, DA = BC and AB = CD (Opposite sides of a parallelogram)
A parallelogram whose adjacent sides are equal, is a rhombus.
Thus, ABCD is a rhombus.
Hence, proved.
4. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(a) ABCD is a square.
(b) Diagonal BD bisects ∠B as well as ∠D
DC
AB
Sol. (a) It is given that ABCD is a rectangle.
11
(d) AQ = CP
(e) APCQ is a parallelogram
Sol. (a) In ΔAPD and ΔCQB,
DP = BQ ( Given)
∠ADP =∠CBQ ( Alternate angles)
AD = BC ( Opposite sides of a parallelogram)
Thus, ΔAPD ≅ΔCQB ( SAS Congruency rule) Hence, proved.
(b) ΔAPD ≅ΔCQB
⇒ AP = CQ (1)
⇒∠=∠
AC AC ∠=∠
22
⇒∠DAC =∠DCA (Given: AC bisects ∠A and ∠C)
In ACD∆ ,
CDDA = (Sides opposite to equal angles are also equal)
However, DABC = and ABCD = (Opposite sides of a rectangle are equal)
ABBCCDDA ===
ABCD is a rectangle and all the sides are equal.
Thus, ABCD is a square.
(b) Let us join BD
In BCD∆ ,
BCCD = (Sides of a square are equal to each other)
CDBCBD∠=∠ (Angles opposite to equal sides are equal)
However, CDBABD∠=∠ (Alternate interior angles for ABCD ‖ )
CBDABD∠=∠
BD bisects B∠ .
Also, CBDADB∠=∠ (Alternate interior angles for BCAD ‖ )
CDBABD∠=∠ Therefore, BD bisects both D∠ and B∠
5. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that:
(a) ΔAPD ≅ΔBQC
(b) AP = CQ
(c) ΔAQB ≅ΔCPD A P D BC Q
( Congruent parts of Congruent triangles) Hence, proved.
(c) In ΔAQB and ΔCPD,
QB = DP ( Given)
∠ABQ =∠CDP ( Alternate angles)
AB = CD ( Opposite sides of a parallelogram)
Hence, ΔAQB ≅ΔCPD ( SAS Congruency rule) Hence, proved.
(d) ΔAQB ≅ΔCPD
⇒ AQ = CP (2)
( Congruent parts of Congruent triangles) Hence, proved.
(e) In APCQ,
AP = CQ ( From (1))
AQ = CP ( From (2))
The opposite sides of quadrilateral APCQ are equal.
Thus, APCQ is a parallelogram.
Hence, proved.
6. ABCD is a parallelogram, and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Figure). Show that:
(a) APBCQD∆≅∆
(b) APCQ =
Sol. (a) In APB∆ and CQD∆ :
90 APBCQD ∠=∠=° (Both are perpendiculars)
ABPCDQ
∠=∠ (Alternate angles since || ABCD and BD is the transversal)
ABCD = (Opposite sides of a parallelogram are equal)
Hence, by the AAS (Angle-Angle-Side) congruence rule, APBCQD∆≅∆ Hence, proved.
(b) Since APBCQD∆≅∆ , the corresponding sides are equal.
Therefore, APCQ = Hence, proved.
7. ABCD is a trapezium in which || ABCD and ADBC = (see Figure). Show that:
(a) AB∠=∠
(b) CD∠=∠
(c) ABCBAD∆≅∆
(d) Diagonal AC = Diagonal BD
(Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E .)
Sol. Let us extend AB. Then, draw a line through C, which is parallel to AD intersecting AB at point E. It is clear that AECD is a parallelogram.
(a) ADCE = (Opposite sides of parallelogram AECD)
However, ADBC = (Given)
Therefore, BCCE = CEBCBE∠=∠
(Angle opposite to equal sides are also equal)
Consider parallel lines AD and CEAE is the transversal line for them.
180 ACEB ∠+∠= (Angles on the same side of transversal)
From Equations (1) and (2), we obtain AB∠=∠ Hence, proved.
(b) Since || ABCD , 180 AD ∠+∠=° and 180. BC ∠+∠=°
But AB∠=∠ (Proved in part (i)), so DC∠=∠ Hence, proved.
(c) In ABC∆ and BAD∆ : ABBA = (Common side)
ADBC = (Given)
AB∠=∠ (Proved in part (i))
ABCBAD∆≅∆ (By SAS congruence rule) Hence, proved.
(d) Since ABCBAD∆≅∆ , the corresponding sides are equal.
Therefore, AC = BD . (Congruent part of congruent triangle) Hence, proved.
NCERT Exercise 8.2
1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides , , ABBCCD and DA respectively. (see figure). AC is a diagonal. Show that:
(a) || SRAC and 1 2 SRAC = (b) PQSR = (c) PQRS is a parallelogram.
Sol. (a) In , ADCS∆ and R are the mid-points of sides AD and CD respectively. In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallelto the third side and is half of it.
SRAC ∴ ‖ and 1 2 SRAC = (1) Hence, proved.
(b) In , ABCP∆ and Q are mid-points of sides AB and BC respectively. Therefore, by using mid-point theorem,
Therefore, OMQN is a parallelogram.
∴∠=∠
∴∠=∠
MQNNOM PQRNOM
1 and 2
PQACPQAC = ‖ (2)
Using Equations (1) and (2), we obtain PQSR ‖ and PQSR = (3) PQSR∴= Hence, proved.
(c) From Equation (3), we obtained PQSR ‖ and PQSR =
Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal. Thus, PQRS is a parallelogram. Hence, proved.
2. ABCD is a rhombus and , , PQR and S are the mid-points of the sides AB , , BCCD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
SOQ
APB
Sol. Construction: Join , ACPR and SQ .
In , ABCP∆ and Q are the mid-points of sides AB and BC respectively.
However, 90 NOM ∠= (Diagonals of a rhombus are perpendicular to each other)
90 PQR ∴∠=
Clearly, PQRS is a parallelogram having one of its interior angles as 90. Hence, PQRS is a rectangle. Hence, proved.
3. ABCD is a rectangle and , , PQR and S are midpoints of the sides AB , BC , CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
DR O C Q APB S
Sol. Construction: Join AC
Proof: In ABC∆ , P and Q are the mid-points of the sides AB and BC .
∴ || PQAC and 1 2 PQAC = (Mid point theorem) (i)
Similarly, in ΔADC, || SRAC and 1 2 SRAC = (ii)
PQACPQAC∴= ‖ (1)
1 and 2
(Using mid-point theorem) In ADC∆ , R and S are the mid-points of CD and AD respectively.
From (i) and (ii), we get || PQSR and PQSR = (iii)
Now in quadrilateral PQRS , its one pair of opposite sides PQ and SR is parallel and equal (From (iii))
∴PQRS is a parallelogram.
1 and 2
RSACRSAC∴= ‖ (2)
(Using mid-point theorem)
From Equations (1) and (2), we obtain and PQRSPQRS = ‖
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.
Let the diagonals of rhombus ABCD intersect each other at point O
From (iii) and (v), we have PQRS is a rhombus Hence, Proved.
4. ABCD is a trapezium in which ||, ABDCBD is a diagonal and E is the mid-point of AD . A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC .
AB
Sol. By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side.
In ABD∆ , EFAB ‖ and E is the mid-point of AD. Therefore, G will be the mid-point of DB
As EFAB ‖ and ABCD ‖ , EFCD ∴ ‖
( Two lines parallel to the same line are parallel to each other)
In , BCDGFCD∆ ‖ and G is the mid-point of line BD. Therefore, by using converse of mid-point theorem, F is the mid-point of BC Hence, proved.
5. In a parallelogram ABCD , E and F are the midpoints of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD .
Sol. Since E and F are mid-points of AB and DC respectively.
⇒ 1 2 AEAB = and 1 2 CFDC = (i)
But, ABDC = and || ABDC (ii) (Opposite sides of a parallelogram)
∴ AECF = and || AECF
⇒ AECF is a parallelogram.
(One pair of opposite sides is parallel and equal)
In BAP∆ , E is the mid-point of AB || EQAP
⇒ Q is mid-point of PB (Converse of mid-point theorem)
⇒ PQQB = (iii)
Similarly, in DQC∆ , P is the mid-point of DQ
DPPQ = (iv)
From (iii) and (iv), we have DPPQQB ==
or line segments AF and EC trisect the diagonal BD Hence, Proved.
6. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D . Show that
(a) D is the mid-point of AC
(b) MDAC ⊥
(c) 1 2 CMMAAB ==
Sol. Construction: Join CM .
(a) In ABC∆ , M is the mid-point of AB . (Given) || BCDM (Given)
D is the mid-point of AC (Converse of mid-point theorem) Hence, proved.
(b) ADMACB∠=∠ ( Corresponding angles)
But 90 ACB ∠=° (Given)
∴ 90 ADM ∠=°
But 180 ADMCDM ∠+∠=° (Linear pair)
∴ 90 CDM ∠=°
Thus, MDAC ⊥ Hence, proved.
(c) ADDC = ( D is the mid-point of AC ) (1)
Now, in ADM∆ and CMD∆ , we have
ADMCDM∠=∠ (Each = 90°)
ADDC = (From (1))
DMDM = (Common)
CMMA = (2)
(Congruent parts of Congruent triangles)
Since M is mid-point of AB,
Multiple Choice Questions
1. Diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50°, then ∠OAB is (a) 90º
(b) 50º (c) 40º (d) 10º
2. A diagonal of a rectangle is inclined to one side of the rectangle at 25º. The acute angle between the diagonals is
(a) 55º (b) 50º (c) 40º (d) 25º
3. ABCD is a rhombus such that ∠ACB = 40º. Then ∠ADB is
(a) 40º (b) 45º (c) 50º (d) 60º
4. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if (a) PQRS is a rectangle (b) PQRS is a parallelogram (c) diagonals of PQRS are perpendicular (d) diagonals of PQRS are equal.
5. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if (a) PQRS is a rhombus
(b) PQRS is a parallelogram
(c) diagonals of PQRS are perpendicular (d) diagonals of PQRS are equal.
6. If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 1 : 1 : 1 : 1, then ABCD is a (a) rhombus (b) parallelogram
(c) rectangle (d) kite
Hence, 1 2 CMMAAB == (From (2) and (3))
Hence, proved.
7. If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S, then PQRS is a
(a) rectangle
(b) rhombus
(c) parallelogram
(d) quadrilateral whose opposite angles are supplementary
8. If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form
(a) a square
(b) a rhombus
(c) a rectangle
(d) any other parallelogram
9. The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is
(a) a rhombus
(b) a rectangle
(c) a square
(d) any parallelogram
10. D and E are the mid-points of the sides AB and AC of ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is
(a) a square (b) a rectangle
(c) a rhombus (d) a parallelogram
11. The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if,
(a) ABCD is a rhombus
(b) diagonals of ABCD are equal
(c) diagonals of ABCD are equal and perpendicular
(d) diagonals of ABCD are perpendicular
12. D and E are the mid-points of the sides AB and AC respectively of ABC. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is
(a) ∠DAE =∠EFC (b) AE = EF
(c) DE = EF (d) ∠ADE =∠ECF
13. Which of the following is not true for a parallelogram?
(a) opposite sides are equal (b) opposite angles are equal
(c) opposite angles are bisected by the diagonals
(d) diagonals bisect each other.
14. The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32º and ∠AOB = 70º, then ∠DBC is equal to
(a) 24º (b) 86º
(c) 38º (d) 32º
15. Three angles of a quadrilateral are 80°, 95° and 112°. Its fourth angle is
(a) 78° (b) 73°
(c) 85° (d) 100°
16. ABCD is a rhombus such that ∠ACB = 50°. Then ∠ADB =?
(a) 40° (b) 25°
(c) 65° (d) 130°
17. Three angles of a parallelogram are 85,95 and 85 The fourth angle is (a) 90° (b) 95° (c) 105° (d) 120°
18. A diagonal of a parallelogram divides into two triangles. The triangles formed are (a) congruent (b) similar (c) does not have equal area
(d) None of these
19. If the perimeter of a parallelogram ABCD is 32 cm and one of its side 7 cm AD = , then what is the length of AB ?
(a) 7 cm (b) 8 cm (c) 9 cm (d) 10 cm
20. Which of the following relation is correct, if S is the mid-point of PR such that TSQR ‖ ? P TS
21. Angles ,,, ABCD of a quadrilateral ABCD are in the ratio 3:4:4:7 . If the bisectors of angles A and B intersect at O , then AOB ∠= (a) 70 (b) 80 (c) 110 (d) 100
22. The bisectors of any two adjacent angles of parallelogram intersect at (a) 30 (b) 45 (c) 60 (d) 90
23. In given figure, ABCD is a parallelogram such that diagonal BD bisects B∠ as well as D∠ and O is the mid-point of BD . Then, AOB ∠=
(a) 30 (b) 45 (c) 60 (d) 90
24. If angles ,, ABC and D of the quadrilateral ABCD , taken in order, are in the ratio 3:7:6:4, then ABCD is a
25. In given figure, D and E are the mid-points of the sides AB and AC respectively of ABC∆ DE is produced to F . To prove that CF is equal and parallel to DA , we need an additional information which is
(a) DAEEFC∠=∠
(b) AEEF =
(c) DEEF =
(d) ADEECF∠=∠
(NCERTExemplar)
Answers
1. (c) 40º
2. (b) 50º
3. (c) 50º
4. (c) diagonals of PQRS are perpendicular
5. (d) diagonals of PQRS are equal.
6. (c) rectangle
7. (d) quadrilateral whose opposite angles are supplementary
8. (c) a rectangle
9. (b) a rectangle
10. (d) a parallelogram
11. (c) diagonals of ABCD are equal and perpendicular
12. (c) DE = EF
13. (c) opposite angles are bisected by the diagonals
Constructed Response Questions
Very Short Answer Questions
1. ABCD is a trapezium in which AB||DC and ∠A =
∠B = 45°. Find angles C and D of the trapezium.
Sol. Quadrilateral ABCD is a trapezium.
∠A and ∠D and ∠B and ∠C are the angles present on the same side of the trapezium.
Angles on the same side of the trapezium are supplementary.
∠A +∠D = 180°.
∴ 45° +∠D = 180° (∠A = 45°)
∴∠D = 180° 45° = 135°
Similarly,
∴∠C = 135°
Therefore, ∠A =∠B = 45° and ∠D and ∠C = 135°
2. The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60º. Find the angles of the parallelogram.
Sol. Given: DP and DQ are perpendicular to side AB and BC respectively.
120 ADCABC ∴∠=∠=° (Opposite angles of parallelogram are equal)
Now, the Adjacent angles of the parallelogram are supplementary.
180 DABABC ∴∠+∠=°
120 180 DAB ∴°+∠=° 18012060 DAB ∴∠=°−°=° and 60 DABDCB ∴∠∠=° (Opposite angles of a parallelogram are equal)
3. D,E and F are the mid-points of the sides BC,CA and AB, respectively of an equilateral triangle ABC. Show that DEF is also an equilateral triangle.
Sol. Given: that ΔABC is an equilateral triangle.
D is the midpoint of BC, E is the midpoint of AC and F is the midpoint of AB. According to midpoint theorem, 1 2 DEAB = Also, 1 2 DFAC = Similarly, 1 2 FEBC = But, AB = BC = AC
∴DF = DE = FE
Thus, ΔDFE is an equilateral triangle. Hence, proved.
4. E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram.
Sol. Given: Quadrilateral ABCD is a parallelogram. Diagonals of a parallelogram bisect each other.
∴AO = OC and DO = OB
According to the diagram, AEOFC
AE + OE = CF + OF Also, AE = CF (given)
∴ EO = OF And OD = OB
Therefore, in quadrilateral BFDE: Both pairs of opposite sides are equal (EO = OF and DO = OB).
Thus, BEDF is a parallelogram. Hence, proved.
5. Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 35°, determine ∠B. D AB C O
Sol. In quadrilateral ABCD the diagonals bisect each other. Thus, the given quadrilateral ABCD is a parallelogram.
∠A = 35°
In a parallelogram the adjacent angles are supplementary.
180 AB ∴∠+∠=°
35180 B ∴°+∠=°
145 B ∠=°
6. In the given figure ABCD and AEFG are two parallelograms. If ∠C = 55°, determine ∠F DC G AEB F
Sol. In parallelogram ABCD, ∠C = 55° Opposite angles of a parallelogram are equal.
∴∠A =∠C = 55°
∠A = 55°
Now, In parallelogram AEFG,
∠A =∠F
55 F ∴∠=°
7. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Sol. Let ABCD is a parallelogram such that AC = BD A DC B
In ΔABC and ΔDCB, AC = DB (Given)
AB = DC (Opposite sides of a parallelogram)
BC = CB (Common)
∴ΔABC ≅ΔDCB (By SSS congruency)
⇒∠ABC =∠DCB (1) (Congruent parts of Congruent triangles)
Now, AB||DC and BC is a transversal.
( ABCD is a parallelogram)
∴∠ABC +∠DCB = 180° (Co-interior angles) (2)
From (1) and (2), we have
∠ABC =∠DCB = 90°
i.e., ABCD is a parallelogram having an angle equal to 90°.
∴ ABCD is a rectangle. Hence, proved.
8. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Sol. Let ABCD be a quadrilateral such that the diagonals AC and BD bisect each other at right angles at O.
∴ In ΔAOB and ΔAOD, we have AO = AO (Common)
OB = OD (O is the mid-point of BD)
∠AOB =∠AOD (Each 90°)
∴ ΔAQB ≅ΔAOD (By SAS congruency.
∴ AB = AD (1) (Congruent parts of Congruent) triangles)
Similarly, AB = BC (2)
BC = CD (3)
CD = DA
∴ From (1), (2), (3) and (4), we have
AB = BC = CD = DA
Thus, the quadrilateral ABCD is a rhombus. Hence, proved.
9. Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.
DC O Q
Sol. In ΔAOP and ΔCOQ
PAOOCQ∠=∠ (Alternate interior angles)
APCQ = (Given)
POAQOC∠=∠ (Vertically opposite angles)
By ASA criteria,
AOPCOQ∆≅∆
OPOQ∴= and OCOA = (Congruent parts of Congruent triangles) Therefore, PQ and AC bisect each other. Hence, proved.
10. Prove that two lines perpendicular to the same line are parallel to each other. 2 m l n 1
Sol. Let lines ,, lmn be such that ln ⊥ and mn ⊥ as shown in given figure. We have to prove that lm ‖ Now, and 190 and 290 12 lnmn ⊥⊥
Thus, the corresponding angles made by the transversal n with lines l and m are equal. Hence, lm ‖ .
11. Angles of a quadrilateral are in the ratio 3: 4: 4: 7. Find all the angles of the quadrilateral.
(NCERTExemplar)
Sol. The angles of a quadrilateral are in the ratio 3 : 4 : 4 : 7.
Let x be the common multiple. So, the angles will be 3x, 4x, 4x and 7x.
The sum of all angles of a quadrilateral is 360º.
3447360 xxxx ∴+++=°
18360 x ⇒=°
20 x ⇒=°
The angles are, 332060 x =×°=°
442080 x =×°=°
7720140 x =×°=°
So the angles of the quadrilateral are 60°, 80°, 80° and 140° .
12. In ΔABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are respectively the midpoints of AB and BC, determine the length of DE
(NCERTExemplar)
Sol. Given:
In ΔABC,
AB = 5 cm
BC = 8 cm
AC = 7 cm
Thus, using the midpoint theorem, DE is half of AC. 1 2 DEAC∴=×
121273.5cmDEAC ⇒==×=
Therefore, 3.5 cm. DE =
13. One angle of a quadrilateral is of 108° and the remaining three angles are equal. Find each of the three equal angles.
(NCERTExemplar)
Sol. The sum of all angles of a quadrilateral is 360°. One angle is 108°.
Let the other three angles be x.
108 360 xxx ∴°+++=°
1083360 x °+=°
3252 x =° 252 84 3 x ° ==°
Each of the three angles will be of 84° each.
Short Answer Questions
1. E is the mid-point of the side AD of the trapezium ABCD with AB||DC. A line through E drawn parallel to AB intersects BC at F. Show that F is the mid-point of BC. (Hint: Join AC)
Sol. In trapezium ABCD,AB||CD and FE are drawn parallel to CD.
∴ AB||CD||FE
In ΔADC, E is the midpoint of AD and EO||DC. By converse of the Midpoint theorem, O is the midpoint of AC Now,
In ΔABC, O is the midpoint of AC and FO||AB.
By converse of the Midpoint theorem, F is the midpoint of BC. Hence, proved.
2. Through A,B and C, lines RQ, PR and QP have been drawn, respectively parallel to sides BC, CA and AB of a triangle
ABC as shown in the given figure.
Show that BC = 1 2 QR.
Sol. In parallelogram ACBR,
BC = AR (i)
(Opposite sides of a parallelogram are equal)
Also, In parallelogram ABCQ,
BC = AQ (ii)
(Opposite sides of a parallelogram are equal)
From (i) and (ii)
BC = AR = AQ (iii)
Now, In the above figure, RAQ,
∴RQ = AR + AQ
⇒ RQ = BC + BC (from iii)
Hence, proved.
3. ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus.
AP DC B
Sol. According to the question,
DP is perpendicular to side AB and AP = PB
Now,
In ΔAPD and ΔBPD,
∠DPA =∠DPB = 90°
AP = PB (given)
DP = PD (common side)
∴ΔAPB ≅ΔBPD (SAS congruence)
∴∠DAB =∠DBA (1) (Congruent parts of Congruent triangles)
Also, the diagonals of a rhombus bisect the angles.
∴∠DBC =∠DBA (2)
From (1) and (2)
∠DBC =∠DBA =∠DAB = x°
The sum of adjacent angles in a rhombus is supplementary.
() 180 DABDBCDBA ∴∠+∠+∠=°
3180 x ∴=°
180 60 3 x ° ==°
60 DBCDBADAB ∴∠=∠=∠=°
Sol. In the parallelogram ABCD, BP = PC
∠BAP =∠PAD (Given) (i)
Adjacent angles in a parallelogram are supplementary.
∴∠A +∠B = 180° (ii)
Now, In ΔBAP, The sum of angles in a triangle is 180°
ABCDBCDBA ∠=∠+∠=°+°=°
()()6060120
Opposite angles of a rhombus are congruent.
60 DCBDAB ∴∠=∠=°
120 ABCADC ∠=∠=°
4. In given figure, points M and N are taken on opposite sides AB and CD , respectively of a parallelogram ABCD such that AM = CN
Show that AC and MN bisect each other.
(NCERTExemplar)
O MB DNC 4 3 1 2
Sol. Since, ABCD ‖
Therefore, in AOM∆ and ΔCON, we have 13 (Alternate interior angles) ∠=∠
AMCN = (Given)
24∠=∠ (Alternate interior angles)
AOMCON∴∆≅∆ (By ASA congruence criterion)
⇒ AO = OC (Congruent parts of Congruent triangles) and MO = NO (Congruent parts of Congruent triangles)
Thus, AC and MN bisect each other. Hence, proved.
5. In the given figure, P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP =∠DAP
Prove that AD = 2CD.
∠BAP +∠APB +∠ABP = 180°
1 2
AAPBBAB ∠+∠+∠=∠+∠ (from (i) and (ii))
1 2
APBA∠=∠
∴∠BAP =∠APB
PB = AB (iii) (sides opposite to congruent angles are equal)
AD = BC
(Opposite sides of a parallelogram are equal)
11 22
ADBC∴=
Also, P is the midpoint of BC 1 2
ADBP⇒= From (iii) 1 2
ADAB⇒=
But AD = CD (Opposite sides of a parallelogram are equal)
ADCD⇒=
1 2
⇒ AD = 2CD Hence, proved.
6. In the figure, ABCD is a parallelogram and E is the mid-point of side BC . DE and AB on producing meet at F. Prove that 2 AFAB = .
Sol. In DCE∆ and FBE∆ , (alternate interior angles)
DCEFBE∠=∠
CEBE = (Given)
DECBEF∠=∠ (Vertically opposite angle)
DCEFBE∴∆≅∆ (ASA congruency)
or, DCFB = (Congruent part congruent triangle) ) (opp. sides of gm
7. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Sol. Let ABCD be a quadrilateral, where P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.
Join PQ, QR, RS and SP.
Let us also join PR, SQ and AC
Now, in ΔABC, we have P and Q are the mid-points of its sides AB and BC respectively.
∴ PQ || AC and 1 2 PQAC = (1) (By mid-point theorem)
Similarly, RS || AC and 1 2 RSAC = (2)
∴ By (1) and (2), we get PQ || RS, PQ = RS
∴ PQRS is a parallelogram.
And the diagonals of a parallelogram bisect each other, i.e., PR and SQ bisect each other. Thus, the line segments joining the midpoints of opposite sides of a quadrilateral ABCD bisect each other.
Hence, proved.
9. ABCD is a rectangle and P, Q, R ans S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.
Sol. We have,
Now, in ΔABC, we have 1 2 PQAC = and PQ || AC (1)
(By mid-point theorem)
Similarly, in ΔADC, we have 1 2 SRAC = and SR || AC (2)
From (1) and (2), we get PQ = SR and PQ || SR
∴ PQRS is a parallelogram.
Now, in ΔPAS and ΔPBQ, we have
∠A =∠B (Each 90°)
AP = BP ( P is the mid-point of AB)
AS = BQ ( 11 22 ADBC = )
∴ΔPAS ≅ΔPBQ (By SAS congruency)
⇒ PS = PQ
(Congruent part congruent triangle)
Also, PS = QR and PQ = SR
( Opposite sides of a parallelogram are equal)
So, PQ = QR = RS = SP i.e., PQRS is a parallelogram having all of its sides equal.
Thus, PQRS is a rhombus.
Hence, proved.
8. In ,, ABCDE∆ and F are the mid-points of sides , ABBC and CA If 6 cm,7.2 cm ABBC== and 7.8 cm AC = , find the perimeter of DEF∆ .
9. In the given figure, AX and CY are respectively the bisectors of the opposite angles A and C of a parallelogram ABCD. Show that AX || CY
(NCERTExemplar)
Sol. ABCD is a parallelogram.
AC∴∠=∠ 11 22
AC∠=∠
(AX and CY are the bisectors of ∠A and ∠C )
∠YAX =∠YCX (1)
Now, YA||CX
∠AYC +∠YCX = 180° (Corresponding angles are supplementary)
∠AYC +∠YAX = 180° (from 1)
We know that,
Interior angles of the same side of the transversal are supplementary.
Therefore, AX||CY
Hence, proved.
10. In a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD
DFC
AEB
Sol. Since the opposite sides of a parallelogram are parallel and equal.
∴ AB || DC
⇒ AE || FC (1) and AB = DC 1 1 22 ABDC⇒=
⇒ AE = FC (2)
From (1) and (2), we have
AE || PC and AE = PC
∴ΔECF is a parallelogram.
Now, in ΔDQC, we have F is the mid-point of DC and FP || CQ ( AF || CE )
⇒ DP = PQ (By converse of mid-point theorem) (3)
Similarly, in ΔBAP, E is the mid-point of AB and EQ || AP ( AF || CE)
⇒ BQ = PQ (4)(By converse of mid-point theorem)
∴ From (3) and (4), we have
DP = PQ = BQ
So, the line segments AF and EC trisect the diagonal BD. Hence, proved.
Long Answer Questions
1. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Sol. Let ABCD be a quadrilateral such that diagonals AC and BD are equal and bisect each other at right angles.
DC AB O 1 90° 2 90°
Now, in ΔAOD and ΔAOB, We have
∠AOD =∠AOB (Each 90°)
AO = AO (Common)
OD = OB ( O is the midpoint of BD)
∴ΔAOD ≅ΔAOB (By SAS congruency)
⇒AD = AB (Congruent part congruent triangle) (1)
Similarly, we have
AB = BC (2)
BC = CD (3)
CD = DA (4)
From (1), (2), (3) and (4), we have
AB = BC = CD = DA
∴ Quadrilateral ABCD has all sides equal.
In ΔAOD and ΔCOB, we have AO = CO (Given)
OD = OB (Given)
∠AOD =∠COB (Vertically opposite angles)
So, ΔAOD ≅ΔCOB (By SAS congruency)
∴∠1 =∠2 (Congruent part congruent triangle)
But, they form a pair of alternate interior angles.
∴ AD || BC
Similarly, AB || DC
∴ ABCD is a parallelogram.
∴ A parallelogram having all its sides equal is a rhombus.
∴ ABCD is a rhombus.
Now, in ΔABC and ΔBAD, we have
AC = BD (Given)
BC = AD (Proved)
AB = BA (Common)
∴ΔABC ≅ΔBAD (By SSS congruency)
∴∠ABC =∠BAD (5)
(Congruent part congruent triangle)
Since AD || BC and AB are transversals,
∴∠ABC +∠BAD = 180° (Co-interior angles) (6)
⇒∠ABC =∠BAD = 90° (By(5) & (6))
So, rhombus ABCD has one angle equal to 90°. Thus, ABCD is a square.
Hence, proved.
2. Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that
(a) it bisects ∠C also, (b) ABCD is a rhombus.
DC AB
Sol. We have a parallelogram ABCD in which diagonal AC bisects ∠A
⇒∠DAC =∠BAC 3 4 1 2 AB DC
(a) Since, ABCD is a parallelogram.
∴ AB || DC and AC is a transversal.
∴∠1 =∠3 (1)
(
Alternate interior angles are equal)
Also, BC || AD and AC is a transversal.
∴∠2 =∠4 (2)
( Alternate interior angles are equal)
Also, ∠1 =∠2 (3)
( AC bisects ∠A)
From (1), (2) and (3), we have
∠3 =∠4
⇒ AC bisects ∠C.
(b) In ΔABC, we have
∠1 =∠4 (From (2) and (3))
⇒ BC = AB (4)
( Sides opposite to equal angles of a Δ are equal)
Similarly, AD = DC (5)
But, ABCD is a parallelogram. (Given)
∴ AB = DC (6)
From (4), (5) and (6), we have AB = BC = CD = DA
Thus, ABCD is a rhombus.
Hence, proved.
3. In ΔABC and ΔDEF, AB = DE, AB || DE, BCEF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see figure).
Show that.
(a) quadrilateral ABED is a parallelogram
(b) quadrilateral BEFC is a parallelogram
(c) AD || CF and AD = CF
(d) quadrilateral ACFD is a parallelogram
(e) AC = DF
(f) ΔABC ≅ΔDEF
Sol. (a) We have AB = DE (Given) and AB || DE (Given)
i. e., ABED is a quadrilateral in which a pair of opposite sides (AB and DE) are parallel and of equal length.
∴ ABED is a parallelogram. Hence, proved.
(b) BC = EF (Given) and BC || EF (Given)
i.e. BEFC is a quadrilateral in which a pair of opposite sides (BC and EF) are parallel and of equal length.
∴ BEFC is a parallelogram. Hence, proved.
(c) ABED is a parallelogram (Proved)
∴ AD || BE and AD = BE (1)
(
Opposite sides of a parallelogram are equal and parallel) Also, BEFC is a parallelogram. (Proved)
BE || CF and BE = CF (2)
( Opposite sides of a parallelogram are equal and parallel)
From (1) and (2), we have
AD || CF and AD = CF Hence, proved.
(d) Since, AD || CF and AD = CF (Proved)
i.e., In quadrilateral ACFD, one pair of opposite sides (AD and CF) are parallel and of equal length.
∴Quadrilateral ACFD is a parallelogram. Hence, proved.
(e) Since, ACFD is a parallelogram. (Proved)
So, AC = DF ( Opposite sides of a parallelogram are equal)
Hence, proved.
(f ) In ΔABC and ΔDFF, we have
AB = DE (Given)
BC = EF (Given)
AC = DE (Proved in (v) part)
ΔABC ≅ΔDFF (By SSS congruency)
Hence, proved.
4. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure).
Show that,
AD
B Q C P
(a) ΔAPD ≅ΔCQB (b) AP = CQ
(c) ΔAQB ≅ΔCPD (d) AQ = CP
(e) APCQ is a parallelogram.
Sol. We have a parallelogram ABCD, BD is the diagonal and points P and Q are such that PD = QB.
(a) Since, AD || BC and BD is a transversal.
∴∠ADB =∠CBD ( Alternate interior angles are equal)
⇒∠ADP =∠CBQ
Now, in ΔAPD and ΔCQB, we have
AD = CB (Opposite sides of a parallelogram
ABCD are equal)
PD = QB (Given)
∠ADP =∠CBQ (Proved)
∴ΔAPD ≅ΔCQB (By SAS congruency) Hence, proved.
(b) Since, ΔAPD ≅ΔCQB (Proved)
⇒ AP = CQ (Congruent part congruent triangle) Hence, proved.
(c) Since, AB || CD and BD is a transversal.
∴∠ABD =∠CDB
⇒∠ABQ =∠CDP
Now, in ΔAQB and ΔCPD, we have
QB = PD (Given)
∠ABQ =∠CDP (Proved)
AB = CD (Y Opposite sides of a parallelogram ABCD are equal)
∴ΔAQB =ΔCPD (By SAS congruency) Hence, proved.
(d) Since, ΔAQB =ΔCPD (Proved)
⇒ AQ = CP (Congruent part congruent triangle) Hence, proved.
(e) In a quadrilateral ΔPCQ, Opposite sides are equal. (Proved)
∴ΔPCQ is a parallelogram. Hence, proved.
5. ABCD is a rhombus and P, Q, R and S are the midpoints of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.
Sol. We have a rhombus ABCD and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Join AC 1 2 4 5 3 E R D S A FP B Q C
In ΔABC, P and Q are the mid-points of AB and BC respectively.
∴ PQ = 1 2 AC and PQ || AC (1) (By mid-point theorem)
In ΔADC, R and S are the mid-points of CD and DA respectively.
∴ SR = 1 2 AC and SR || AC (2) (By mid-point theorem)
From (1) and (2), we get PQ = 1 2 AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
∴ PQRS is a parallelogram. (3)
Now, in ΔERC and ΔEQC,
∠1 =∠2
( The diagonals of a rhombus bisect the opposite angles)
CR = CQ 22 CDBC = CE = CE (Common)
∴ΔERC ≅ΔEQC (By SAS congruency)
⇒∠3 =∠4 (Congruent part congruent triangle) (4)
But ∠3 +∠4 = 180° (Linear pair) (5)
From (4) and (5), we get
⇒∠3 =∠4 = 90°
Now, ∠RQP = 180° ∠b (Y Co-interior angles for PQ || AC and EQ is transversal)
But ∠5 =∠3 ( Vertically opposite angles are equal)
∴∠5 = 90°
So, ∠RQP = 180° ∠5 = 90°
∴ One angle of parallelogram PQRS is 90°.
Thus, PQRS is a rectangle.
Hence, proved.
6. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(a) D is the mid-point of AC
(b) MD ⊥ AC
(c) CM = MA = 1 2 AB
Sol. we have
(a) In ΔACB, We have M is the mid-point of AB. (Given)
MD || BC (Given)
∴ Using the converse of the mid-point theorem, D is the mid-point of AC. Hence, proved.
(b) Since MD || BC and AC are transversals,
∠MDA =∠BCA
( Corresponding angles are equal)
As ∠BCA = 90° (Given)
∠MDA = 90°
⇒ MD ⊥ AC Hence, proved.
(c) In ΔADM and ΔCDM, we have
∠ADM =∠CDM (Each equal to 90°)
MD = MD (Common)
AD = CD ( D is the mid-point of AC )
∴ΔADM ≅ΔCDM (By SAS congruency)
⇒ MA = MC (1)
(Congruent part congruent triangle)
M is the mid-point of AB (Given)
1 2 MAAB = (2)
From (1) and (2), we have 1 2 CMMAAB == Hence, proved.
7. In the given figure, ABCD is a parallelogram and ∠DAB = 60°. If the bisector of angles. A and B meet at M on CD, prove that M is the mid-point of CD A
B 60° 60° 30° 30°
Sol. We have, ∠DAB = 60°
Since, AD || BC and AB is the transversal.
∴∠A +∠B = 180°
60° +∠B = 180°
∠B = 120°
Also, AM and BM are angle bisectors of ∠A and ∠B respectively.
∴∠DAM =∠MAB = 30° and ∠CBM =∠MBA = 60°
Now, AB || DC and transversal AM cuts them.
∴∠MAB =∠DMA (Alternate interior angles)
⇒∠DMA = 30°
Thus, in ΔAMD, we have
∠MAD =∠AMD (Each equals to 30°)
⇒ MD = AD (Sides opposite to equal angles) (i)
Again, AB || DC and transversal BM cuts them.
∴∠CMB =∠MBA (Alternate interior angles)
⇒∠CMB = 60°
Thus, in ΔCMB, we have
∠CBM =∠CMB (Each equals to 60°)
⇒ CM = BC (Sides opposite to equal angles)
⇒ CM = AD ( BC = AD) (ii)
From (i) and (ii), we get MD = CM
⇒ M is the mid-point of CD. Hence, proved.
8. Prove that the diagonal divides a parallelogram into two congruent triangles.
Sol. Given: A parallelogram ABCD
To prove: ΔABC ≅ΔCDA
Since ABCD is a parallelogram.
Therefore, AB || DC and AD || BC
Now, AB || DC and transversal AC cuts them at A and C respectively.
∴∠DCA =∠BAC (Alternate interior angles)
Again AD || BC and AC is the transversal.
∴∠DAC =∠ACB (Alternate interior angles)
Now, in ΔABC and ΔCDA; we have
∠BAC =∠DCA (Shown above)
AC = AC (Common)
∠ACB =∠DAC (Shown above)
∴ΔABC ≅ΔCDA (By ASA congruence criterion)
Hence, proved.
9. If two parallel lines are intersected by a transversal, then prove that bisectors of the interior angles form a rectangle.
Sol. ABCD ‖ and EF cuts them () Alternate 11
AGHDHGs AGH ∠=∠∠
⇒∠=∠
Multiplying both sides by 22 12
Thus, lines GM and HL, are intersected by a transversal GH at G and H respectively such that pair of alternate angles are equal, i.e., 12∠=∠ || GMHL∴
Similarly, we can prove that GLHM ‖ So, GMHL is a parallelogram.
Since ABCD ‖ and EF is a transversal, therefore, 180 BGHDHG∠∠+=
[∴ Sum of interior ∠s on the same side of a transversal = 180°]
111 180 222 BGHDHG∠∠ ⇒+=× 1
Multiplying both sides by 2
3290∠∠ ⇒+= (1) 11 3 and 2 22 BGHDHG∠∠∠∠
But, 32180 GLH∠∠∠++=
() Sum of the s of a 180 ∠∆= () () 90 180 using 1 GLH∠ ⇒+= 1809090 GLH∠ ⇒=−=
Thus, in the parallelogram GMHL, we have 90 GLH ∠=
Thus, GMHL is a rectangle. Hence, proved.
10. Prove that the straight line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and is equal to half the difference of these sides.
Competency Based Questions
Multiple Choice Questions
1. In the given diagram, ABCD is a rhombus, AEC and BED are straight lines. pqrst++++= ?
Sol. Construction: Join CN and produce it to meet AB at E . In CDN∆ and ΔEBN, = 90°
DCNBEN∠∠=() Alternate interior angles
CDNEBN∠∠=() Alternate interior angles and DNB = (N is the mid-point of BD )
CDN∠ ∴≡ ∠EBN = 90°. (By AAS−congruence criterion) and CNENDCEB⇒== (Congruent part of congruent triangle)
Now, in , ACEM∆ and N are the mid-points of CA and CE respectively.
Also 11 and 22 MNAEMNAB ABDC MNABDC MNAEMNABEB
1 2 MNABDC ⇒=−() , proved above EBDC∴=
Hence, proved.
3. What is the value of x in the given figure?
C c b x (a) bac (b) bac −+ (c) bac +− (d) abc ++
2. Under what conditions must PQRS be a parallelogram?
(a) 45 x
(b) 16 y
(c) 45,16xy
(d) 16,45xy
4. ABCD is a parallelogram. If P be a point on CD such that APAD = , then the measure of PABBCD∠+∠ is: (a) 180 (b) 225 (c) 240 (d) 135
5. Given that ABCD is a parallelogram whose diagonals intersect at point .110,35 OABCACB ∠=∠= and ADB∠ 55 = . The term that best describes ABCD is: (a) Rectangle (b) Rhombus (c) Square (d) Kite
6. In a trapezium ABCD , if ABCD ‖ , then 22ACBD+=
(a) 22 2 BCADBCAD ++×
(b) 22 2 ABCDABCD ++×
(c) 22 2 ABCDADBC ++×
(d) 22 2 BCADABCD ++×
7. In given figure, P is the mid-point of side BC of parallelogram ABCD . If AP is the bisector of BAD∠ , then : ADCD =
(a) 1:2
(b) 2:1
(c) 2:3
(d) 1:1
8. In a quadrilateral , ABCDAC ∠+∠ is 2 times BD∠+∠ . If 140 A ∠= and 60 D ∠= , then B ∠=
(a) 60
(b) 80
(c) 120
(d) None of these
9. In given figure, D is the mid-point of AB and 1 3 cm 2 PCAP== . If 4 cm ADDB== and DEBP ‖ , then AE =
Assertion-Reason Based Questions
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): ABCD and PQRC are rectangles and Q is a midpoint of AC, then DP = PC.
Reason (R): The line segment joining the midpoint of any two sides of a triangle is parallel to third side and is equal to half of it.
2. Assertion (A): In ABC∆ , median AD is produced to X such that ADDX = . Then ABXC is a parallelogram.
Reason (R): Diagonals AX and BC bisect each other at right angles. A
(a) 4 cm (b) 3 cm
(c) 2 cm (d) 5 cm
10. From the adjoining parallelogram, what are the value of x, y and z? D zy C F AEB x 50°
(a) 40, 50, 50 xyz=°=°=°
(b) 40, 40, 40 xyz=°=°=°
(c) 50, 40, 30 xyz=°=°=°
(d) 40, 50, 60 xyz=°=°=°
3. Assertion (A): In given figure, if AD and BE are medians of ABC∆ such that BEDF ‖ , then 1 4 CFAC =
F
Reason (R): The line drawn through the mid-point of one side of a triangle, parallel to another side, intersects the third side at its mid-point.
Case Study Based Questions
1. Manish had a plot of land in the shape of a quadrilateral. To efficiently utilize the land, he decided to construct his house at the center by joining the midpoints of the four sides of the land.
The remaining four triangular plots were used for gardening, parking, a playground, and a storage area.
Based on this information, answer the following questions:
(a) (i) Manish plans to build a boundary wall around the entire plot. If the adjacent sides of the plot are
Answers
Multiple Choice Questions
1. (b)
By angel sum property in
2. (c)
45,16xy==
We know that opposite angles of parallelogram are equal and PRQS ∴∠=∠∠=∠ 6516 yy=+ 16 y = () 34243 xxQS +=−∠=∠
3. (a) bac
exterior angle theorem
in the ratio of 1 : 2 and the total perimeter is 180 meters, what are the lengths of the adjacent sides?
(ii) If Manish had chosen a plot where all the angles were equal, what type of quadrilateral would it be?
(b) While designing a decorative pathway around a parallelogram-shaped garden PQRS, Manish noted some angles:
(i) If ∠PSQ = 30° and ∠QRS = 110°, find the value of ∠SQP
(ii) What will be the measure of the angle formed by the bisectors of any two adjacent angles of this parallelogram-shaped garden?
adding
and
ACEACEaxc ∴∠+∠=++ DE and BF intersect at C and ECFBCD∠=∠ (Opposite angles) but BCDb∠= ECFACFACE baxc xbac ∠=∠+∠ =++ =−−
4. (a) 180 DpC AB
ABCD ‖ Angles opposite to equal sides are equal () Alternate angles APAD APDADP
ABCD PABAPD ∠∠α ∠∠α = ∴== ∴== ‖
ADBC ‖ , and CD is transversal
ADC∠ and BCD∠ are consecutive co-interior an gees
180 ADCBCD ∴∠+∠= (Consecutive Co-interior angles are supplementary)
5. (b) Rhombus D O 55° 110° 35° C
AB , ADBCDB ‖ is the transversal 55 CBDADB ∠=∠=
In BOC∆
BDAC ⊥
In ∆ABC :
1801103535 BAC ∠=−+⇒
if , ACBBAC∠=∠
ABBC⇒=
ABCD ∴ is Rhombus
6. (d) 22 2 BCADABCD ++×
In , ABCB∆∠ is acute angle.
222 2 ACABBCABBF ∴=+−× (1)
In , ABDA∆∠ is acute angle.
222 2 BDABADABAE ∴=+−× (2)
Adding (1) and (2), we get
BCADABEFBCADABCD
7. (b) 2:1
In parallelogram ABCD , we find that ADBC ‖ and transversal AP cuts them at A and P respectively. Therefore, 23∠=∠
But, AP is the bisector of BAD∠ . Therefore, 12∠=∠
From (i) and (ii), we obtain ∠1 =∠3 ⇒ BP = AB (Sides opposite to equal angles)
8. (a) 60°
We know that in a quadrilateral,
∠A +∠B +∠C +∠D = 360°
Also, it’s given:
∠A +∠C = 2 × (∠B +∠D)
Substitute the known values:
140° +∠C = 2 × (∠B + 60°)
140° +∠C = 2∠B + 120°
∠C = 2∠B + 120° 140°
∠C = 2∠B 20°
Now, using the angle sum property:
140° +∠B +∠C + 60° = 360°
∠B +∠C = 160°
Substitute ∠C:
∠B + (2∠B 20°) = 160°
3∠B = 180°
∠B = 60°
9. (b) 3 cm
We have, 1 2 PCAP = and 3 cm PC = . Therefore, 6 cm AP = .
In ∆ABP , it is given that DEBP ‖ and D is the mid-point of AB . Therefore, E is the mid-point of AP
Hence, 1 3 cm 2 AEAP==
10. In ΔCEB,
180 CEBECBCBE ∠+∠+∠=° (Sum of angles in a triangle)
9050180 x °+°+=° ( 90,50CEBECB ∠=°∠=° given)
140180 x °+=°
40 x =° (1)
ABCADC∠=∠ (Opposite angles of a parallelogram are congruent)
From (1)
40 zx==° (2)
In DFC∆ ,
180 DFCFDCFCD ∠+∠+∠=° (Sum of angles in a triangle)
9040180 FCD °+°+∠=° ( 90,50CEBECB ∠=°∠=° given)
130180 FCD °+∠=°
50 FCD ∠=° (3)
Sum of any two adjacent angles of a parallelogram is equal to 180°
180 ADCDCB ⇒∠+∠=°
180 ADCDCFFCEECB ⇒∠+∠+∠+∠=° From (2) and (3)
405050180 y ⇒°+°++°=°
40 y ⇒=°
Assertion-Reason
Based Questions
1. (b) Both A and R are true, but R is not the correct explanation of A.
In case of assertion (A): Q is a midpoint of AC.
So, P is also a midpoint of DC.
(According to converse of midpoint theorem)
∴ Assertion is true.
In case of reason (R): It is a midpoint theorem.
∴ Reason is true but not correct explanation of Assertion.
2. (c) A is true, but R is false.
In case of assertion (A): ABXC is a || gm
∴ In a parallelogram Diagonals bisects each other.
∴ Assertion is true.
In case of reason (R): Diagonals of a m gm‖ do not bisect each other at right angles.
∴ Reason is false.
Hence, Assertion is true but Reason is false.
3. (a) Both A and R are true, and R is the correct explanation of A.
Additional Practice Questions
1. If three angles of a quadrilateral are each equal to 75°, the fourth angle is:
(a) 150° (b) 135°
(c) 45° (d) 75°
In case of assertion (A): F is the mid-point of CE . 1111 2224
∴ Assertion is true.
In case of reason (R): It is a midpoint theorem.
∴ Reason is true and correct explanation of Assertion.
Case Study Based Questions
1. (a) (i) Let the adjacent sides be x and 2x .
Perimeter of parallelogram () 2 ( aba=+ and b are adjacent sides of parallelogram )
Then () 22180 xx+= 180 6180 30 6 xxm⇒=⇒==
Hence, the lengths of adjacent sides are 30 m and 60 m
(ii) All angles of a square and rectangle are equal. (b) (i) 110 PR ∠=∠= (Opposite angles of a parallelogram are equal)
1801103040 SQP ∠=−+=
(Using angle sum property in PQS∆ ) (ii) In parallelogram ,180ABCDCD∠+∠=
(Using angle sum property in COD∆ ).
2. What is the maximum number of obtuse angles that a quadrilateral can have?
(a) 1 (b) 2
(c) 3 (d) 4
3. ABCD is a rhombus such that ∠ACB = 40°. Then ∠ADB is: (a) 40° (b) 45° (c) 50° (d) 60°
4. If PQRS is a parallelogram, then ∠P ∠R is: (a) 90° (b) 45° (c) 60° (d) 0°
5. In a quadrilateral , ABCDAC ∠+∠ is two times BD∠+∠ . If 140 A ∠= and 60 D ∠= , then B∠ equals (a) 60 (b) 80 (c) 120 (d) None of these
6. In given figure, , lm and n are parallel lines intersected by transversal p at , XY and Z respectively. Find 1,2∠∠ and 3∠ . l p X Y z 1 2 120° 3 m n
7. In given figure, rd 2 160 and 2 3
of a right angle. Prove that lm ‖ l 1 2 3 m n
8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that ABCD is a square.
9. P is the midpoint of the side CD of a parallelogram. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and CQ = QR.
10. ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.
11. ABCD is a parallelogram, E and F are the mid points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH
12. Show that the bisectors of angles of a parallelogram form a rectangle.
13. l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). Show that l, m and n cut off equal intercepts DE and EF on q also.
14. ABCD is a parallelogram on which P and Q are mid points of opposite sides AB and CD ( see figure) If AQ intersects DP at S and BQ intersects CP at R, show that:
(a) APCQ is a parallelogram
(b) DPBQ is a parallelogram.
(c) PSQR is a parallelogram.
15. In the adjoining trapezium ,102ABCDC∠= . Find all the remaining angles of the trapezium.
16. Show that each angle of a rectangle is a right angle.
17. Show that the diagonals of a rhombus are perpendicular to each other.
18. Show that the line-segments joining the mid-points of the opposite sides of quadrilateral bisect each other.
19. ABCD is a rectangle in which diagonal AC bisects A∠ as well as C∠ . Show that ABCD is square.
20. In the figure & AXCY are respectively the bisectors of the opposite angles A & C of a parallelogram
ABCD. Show that AXCY ‖ .
21. In a parallelogram ABCD, the diagonals AC & BD intersect each other at O. Through O, a line is drawn to intersect AD at X and BC at Y Show that OXOY =
22. In a quadrilateral, & CODO are the internal bisectors of & CD∠∠ respectively. Prove that 2 ABCOD∠+∠=∠ .
23. L,M,N,K are mid points of sides BC,CD,DA, and AB respectively of a square ABCD. Prove that DL, DK,BM and BN enclose a rhombus.
27. In given figure, ABCD ‖ and P is any point shown in the figure. Prove that: 360 ABPBPDCDP ∠+∠+∠= .
24. The sides AD and BC of a quadrilateral are produced as shown in the given figure. Prove that x = 2 ab + . AB a xx
25. In the figure, ABCD is a parallelogram and E is the mid-point of side BCDE and AB on producing meet at F. Prove that AF = 2AB.
28. A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse. (NCERTExemplar)
29. In given figure, show that . ABEF ‖
26. In the figure ABCD is a parallelogram. If & XY are the points on the diagonal BD such that DXBY = , prove that AXCY is a parallelogram.
Challenge Yourself
1. In the figure above, if ABAC = and BC is extended to D , then find the value of x γ + .
30. In given figure, AB || CD || EF and GH || KL Find . HKL∠
Scan me
2. In the figure above, KLMN is a rectangle. , PQ , R , and S are the mid-points of KL, LM, MN, and NK, respectively. If 30 KPS ∠= , then find QRS∠ .
3. ABCD is a rhombus and its two diagonals intersect at O, show that AB 2 + BC 2 + CD 2 + DA2 =4 (AO 2 + DO 2).
4. The angle bisectors of ,,& MNOP ∠∠∠∠ of a trapezium MNOP intersect at points W and X respectively. If 50&70MPXNOX ∠=∠= , find the measure of the angles ,, & abcd
Answers
Additional Practice Questions
1. (b) 135
2. (c) 3
3. (c) 50°
4. (d) 0°
5. (a) 60
6. 160,2120 ∠=°∠=° and 360 ∠=°
5. In the figure, ABCD is a square and CDE is an equilateral triangle. Find reflex AEC∠ . DC
1. In given figure, ABCD and PQRS are rectangles where Q is the mid-point of BD. If 5 cm QR = , then AB =
(a) 5 cm
(b) 7.5 cm (c) 10 cm
(d) 15 cm
2. If angles ,, ABC and D of the quadrilateral ABCD , taken in order, are in the ratio 3:7:6:4, then ABCD is a (a) rhombus (b) parallelogram (c) trapezium (d) kite (NCERTExemplar)
3. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a rectangle, if (a) ABCD is a rectangle.
(b) diagonals of ABCD are perpendicular. (c) diagonals of ABCD are equal.
(d) ABCD is a parallelogram.
Assertion-Reason Based Questions
[2×1=2Marks]
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
4. Assertion (A): In given figure, if AD is the median of ABC∆ and E is the mid-point of ADBE produced meets AC in F , then 1 3 AFAC = .
Reason (R): The line through the mid-point of one side of a triangle, parallel to another side, intersects the third side at its mid-point.
5. Assertion (A): If the measures of three angles of a quadrilateral are 130,70 and 60, then the measure of fourth angle is 100
Reason (R): The sum of all the angles of a quadrilateral is 360 .
Very Short Answer Questions
6. PQRS is a parallelogram, in which PQ = 22 cm and its perimeter is 76 cm. What is the length of the side QR?
7. In a trapezium ABCD, AB || CD, if ∠B = 60°, find ∠C
Short Answer Questions
[4×3=12Marks]
8. ABC is a triangle and through A, B, and C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of ΔABC
9. In given figure, . Prove that 180. ABDEABCBCDCDE ∠∠∠ +=+ ‖
10. In given figure, . Find the value of . ABCDx ‖
11. In a parallelogram ABCD , the bisector of A∠ also bisects BC at X . Prove that 2 ADAB =
Long Answer Questions
[3×5=15Marks]
12. In the figure, bisectors of ∠B and ∠D of quadrilateral ABCD meet CD and AB produced at P and Q respectively. Prove that () 1 2 PQABCADC ∠+∠=∠+∠
13. In given figure, ABCD is a trapezium in which side AB is parallel to side DC and EE is the mid-point of side AD. If F is the point on the side BC such that the segment EF is parallel to side DC. Prove that F is the mid-point of BC and EF = 1 2 (AB +DC).
14. Show that the diagonals of a rhombus are perpendicular to each other.
15. The Eiffel Tower in Paris is an architectural wonder known for its strength and beauty. Inspired by this, Arjun, a young architecture student, is studying how truss structures provide stability to tall buildings. He creates a model of a trapezium-shaped section of the Eiffel Tower to understand how forces distribute in such structures.
In his model, ABCD is a trapezium where AB || DC. He marks points P, Q, R, and S as the midpoints of sides AB, BC, CD, and DA, respectively, to analyze the internal framework resembling the criss-crossing beams of the Eiffel Tower.
Help Arjun answer the following questions based on his observations:
(a) Arjun joins the midpoints P, Q, R, and S to form a quadrilateral. Can you help him identify the type of quadrilateral PQRS formed inside the trapezium?
(b) If Arjun observes that PQ = QR in his model, what special type of quadrilateral does PQRS become?
(c) In trapezium ABCD, if Arjun finds that AD = BC and ∠A =∠B = 45°, calculate the measures of the other two angles ∠C and ∠D of the trapezium.
(d) Arjun wonders why the Eiffel Tower uses multiple quadrilateral frames like PQRS with equal opposite sides and angles. Explain to him how this design helps maintain the structural stability of the tower against external forces like wind and vibrations.
9 Circles
Chapter at a Glance
Circle: A circle is the collection of all those points in a plane which are equidistant from a fixed point in the plane.
Radius: A line segment joining the center and a point on the circle is called its radius. The plural of radius is radii.
Chord: A line segment joining any two points on a circle is called its chord. A circle can have infinitely many chords.
Diameter: A chord passing through the center of the circle is called diameter. It is the longest chord.
Arcs: It is a segment of a circle’s circumference.
Minor Arc: An arc measuring less than 180°.
Major Arc: An arc measuring greater than or equal to 180°.
Semicircle: An arc measuring exactly 180°.
Cyclic Quadrilaterals: A quadrilateral whose all the four vertices lie on a circle.
1. The sum of either pair of opposite angles of a cyclic quadrilateral is 180º. i.e ∠A +∠C = 180°, ∠B +∠D = 180°
2. If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.
Important Theorems
1. Equal chords of a circle subtend equal angles at the centre.
2. If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.
3. The perpendicular from the centre of a circle to a chord bisects the chord.
4. The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
5. Equal chords of a circle (or of congruent circles) are equidistant from the center (or their corresponding centers).
6. Chords equidistant from the center (or centers) of a circle (or of congruent circles) are equal in length.
7. If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal.
8. Congruent arcs (or equal arcs)of a circle subtend equal angles at the center.
9. The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
10. Angles in the same segment of a circle are equal.
11. Angle in a semicircle is a right angle.
12. If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).
NCERT Zone
NCERT Exercise 9.1
1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Sol. Given: Two congruent circles with centres O and O'
AB and CD are equal chords of the circles with centres O and O’ respectively.
To Prove: ∠AOB =∠CO'D
Proof: In triangles AOB and COD,
AB = CD (Given)
AO = CO' (Radii of congruent circle)
BO = DO' (Radii of congruent circle)
⇒ΔAOB ≅ΔCO'D (By SSS congruency)
⇒∠AOB =∠CO'D (Corresponding parts of congruent triangles)
Hence, proved.
2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Sol. Given: Two congruent circles with centres O and O'
AB and CD are chords of the circles with centres O and O' respectively such that
To Prove: AB = CD
Proof: In triangles AOB and CO'D,
AO = CO' (Given)
BO = DO' (Radii of congruent circle)
⇒∠AOB =∠CO'D (Given)
⇒ΔAOB ≅ΔCO'D (By SAS congruence)
⇒ AB = CD (Corresponding parts of congruent triangles)
Hence, proved.
NCERT Exercise 9.2
1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord.
Sol. Given that the circles intersect at 2 points, so we can draw the above figure.
∠AOB =∠CO'D
Let AB be the common chord. Let O and O' be the centers of the circles respectively.
O'A = 5 cm, OA = 3 cm, OO' = 4 cm.
Since, the radius of the bigger circle is more than the distance between the 2 centers, we can say that the center of smaller circle lies inside the bigger circle itself. OO' is the perpendicular bisector of AB.
So, OA = OB = 3 cm
AB = 3 + 3 = 6 cm
Length of the common chord is 6 cm.
2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Sol. Given: AB and CD are two equal chords of a circle which meet at E
To prove: AE = CE and BE = DE
Construction: Draw OM ⊥ AB and ON ⊥ CD and join OE.
Proof: In ΔOME and ΔONE
OM = ON (Equal chords are equidistant)
OE = OE (Common)
∠OME =∠ONE (Each equal to 90°)
∴ΔOME ≅ΔONE (By RHS congruence)
⇒ EM = EN (i)
(Corresponding parts of congruent triangles)
Now, AB = CD (Given)
⇒ 1 2 AB = 1 2 CD
⇒ AM = CN (ii)
(Perpendicular from centre bisects the chord)
Adding (i) and (ii), we get
EM + AM = EN + CN
⇒ AE = CE (iii)
Now, AB = CD (iv)
Subtracting (iii) from (iv)
⇒ AB - AE = CD - AE
⇒ BE = CD - CE
Hence, proved.
3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Sol. Given: AB and CD are two equal chords of a circle which meet at E within the circle and a line PQ joining the point of intersection to the centre.
To Prove: ∠AEQ =∠DEQ
Construction: Draw OL ⊥ AB and OM ⊥ CD.
Proof: In ΔOLE and ΔOME, we have
OL = OM (Equal chords are equidistant)
OE = OE (Common)
∠OLE =∠OME (Each = 90°)
∴ΔOLE ≅ΔOME (By RHS congruence)
⇒∠LEO =∠MEO (Corresponding parts of congruent triangles)
⇒∠AEQ =∠DEQ
Hence, proved.
4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see figure).
Sol. Given: A line AD intersects two concentric circles at A, B, C and D, where O is the centre of these circles.
To prove: AB = CD
Construction: Draw OM ⊥ AD
Proof: AD is the chord of larger circle.
∴ AM = DM (OM bisects the chord) (i)
BC is the chord of smaller circle
∴ BM = CM (OM bisects the chord) (ii)
Subtracting (ii) from (i), we get
AM - BM = DM - CM
⇒ AB = CD
Hence, proved.
5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Sol. Let Reshma, Salma and Mandip be represented by R, S and M respectively.
Draw OL ⊥ RS,
OL2 = OR 2 - RL2
OL2 = 52 - 32 (RL = 3 m, because OL ⊥ RS)
= 25 - 9 = 16
OL = 16 = 4
Now, area of triangle ORS = 1 2 × KR × OS
Also, area of ΔORS = 1 2 × RS × OL
= 1 2 × 6 × 4 = 12 m2
⇒ 1 2 × KR × 5 = 12 (OS = 5 m)
⇒ 12224 4.5 m 55 KR × ===
⇒ RM = 2KR (perpendicular from center bisects the chord in equal)
⇒ RM = 2 × 4.8 = 9.6 m Thus, distance between Reshma and Mandip is 9.6 m.
6. A circular park of radius 20 m is situated at a colony. Three boys Ankur, Syed and David are siting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
S O (Ankur) (David) (Syed)
Sol. Let A, D, S denote the positions of Ankur, David and Syed respectively.
ΔADS is an equilateral triangle since all the 3 boys are on equidistant from one another.
Let B denote the mid-point of DS and hence AB is the median and perpendicular bisector of DS. Hence ΔABS is a right-angled triangle with ABS = 90°.
O (centroid) divides the line AB in the ratio 2 : 1.
So, OA : OB = 2 : 1.
OA
OB =
Since, OA = 20 then OB = 10 m
AB = OA + OB = 20 + 10 = 30 m (i)
Let the side of equilateral triangle ΔADS be 2x.
AD = DS = SA = 2x (ii)
Since B is the mid-point of DS, we get
BS = BD = x (iii)
Applying Pythagoras theorem to ΔABS, we get:
AD 2 = AB 2 + BD 2
(2x)2 = 302 + x 2
4x 2 = 900 + x 2
3x 2 = 900 x 2 = 300
x = 10 3
x = 17.32
AD = DS = SA = 2x = 34.64 m
Length of the string = Distance between them = AD or DS or SA = 34.64 m
NCERT Exercise 9.3
1. In the figure, A, B and C are three points on a circle withcentre O such that ∠BOC = 30° and ∠AOB = 60°. If D isa point on the circle other than the arc ABC, find ∠ADC
Sol. We have, ∠BOC = 30° and ∠AOB = 60°
∠AOC =∠AOB +∠BOC = 60° + 30° = 90°
B C A O D 60° 30°
We know that angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc on the remaining part of the circle.
∴ 2∠ADC =∠AOC
⇒∠ADC = 1 2 ∠AOC = 1 2 × 90°
⇒∠ADC = 45°
2. A chord of a circle is equal to the radius of the circle. Find the angle sub-tended by the chord at a point on the minor arc and also at a point on the major arc.
Sol. We have, OA = OB = AB
Therefore, ΔOAB is a equilateral triangle.
⇒∠AOB = 60° O AB D C 60°
We know that the angle subtended by an arc at the center of a circle is double the angle subtended by the same arc on the remaining part of the circle.
∴∠AOB = 2∠ACB
⇒∠ACB = 12∠AOB = 12 × 60°
⇒∠ACB = 30°
Also, ∠ADB = 1 2 reflex ∠AOB
= 1 2 (360° - 60°) = 1 2 × 300° = 150°
Thus, the angle subtended by the chord at a point on the minor arc is 150° and a point on the major arc is 30°.
3. In the figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Sol. Reflex angle POR = 2∠PQR
= 2 × 100° = 200°
Now, angle POR = 360° - 200 = 160°
Also, PO = OR (Radii of a circle)
∠OPR =∠ORP (Opposite angles of isosceles triangle)
In ΔOPR, ∠POR = 160°
∠OPR +∠ORP +∠POR = 180° (Angle sum property of a triangle)
⇒∠OPR +∠OPR + 160° = 180°
⇒ 2∠OPR = 20°
⇒∠OPR = 10°
∴∠OPR =∠ORP = 10°
4. In the figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC. 69° 31° BC A D
Sol. In ΔABC, we have
∠ABC +∠ACB +∠BAC = 180° (Angle sum property of a triangle)
⇒ 69° + 31° +∠BAC = 180° (By RHS congruence)
⇒∠BAC = 180° - 100° = 80°
Also, ∠BAC =∠BDC (Angles in the same segment)
⇒∠BDC = 80°
5. In the figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC BC
E 130° 20°
Sol. ∠BEC +∠DEC = 180° (Linear Pair)
⇒ 130° +∠DEC = 180°
⇒∠DEC = 180° - 130° = 50°
Now, in ΔDEC,
⇒∠DEC +∠DCE +∠CDE = 180° (Angle sum property of a triangle)
⇒ 50° + 20° +∠CDE = 180°
⇒∠CDE = 180° - 70° = 110°
Also, ∠CDE =∠BAC (Angles in same segment)
⇒∠BAC = 110°
6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC = 30°, find ∠BCD. Further, if AB = BC, find ∠ECD
Sol. ∠CAD =∠DBC = 70° (Angles in the same segment)
Therefore, ∠DAB =∠CAD +∠BAC = 70° + 30° = 100°
But, ∠DAB +∠BCD = 180°
(Opposite angles of a cyclic quadrilateral)
So, ∠BCD = 180° - 100° = 80°
Now, we have AB = BC
Therefore, ∠BCA = 30° (Opposite angles of an isosceles triangle)
Again, ∠DAB +∠BCD = 180°
(Opposite angles of a cyclic quadrilateral)
⇒ 100° +∠BCA +∠ECD = 180°
(Q ∠BCD =∠BCA +∠ECD)
⇒ 100° + 30° +∠ECD = 180°
⇒ 130° +∠ECD = 180°
⇒∠ECD = 180° - 130° = 50°
Thus, ∠BCD = 80° and ∠ECD = 50°
7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Sol. Given: ABCD is a cyclic quadrilateral, whose diagonals AC and BD are diameter of the circle passing through A, B, C and D.
To Prove: ABCD is a rectangle.
Proof: In ΔAOD and ΔCOB
AO = CO (Radii of a circle)
OD = OB (Radii of a circle)
∠AOD =∠COB (Vertically opposite angles)
∴ΔAOD ≅ΔCOB (By SAS congruence)
∴∠OAD =∠OCB (Corresponding parts of congruent triangles)
But these are alternate interior angles made by the transversal AC, intersecting AD and BC.
∴ AD || BC
Similarly, AB || CD. Hence, quadrilateral ABCD is a parallelogram.
Also, ∠ABC =∠ADC (i)
(Opposite angles of a || gm are equal) and,
∠ABC +∠ ADC = 180° (ii)
(Sum of opposite angles of a cyclic quadrilateral is 180°)
⇒∠ABC =∠ADC = 90° (From (i) and (ii))
∴ ABCD is a rectangle.
(A parallelogram one of whose angles is 90° is a rectangle)
Hence, proved.
8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Sol. Given: A trapezium ABCD in which AB || CD and AD = BC.
To Prove: ABCD is a cyclic trapezium.
Construction: Draw DE ⊥ AB and CF ⊥ AB
Proof: In ΔDEA and ΔCFB, we have AD = BC (Given)
∠DEA =∠CFB = 90° (DE ⊥ AB and CF ⊥ AB)
DE = CF
(Distance between parallel lines remains constant)
∴ΔDEA ≅ΔCFB (By RHS congruence)
⇒∠A =∠B (i)
(Corresponding parts of congruent triangles) and ∠ADE =∠BCF (ii) (Corresponding parts of congruent triangles)
Since, ∠ADE =∠BCF (From (ii))
⇒∠ADE + 90° =∠BCF + 90°
⇒∠ADE +∠CDE =∠BCF +∠DCF
⇒∠D =∠C (iii)
(∠ADE +∠CDE =∠D, ∠BCF +∠DCF =∠C)
(From (i) and (iii))
∴∠A =∠B and ∠C =∠D (iv)
∠A +∠B +∠C +∠D = 360°
(Sum of the angles of a quadrilateral is 360°)
⇒ 2(∠B +∠D) = 360° (Using (iv))
⇒∠B +∠D = 180°
⇒ Sum of a pair of opposite angles of quadrilateral ABCD is 180°.
⇒ ABCD is a cyclic trapezium Hence, proved.
9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D, P and Q respectively. Prove that ∠ACP =∠QCD.
Sol. Given: Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circlesat A, D and P, Q respectively. A P
To Prove: ∠ACP =∠QCD.
Proof: ∠ACP =∠ABP (i) (Angles in the same segment)
10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Sol. Given: Sides AB and AC of a triangle ABC arediameters of two circles which intersect at D
11. ABC and ADC are two right triangles with common hypotenuse AC.
Prove that ∠CAD =∠CBD.
Sol. Given: ABC and ADC are two right triangles with common hypotenuse AC C A BD
To Prove: ∠CAD =∠CBD
Proof: Sum of all angles in a triangle is 180°.
If the sum of pair of opposite angles in a quadrilateral is 180° then it is cyclic quadrilateral Angles in the same segment of a circle are equal.
Consider ΔABC,
∠ABC +∠BCA +∠CAB = 180°
(Angle sum property of a triangle)
90° +∠BCA +∠CAB = 180°
∠BCA +∠CAB = 90° (i)
Consider ΔADC,
∠CDA +∠ACD +∠DAC = 180°
(Angle sum property of a triangle)
90° +∠ACD +∠DAC = 180°
∠ACD +∠DAC = 90° (ii)
Adding Equations (i) and (ii), we obtain
∠BCA +∠CAB +∠ACD +∠DAC = 180°
To Prove: D lies on BC.
Proof: Join AD
∠ADB = 90° (i) (Angle in a semicircle)
Also, ∠ADC = 90° (ii)
Adding (i) and (ii), we get
∠ADB +∠ADC = 90° + 90°
⇒∠ADB +∠ADC = 180°
⇒ BDC is a straight line.
∴ D lies on BC
Thus, the point of intersection of circles lies on the third side BC Hence, proved.
(∠BCA +∠ACD) + (∠CAB +∠DAC) = 180°
∠BCD +∠DAB = 180° (iii)
However, it is given that
∠B +∠D = 90° + 90° = 180° (iv)
From equations (iii) and (iv), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180°.
Therefore, it is a cyclic quadrilateral.
Consider chord CD
∠CAD =∠CBD (Angles in the same segment)
12. Prove that a cyclic parallelogram is a rectangle.
Sol. Given: ABCD is a cyclic parallelogram.
To prove: A cyclic parallelogram is a rectangle.
Proof: Let ABCD be the cyclic parallelogram. We know that opposite angles of a parallelogram are equal.
∠A =∠C and ∠B =∠D
Multiple Choice Questions
1. AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is
(a) 17 cm
(b) 15 cm
(c) 4 cm
(d) 8 cm (NCERTExemplar)
2. In the given figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to
D C O
(a) 2 cm (b) 3 cm
(c) 4 cm (d) 5 cm (NCERTExemplar)
3. In the given figure, if ∠ABC = 20°, then ∠AOC is equal to
C O 20°
(a) 20° (b) 40°
(c) 60° (d) 10° (NCERTExemplar)
We know that the sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
∠A +∠C = 180°
Substituting (i) in (ii),
∠A +∠C = 180°
∠A +∠A = 180°
2∠A = 180°
∠A = 90°
We know that if one of the interior angles of a parallelogram is 90°, all the other angles will also be equal to 90°.
Since, all the angles in the parallelogram is 90°, we can say that parallelogram ABCD is a rectangle.
4. In the given figure, two congruent circles have centres O and O′. Arc AXB subtends an angle of 75° at the centre O and arc A′YB′ subt ends an angle of 25° at the centre O′. Then the ratio of arcs AXB and A′YB′ is:
(a) 2 : 1
(b) 1 : 2
(c) 3 : 1
(d) 1 : 3
5. In given figure, AB and CD are two equal chords of a circle with centre OOP and OQ are perpendiculars on chords AB and CD, respectively. If ∠POQ = 150°, then ∠APQ is equal to
AC PQ O 150°
(a) 30º (b) 75º
(c) 15º (d) 60º
6. In the given figure, OD is perpendicular to the chord AB of a circle whose center is O. If BC is a diameter and OD = 4 cm, then CA equals
(a) 6 cm (b) 8 cm
(c) 10 cm (d) 12 cm
7. In the given figure, a line intersect two concentric circle with center O at A, B, C and D. If AB = 10 cm, find CD
10. In the given figure, BC is the diameter of the circle, and ∠BAO = 60°. Then, ∠ADC is equal to
(a) 5 cm (b) 10 cm
(c) 15 cm (d) 20 cm
8. In the given figure, if ∠AOB = 120°, then find the value of ∠ACB.
(a) 100° (b) 110°
(c) 130° (d) 120°
9. In the given figure, if ∠DAB = 60°, ∠ABD = 50°, then ∠ACB is equal to
60° 50°
(a) 60° (b) 50° (c) 70° (d) 80°
(a) 30° (b) 45°
(c) 60° (d) 120°
11. In the given figure, O is the center of the circle, and chord AC and BD intersect at Q such that: ∠AQB = 140° and ∠QBC = 25°. Find the value of ∠ADB
O Q 25° 140°
(a) 105° (b) 140°
(c) 115° (d) 135°
12. In figure, O is the center of the circle. If ∠ACB = 35°, then ∠ABC is equal to 35° C O B A
(a) 35° (b) 55°
(c) 15° (d) None of these
13. In figure, O is the center of the circle, the value of x is O PQ R x 84°
(a) 84° (b) 96°
(c) 42° (d) None of these
14. In figure, PQRS is a cyclic quadrilateral. If ∠QRS = 70°, then ∠SPQ is equal to
70°
(a) 110° (b) 60° (c) 70° (d) None of these
15. In figure, PQ is a chord of a circle with center O and radius 5 cm. If OR ⊥ PQ and OR = 4 cm, then the length of chord PQ is equal to
19. In the given figure, O is the center of the circle. The value of x is:
O x C A 35° 25°
(a) 140° (b) 60° (c) 120° (d) 300°
20. In the figure, if ∠ABC = 110°, ∠ACB = 40°, then ∠BDC is:
(a) 3 cm (b) 5 cm (c) 6 cm (d) 8 cm
16. In figure, BOC is a diameter of the circle and AB = AC, then ∠ABC is equal to
A
(a) 60° (b) 30° (c) 45° (d) None of these
17. If a circle is divided into eight equal parts, the angle subtended by each arc at the center is equal to (a) 90° (b) 60° (c) 45° (d) 30°
18. In figure, if O is the center of the circle, then ∠AOB is
(a) 10° (b) 20° (c) 30° (d) 40°
21. A chord is at a distance of 8 cm from the center of a circle of radius 17 cm. The length of the chord is:
(a) 25 cm (b) 12.5 cm
(c) 30 cm (d) 9 cm
22. In the given figure, O is the center of a circle and ∠BOD = 150°, then the value of x is:
(a) 110° (b) 120° (c) 130° (d) 100°
(a) 105° (b) 115° (c) 100° (d) 110°
23. In the given figure, O is the center of a circle and diameter AB bisects the chord CD at a point P such that CP = PD = 10 cm and PB = 6 cm, then the radius of the circle is
(a) 12.3 cm (b) 11.3 cm
(c) 13 cm (d) 10.3 cm
24. A circle with center O is given in figure.
What is the measure of ∠AOB?
(a) 30° b) 40°
(c) 80° (d) 140°
25. In figure, O is the center of circle, if ∠BAC = x, OP bisects ∠BOC and ∠BOP = y, then the relation between x and y is
A x y
(a) x = 2y (b) x = y
(c) 2 xy = (d) none of these
26. In the given figure, ∠AOB = 90° and ∠ABC = 30°, then ∠CAO is equal to:
27. A circle with centre O is given in the figure. Q O S R y x P
Which of these represents the measure of ∠QRP?
(a) 1 2 x (b) 1 2 y
(c) 2 xy + (d) 2 x y +
28. A quadrilateral ABCD in a circle with centre P is given in the figure. A B P yO Dx C 100°
What are the measures of x and y?
(a) x = 40° and y = 50°
(b) x = 50° and y = 40°
(c) x = 80° and y = 100°
(d) x = 100° and y = 80° (CBSEQuestion)
29. In the given figure, PQ and MN are equal chords of a circle with centre O and diameter 20 cm. 16 cm O P Q N M
What is the distance between the two chords?
(a) 6 cm (b) 10 cm
(c) 12 cm (d) 25 cm (CBSEQuestion)
30. In the given figure, O is the centre of a circle and ∠OAB = 500°, then ∠BOD equals: 50° O D C
(a) 30° (b) 45° (c) 90° (d) 60°
(a) 50° (b) 80° (c) 100° (d) 130°
Answers
1. (d) 8 cm
3. (b) 40°
5. (b) 75°
7. (b) 10 cm
9. (c) 70°
11. (c) 115°
13. (c) 42°
15. (c) 6 cm
2. (b) 3 cm
4. (c) 3 : 1
6. (b) 8 cm
8. (d) 120°
10. (c) 60°
12. (b) 55°
14. (a) 110°
16. (c) 45°
Constructed Response Questions
Very Short Answer Questions
1. In the given figure, if ∠POQ = 80°, then find ∠PAQ and ∠PCQ
17. (c) 45°
19. (c) 120°
21. (c) 30 cm
23. (b) 11.3 cm
80°
Sol. Given, ∠POQ = 80°
∠POQ = 2 ×∠PAQ
(Angle at the center is twice the angle at the circumference.)
⇒ 80° = 2 ×∠PAQ
⇒ 80 40 2 PAQ ° ∠==°
Now, ∠PCQ =∠PAQ (Angle in the same segment are equal)
∠PCQ = 40°
2. If ABCD is a cyclic quadrilateral, in which AD || BC, then prove that ∠B =∠C
Sol. Given: ABCD is a cyclic quadrilateral and AD || BC
∴∠A +∠B = 180° (i)
18. (c) 130°
20. (c) 30°
22. (a) 105°
24. (c) 80°
25. (b) x = y 26. (b) 45°
27. (b) 1 2 y
28. (d) x = 100° and y = 80°
29. (c) 12 cm
30. (c) 100°
(Co-interior angles on the same side of transversal AB)
Also, ∠A +∠C = 180° (ii)
(Sum of opposite angles of a cyclic quadrilateral)
On subtracting equation (ii) from equation (i), we get
∠A +∠B -∠A -∠C = 0
∴∠B =∠C Hence, proved.
3. In the figure, find the value of x and y where O is the center of the circle.
Sol. y = 1 2 × 70° = 35°
(Angle at the center is twice the angle at the circumference.)
Also, ∠x =∠y (Angles in the same segment are equal)
⇒ x = 35°
4. Find x in the given figure.
Sol. Here O is the center of the circle.
∴∠BAC = 1 2 ∠y
(Angle at the center is twice the angle at the circumference)
⇒ 50 = 1 2 ∠y
⇒∠y = 100°
Also, ∠x +∠y = 360°
(Angle at the centre of a circle)
⇒∠x + 100° = 360°
⇒∠x = 360° - 100° = 260°
5. In the given figure, ΔABC is equilateral. Find ∠BDC and ∠BEC.
Sol. ∠BAC = 60° (ΔABC is equilateral triangle)
∴∠BAC =∠BDC (Angles in the same segment of a circle are equal)
⇒∠BDC = 60°
Now, DBEC is a cyclic quadrilateral
∴∠BDC +∠BEC = 180° (opposite angles of a cyclic quadrilateral are supplementary)
60° +∠BEC = 180°
⇒∠BEC = 180° - 60° = 120°
6. In the given figure, find the value of x.
4°
Sol. In a cyclic quadrilateral,
∠A +∠C = 180° (opposite angles of cyclic quadrilateral are supplementary)
2x + 4° + 4x - 64° = 180°
6x - 60° = 180°
6x = 180° + 60° = 240°
240 40 6 x ° ==°
7. ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, Find ∠BCD.
∠DBC +∠BCD +∠CDB = 180°
(Angle sum property of triangle)
55° +∠BCD + 45° = 180°
∠BCD = 80°
8. In the given figure, ∠ABC = 45°, prove that OA ⊥ OC C B A O
Sol. ∠ABC = 1 2
∠AOC
(Angle at the center is twice the angle at the circumference)
i.e., ∠AOC = 2∠ABC = 2 × 45° = 90° or OA ⊥ OC
Hence, proved.
9. In the given figure, ∠BAC = 55° and the altitude BE produced, meet the circle at D. Also, ∠AOC = 100°. B O AC D E
Determine: (a) ∠ABE (b) ∠ABC
Sol. In ΔBAE
∠ABE +∠BAE +∠AEB = 180° (angle sum property)
∠ABE = 180° - 90° - 55° = 35°
Now, ∠ABC = 1 2 ∠AOC = 1 2 × 100° = 50°
10. Find the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm.
Sol. Let AB be a chord of circle with centre O and radius 13 cm. Draw OM ⊥ AB and join OA
In the right triangle OMA, we have
OA 2 = OM 2 + AM 2
⇒ 132 = 122 + AM 2
⇒ AM 2 = 169 - 144 = 25
⇒ AM = 5 cm.
∠BAC =∠BDC = 45° (Angles in the same segment)
Now, in ΔDBC,
cm O AB M
Since the perpendicular from the center to a chord bisects the chord, we have:
AB = 2 × AM = 2 × 5 = 10 cm. 12 cm
11. A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC = 130°.
Find ∠BAC. (NCERTExemplar)
Sol. Draw a quadrilateral ABCD inscribed in a circle having centre O 130° O AB E DC
Given, ∠ADC = 130°
Since, ABCD is a quadrilateral inscribed in a circle, therefore ABCD becomes a cyclic quadrilateral.
Q Since, the sum of opposite angles of a cyclic quadrilateral is 180°.
∴∠ADC +∠ABC = 180°
⇒ 130° +∠ABC = 180°
⇒∠ABC = 50°
Since, AB is a diameter of a circle, then AB subtends an angle to the circle is right angle.
∴∠ACB = 90°
In ΔABC,
∠BAC +∠ACB +∠ABC = 180°
(By angle sum property of a triangle)
⇒∠BAC + 90° + 50° = 180°
⇒∠BAC = 180° - (90° + 50°)
⇒∠BAC = 180° - 140°
⇒∠BAC = 40°
12. In the following figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD +∠BED. AB
Sol. Since, A, C, D and E are four point on a circle, then ACDE is a cyclic quadrilateral.
∠ACD +∠AED = 180° (i)
(Sum of opposite angles in a cyclic quadrilateral is 180°)
Now, ∠AEB = 90° (ii)
We know that, diameter subtends a right angle to the circle.
On adding equations (i) and (ii), we get
(∠ACD +∠AED) +∠AEB = 180° + 90° = 270°
⇒∠ACD +∠BED = 270°
13. In figure, AB and AC are two equal chords of a circle whose centre is O. If OD ⊥ AB and OE ⊥ AC, prove that ΔADE is an isosceles triangle. B O DE C A
Sol. We have,
AB = AC
⇒ OD = OE
(Q Equal chords are equidistant from the centre)
In Δ’s AOD and AOE, we have
OA = OA (Common)
OD = OE (From (i))
So, by R.H.S congruence criterion, we obtain
ΔAOD ≅ΔAOE
⇒ AD = AE
Thus, ΔADE is an isosceles triangle. Hence, proved.
14. If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the center of the circle, then find the radius of the circle.
Sol. Let radius OA = r
∴ 8 2 AB AL == and
OL = 15 cm
Now, OA2 = AL2 + OL2
r 2 = 82 + 152 = 64 + 225 = 289
28917 cm r ==
8 cm 15 cm O
AB L
15. In given figure, two equal chords AB and CD of a circle with centre O, intersect each other at E.
Prove that AD = CB
2. Bisector AD of ∠BAC of ΔABC passes through the center O of the circumcircle of ΔABC, as shown in figure. Prove that AB = AC
Sol.
Sol. We have, Chord AB = Chord CD
⇒ Minor arc AB = Minor arc CD
⇒
ABCD ≅
Subtracting BD from both sides
⇒
ABBDCDBD −≅−
⇒
ADCB ≅
⇒ Chord AD = Chord CB
Short Answer Questions
1. In given figure, chord ED is parallel to the diameter AC of the circle.
Given ∠CBE = 65°, calculate ∠DEC. B AC ED O 65°
Sol. Consider the arc CDE.
We find that ∠CBE and ∠CAE are the angles in the same segment of arc CDE.
∴∠CAE =∠CBE
⇒∠CAE = 65° (Q∠CBE = 65°)
Since, AC is the diameter of the circle and the angle in a semi-circle is a right angle.
Therefore, ∠AEC = 90°.
Using angle sum property in ΔACE, we obtain
∠ACE +∠AEC +∠CAE = 180°
∠ACE + 90° + 65° = 180°
⇒∠ACE = 25°
But, ∠DEC and ∠ACE are alternate angles, because AC || DE
∴∠DEC =∠ACE = 25°
PQ
O A
BDC
Draw OP ⊥ AB and OQ ⊥ AC
In ΔAOP and ΔAOQ
⇒∠OPA =∠OQA = 90° (by construction)
⇒∠OAP =∠OAQ (AD bisect ∠BAC)
⇒ OA = AO (common side)
∴ΔAOP ≅ΔAOQ (By AAS congruence)
⇒ OP = OQ (Corresponding parts of congruent triangles)
∴ AB = AC (Chords equidistant from the center of circle are equal in length)
Hence, proved.
3. In the given figure, ∠OAB = 30° and ∠OCB = 57°.
Find ∠BOC and ∠AOC
30°
AB O 57°
C
Sol. In triangle OAB, ∠OAB = 30°, and OA = OB (radii of the same circle), so it is an isosceles triangle.
Therefore, ∠OBA =∠OAB = 30°
⇒∠AOB = 180° - (∠OAB +∠OBA)
⇒∠AOB = 180° - (30° + 30°) = 120° (i)
Now, in triangle OCB:
∠OCB = 57°, and OC = OB (Radii of same circle)
So, ∠OBC =∠OCB = 57°
⇒∠BOC = 180° - (57° + 57°) = 66° (ii)
Now, to find ∠AOC:
∠AOC =∠AOB -∠BOC
Using (i) and (ii)
⇒∠AOC = 120° - 66° = 54°
4. In figure, ABCD is a cyclic quadrilateral.
If ∠ABC = 100° and ∠ACD = 70°. Find ∠CAD
Sol. ∠B +∠D = 180° (opposite angles of cyclic quadrilateral)
⇒ 100° +∠D = 180°
⇒∠D = 180° - 100° = 80°
In ΔACD,
∴∠CAD +∠ACD +∠ADC = 180° (Angle sum property)
∠CAD + 70° + 80° = 180°
∠CAD = 180° - 150° = 30°
5. In figure, if OA = 10 cm, AB = 16 cm and OD is perpendicular to AB. Find the value of CD 10 cm O AB C D
Sol. As OD is perpendicular to AB
⇒ AC = CB (Perpendicular from the center to the chord bisects the chord)
ACAB
==
∴ 8 cm 2
In right ΔOCA,
OA2 = AC 2 + OC 2
⇒ (10)2 = 82 + OC 2
⇒ OC 2 = 100 - 64
⇒ OC 2 = 36
⇒ OC = 36 = 6 cm
Now, CD = OD - OC = 10 - 6 = 4 cm.
6. In the given figure, AB is a chord equal to the radius of the given circle with center O.
Find the values of a and bB
Sol. OB = OA (Radii of circle)
OA = OB = AB (Given)
∴ΔOAB is an equilateral triangle.
∴∠AOB = 60° (Angles of equilateral Δ are 60° each)
∴ a +∠AOB = 180° (Linear pair)
∴ a + 60° = 180°
∴ a = 120°
Reflex angle BOD = 2∠BCD
(Angle subtended by at center is twice at the circumference)
360° - a = 2b
360° - 120° = 2b
2b = 240°
b = 120°
7. ABCD is a parallelogram. The circle through A, B and C intersect CD at E
Prove that AE = AD.
EDC AB
Sol. ∠ABC +∠AEC = 180° (i)
(Opposite angles of cyclic quadrilateral)
∠ADE +∠ADC = 180° (linear pair)
But ∠ADC =∠ABC
(Opposite angles of parallelogram)
∠ADE +∠ABC = 180° (ii)
From (i) and (ii)
ΔABC +∠AEC =∠ADE +∠ABC
∠AEC =∠ADE
⇒ AD = AE (sides opposite to equal angles)
8. In the given figure, O is the center of the circle and BA = AC. If ∠ABC = 50°.
Find ∠BOC and ∠BDC.
50°
Sol. AB = AC (Given)
∴∠ABC =∠ACB = 50° (AB = AC)
By angle sum property of a triangle
∠BAC = 180° -∠ABC -∠ACB
∠BAC = 180° - 50° - 50° = 80°
∠BOC = 2∠BAC (Angle at center is twice the angle at the circumference)
= 2 × 80° = 160°
∠BDC +∠BAC = 180°
(Opposite angles of cyclic quadrilateral)
∴∠BDC = 180° - 80° = 100°
9. In the given figure, AOB is a diameter of the circle, and C, D, E are any three points on the semi-circle. We need to find the value of ∠ACD +∠BED. AB O E D C
Sol. Join BC, Then, ∠ACB = 90° (Angle in the semi-circle).
Since, DCBE is a cyclic quadrilateral
∠BCD +∠BED = 180°
Add ∠ACB to both sides of the equation:
∠BCD +∠BED +∠ACB = 180° +∠ACB
⇒∠BCD +∠BED +∠ACB = 180° + 90°
⇒ (∠BCD +∠ACB) +∠BED = 270°
⇒∠ACD +∠BED = 270°
10. In the given figure, ∠BAC = 30°. Show that BC is equal to the radius of the circumcircle of ΔABC whose centre is O. O
BC A 30°
Sol. ∠BOC = 2∠BAC
(Angle at center is twice the angle at the circumference)
⇒∠BOC = 2 × 30° = 60°
Also, OC = OB (Radii of the same circle)
∴∠OCB =∠OBC
In ΔOBC, we have
∠OBC +∠OCB +∠BOC = 180°
∠OBC +∠OBC + 60° = 180°
⇒ 2∠OBC = 120°
⇒∠OBC = 60°
So, ∠OBC =∠OCB =∠BOC = 60°
∴ΔBOC is an equilateral triangle.
∴ OB = BC = OC
Thus, BC is equal to the radius of the circumcircle. Hence, proved.
11. The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Sol. QP R S DC A B
In ΔAPD,
∠PAD +∠ADP +∠APD = 180° (i)
Similarly, in ΔBQC,
∠QBC +∠BCQ +∠BQC = 180° (ii)
Adding (i) and (ii), we get
∠PAD +∠ADP +∠APD +∠QBC +∠BCQ +∠BQC
= 180° + 180°
⇒∠PAD +∠ADP +∠QBC +∠BCQ +∠APD +∠BQC
= 360° (iii)
But ∠PAD +∠ADP +∠QBC +∠BCQ
= 1 2 (∠A +∠B +∠C +∠D)
= 1 2 × 360° = 180°
∴∠APD +∠BQC = 360° - 180° = 180° (From (iii))
But these are the sum of opposite angles of quadrilateral PRQS.
∴ Quad. PRQS is a cyclic quadrilateral. Hence, proved.
12. AB is a chord of a circle having centre O. If ∠AOB = 60°, then prove that the chord AB is of radius length.
Sol. Let O be the centre and r be the radius of the circle.
B O rr A 60°
Chord AB subtends ∠AOB = 60° at the centre of the circle.
Here, OB = OA = r (radii of same circle)
⇒∠OAB =∠OBA
i.e., ∠A =∠B (i)
(Q angles opposite to equal sides of a triangle are also equal)
In ΔOAB,
∠O +∠A +∠B = 180° (angle sum property)
⇒ 60° +∠A +∠B = 180°
⇒∠A +∠B = 120°
From equation (i), we get
2∠A = 120°
⇒∠A = 1 2
∠B = 60°
× 120° = 60° and
Thus, ∠O =∠A =∠B = 60°
Hence, ΔOAB is an equilateral triangle.
∴ AB = OA = OB = r
i.e. AB = r
Hence, proved.
13. If O is the centre of a circle and points
A, B, C, D, E, F, G and H are on the circle such that AB = BC = CD = DE = EF = FG = GH = HA.
Find ∠AOB, ∠AOC, ∠DOF and ∠EOH.
Sol. We know that the equal chords subtend equal angles. Given, AB = BC = CD = DE = EF = FG = GH = HA
⇒∠AOB =∠BOC =∠COD =∠DOE
=∠EOF =∠FOG =∠GOH =∠AOH (i)
∴ 8 ×∠AOB = 360°
(Q sum of angles at a point is 360° and from Eq. (i))
14. If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD O CD AB X Y
Sol. Let AXB and CYD be arcs of a circle whose centre and radius are O and r units, respectively.
So, OA = OB = OC = OD = r (i)
∴ arc AXB ≅ arc CYD
∴∠AOB =∠COD (ii)
(Congruent arcs of a circle subtend equal angles at the centre)
In ΔAOB and ΔCOD, AO = CO ( Radii of circle)
BO = DO ( Radii of circle) and ∠AOB =∠COD (From Eq. (ii))
∴ΔAOB ≅ΔCOD (By SAS congruence)
⇒ AB = CD (Corresponding parts of congruent triangles)
⇒ 1 1 AB CD =
15. In given figure, ∠ADC = 130° and chord BC = chord
BE. Find ∠CBE. A D C B E O′O
130° (NCERTExemplar)
Sol. Let O' be centre, O'C and O'E are joined.
In ΔO'CB and ΔO'EB,
BC = BE (Given)
O'C = O'E (Radius)
O'B = O'B (Common)
⇒ΔO'CB ≅ΔO'EB
⇒∠CBO' ≅∠EBO' =∠EB
Also, ABCD is a cyclic quadrilateral.
∴∠ADC +∠CBA = 180° (Opposite angles of cyclic quadrilateral)
⇒ 130° +∠CBA = 180°
⇒∠CBA = 180° - 130° = 50°
⇒∠CBO' = 50°
⇒∠EBO' = 50° (From (i))
Now, ∠CBE =∠CBO' +∠EBO' = 50° + 50° = 100°
16. In the given figure, P is the centre of the circle.
Prove that ∠XPZ = 2(∠XZY +∠YXZ). P XZ Y
Sol. Since, arc XY subtends at the centre and ∠XZY at a point Z in the remaining part of the circle.
∴∠XPY = 2∠XZY (i)
Similarly, arc YZ subtends ∠YPZ at the centre and ∠YXZ at a point X in the remaining part of the circle.
∴∠YPZ = 2∠YXZ (ii)
On adding (i) and (ii), we get
∠XPY +∠YPZ = 2∠XZY + 2∠YXZ
⇒∠XPZ = 2(∠XZY +∠YXZ)
Hence, proved.
Long Answer Questions
1. In the following figure, O is the center of the circle and ∠BCO = 30°. Find x and y. x y OD BEC 30°
Sol. In ΔBOC,
BO = CO (Radii of the circle)
∴∠OCB =∠OBC = 30°
(Equal sides have corresponding equal angles)
∴∠BOC = 180° - (∠OBC +∠OCB)
(Angle sum property of a triangle)
= 180° - (30° + 30°) = 120° (i)
Also, ∠BOC = 2∠BAC
(Angle subtended by at center is twice at the circumference)
∴∠BAC = 120 60 2 ° =°
Also, ∠BAE = 1 2 ∠BAC = 30°
⇒∠BAE = x = 30°
Then, ∠COE =∠BOE = 1 2 ∠BOC = 1 2 × 120° = 60°
Now, ∠DOE = 90°
⇒∠DOC + ∠COE = 90°
⇒∠DOC + 60° = 90°
⇒∠DOC = 30°
We know that,
∠DOC = 2∠DBC (angle subtended by same arc)
⇒∠DBC = 1 2 ∠DOC = 1 2 × 30° = 15° = y
2. In the given figure, AB is a diameter of the circle with center O. If AC and BD are perpendiculars on a line PQ and BD meets the circle at E, then prove that AC = ED.
Sol. ∠AEB = 90° (Angle in semi-circle)
∠AEB +∠AED = 180° (Linear pair)
⇒∠AED = 90°
∠EAC +∠ACD +∠CDE +∠AED = 360°
(sum of angles of a quadrilateral)
∠EAC + 90° + 90° + 90° = 360°
∠EAC = 360° - 270° = 90°
Hence, each angle of quadrilateral is 90°.
∴ EACD is a rectangle.
∴ AC = ED
Hence, proved.
3. PQ and RS are two parallel chords of a circle whose centre is O and radius is 10 cm. If PQ = 16 cm and RS = 12 cm. Then the distance between PQ and RS, if they lie.
(a) on the same side of the centre O and (b) on the opposite of the centre O are respectively
Sol. Given, OP = OR = 10 cm (radii of same circle)
PQ = 16 cm,
RS = 12 cm
Draw OL ⊥ PQ and OM ⊥ RS.
Since, perpendicular from the center to the chord bisects the chord.
∴ PL ⊥ LQ = 1 2 PQ = 8 cm
RM = MS = 1 2 RS = 6 cm
In right triangle OLP,
OP 2 = OL2 + PL2 (by Pythagoras theorem)
100 = OL2 + 64
OL =−==
10064366 cm
In right triangle OMR
OR 2 = OM 2 + RM 2 (by Pythagoras theorem)
100 = OM 2 + 36
10036648 cm OM =−==
(a) If PQ and RS lie on same side of center O. O P RMS Q L 8 8 6 6 10 10
Distance between PQ and RS
LM = OM - OL = 8 - 6 = 2 cm
(b) If PQ and RS lie on opposite sides of center O O RMS PLQ
Distance between PQ and RS = LM = OM + OL = 8 + 6 = 14 cm
4. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Sol. Let, r be the radius of given circle and its centre be O. O
x cm 6 - x cm
Draw OM ⊥ AB and ON ⊥ CD
Since, OM ⊥ AB, ON ⊥ CD and AB || CD
Therefore, points M, O and N are collinear.
So, MN = 6 cm
Let, OM = x cm.
Then, ON = (6 - x) cm.
Join OA and OC.
Then, OA = OC = r
As the perpendicular from the centre to a chord of the circle bisects the chord:
AM = BM = 1 2 AB = 1 2 × 5 = 2.5 cm
CN = DN = 1 2 CD = 1 2 × 11 = 5.5 cm
In right triangles OAM and OCN, we have OA2 = OM 2 + AM 2 and OC 2 = ON 2 + CN 2
r 2 = x2 + 2 5 2
r 2 = (6 - x)2 + 2 11 2
From (i) and (ii), we have x2 + 2 5 2 = (6 - x)2 + 2 11 2
x2 + 25 4 = 36 + x 2 - 12x + 121 4
⇒ 4x2 + 25 = 144 + 4x2 - 48x + 121
⇒ 48x = 240
⇒ x = 240 48
⇒ x = 5
Putting the value of x in equation (i), we get
r 2 = 52 + 2 5 2
⇒ r2 = 25 + 25 4
⇒ r 2 = 125 4 ⇒ r = 55 cm 2
(i)
(ii)
5. AC and BD are chords of a circle that bisect each other. Prove that AC and BD are diameters and ABCD is a rectangle.
Sol. Let AC and BD bisect each other at point O.
Then, OA = OC and OB = OD (i)
In triangles AOB and COD, we have (Given):
OA = OC (From (i))
OB = OD (Given)
∠AOB =∠COD (Vertically opposite angles
∴ΔAOB ≅ΔCOD (SAS congruence criterion)
⇒ AB = CD (By Corresponding parts of congruent triangles)
⇒ ABCD ≅ (ii)
Similarly, BC = DA ⇒ BCDA ≅ (iii)
By adding (ii) and (iii), We have ABBCCDDA +≅+
⇒ ABCCDA ≅
⇒ AC is the diameter of the circle.
Similarly, we can prove that BD is also a diameter of the circle.
Since, AC and BD are diameters of the circle,
∴∠ABC = 90° =∠ADC (Angle in a semicircle is a right angle)
Also, ∠BAD = 90° =∠BCD
AB = CD and BC = DA (Proved above) Thus, ABCD is a rectangle. Hence, proved.
6. AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, prove that 4q 2 = p 2 + 3r2 . (NCERTExemplar)
Sol. Draw OM ⊥ AB and ON ⊥ AC Join OA
In right ΔOAM,
OA2 = OM 2 + AM 2
⇒ r2 = p 2 + 2 1 2 AB
(Q OM ⊥ AB, ∴ OM bisects AB) C
1 4 AB 2 = r 2 - p 2 or
In right ΔOAN, OA 2 = ON 2 + AN 2
⇒ r 2 = q 2 + 2 1 2 AC (Q ON ⊥ AC, ∴ ON bisects AC)
⇒ 1 4 AC 2 = r 2 - q 2
⇒ 1 4 2 1 2 AB = r 2 - q 2 (Q AB = 2AC)
⇒ 1 16 AB 2 = r 2 - q 2
AB 2 = 16r 2 - 16q 2 (ii)
From (i) and (ii), we have 4r 2 - 4p 2 = 16r 2 - 16q 2
Or r 2 - p 2 = 4r 2 - 4q 2
Or 4q 2 = 3r 2 + p 2
Hence, proved.
7. Bisectors of angles A, B, and C of a triangle ABC intersect its circumcircle at D, E, and F respectively.
Prove that angles of triangle ΔDEF are 90°2 A , 90°2 B and 90°2 C . (NCERTExemplar)
O p q
Sol. We have, ∠BED =∠BAD (Angles in the same segment)
∠BAD = 1 2 ∠A (AD is angle bisector)
⇒∠BED = 1 2 ∠A (i)
Also, ∠BEF =∠BCF (Angles in the same segment)
∠BCF = 1 2 ∠C (AD is angle bisector)
⇒∠BEF = 1 2 ∠C (ii)
Adding (i) and (ii),
∠BED +∠BEF =∠DEF = 1 2 ∠A + 1 2 ∠C
⇒∠DEF = 1 2 (180° -∠B) (Q ∠A +∠B +∠C = 180°)
⇒∠DEF = 90°1 2 ∠B
Similarly, ∠DFE = 90°1 2 ∠C and ∠EDF = 90°1 2 ∠A
Hence, proved.
8. If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle circumscribing it at the points P and Q, then prove that PQ is a diameter of the circle.
Q C AB P 2 3 1
Sol. Since, ABCD is a cyclic quadrilateral,
∴∠CDA +∠CBA = 180°
(Opposite angles of cyclic quadrilateral)
Dividing both sides by 2
⇒ 1 2 ∠CDA + 1 2 ∠CBA = 1 2 × 180° = 90°
∠1 = 1 2 ∠CDA, ∠2 = 1 2 ∠CBA
⇒∠1 +∠2 = 90°
But ∠2 =∠3 (Angles in the same segment)
∴∠1 +∠3 = 90°
⇒∠PDQ = 90°
Hence, PQ is a diameter.
(Since, diameter subtends angle 90° in semi-circle) Hence, proved.
9. In a ΔABC, if ∠A = 60° and the altitudes from B and C meet AC and AB at P and Q, respectively and intersect each other at I
(a) Prove that APIQ and PQBC are cyclic quadrilaterals.
(b) Find the measure of ∠BIC. Sol. P
(a) ∴∠A = 60° (Given)
∠APB = 90° [Q PB ⊥ AC] And ∠AQC = 90° [Q CQ ⊥ AB]
In quadrilateral APIQ,
∠P +∠Q +∠A +∠I = 360°
∠P = 90°, ∠Q = 90°, ∠A = 60° (Given)
∴∠I = 360° - (90° + 90° + 60°) = 120° (i)
Now we get, ∠P +∠Q = 180° and ∠A +∠I = 180°
Hence, APIQ is a cyclic quadrilateral.
(Since, sum of either pair of opposite angles is 180°)
Also, ∠BPC =∠BQC (each 90°)
∴ Points P,Q, B and C are concyclic. (Converse of angles in the same segment)
⇒ PQBC is a cyclic quadrilateral.
Hence, proved.
(b) From quadrilateral APIQ,
∠A +∠I = 180°
⇒∠I = 180° -∠A
= 180° - 60° = 120°
∴∠BIC =∠PIQ = 120° (vertically opposite angles)
10. The lengths of two parallel chords of a circle are 6 cm and 8 cm . If the smaller chord is at a distance of 4 cm from the centre, what is the distance of other chord from the centre?
Sol. Let, AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 8 cm. O C A M B D N
Draw OM ⊥ AB and ON ⊥ CD
As AB || CD and OM ⊥ AB, ON ⊥ CD. Therefore, points O, N and M are collinear.
As the perpendicular from the centre of a circle to the chord bisects the chord. Therefore
AM = 1 2 AB = 1 2 × 6 = 3 cm
CN = 1 2 CD = 1 2 × 8 = 4 cm
In right triangle OAM, we have
OA2 = OM 2 + AM 2
OA2 = 42 + 32
⇒ OA2 = 25 ⇒ OA = 5 cm
Also, OA = OC (Radii of the same circle)
⇒ OC = 5 cm
In right triangle OCN, we have
OC 2 = ON 2 + CN 2 ⇒ 52 = ON 2 + 42
⇒ ON 2 = 52 - 42 ⇒ ON 2 = 9
⇒ ON = 3 cm
Competency Based Questions
Multiple Choice Questions
1. In given figure, AB and CD are two chords of a circle intersecting at P. Choose the correct statement from the following:
2. In the given figure, if chords AB and CD of the circle intersect each other at right angles, then x + y =
5. In the given figure, if PQRS is a cyclic quadrilateral with respective angles. Then, the ratio of x and y is ____________.
(a) 45° (b) 60° (c) 75° (d) 90°
3. In the given figure, O is the centre of the circle and CD = DE = EF = FG. If the ∠COD = 40°, then reflex ∠COG is equal to
(a) 1 : 3 (b) 5 : 6
(c) 2 : 3 (d) None of these
6. A, B, C and D are points on the circumference of the circle, centre O. DOB is a straight line and angle DAC = 58°. Angle CDB = D O
C
(a) 32° (b) 58°
(c) 90° (d) 122°
7. In given figure, O and O' are centres of two circles intersecting at B and C. If ACD is a straight line, find x
(a) 160° (b) 200° (c) 240° (d) 280°
4. In the given figure, OA and OB are respectively perpendiculars to chords CD and EF of a circle whose centre is O. If OA = OB, then AB O ED
(a) arc CE = arc DF (b) arc CE > arc DF (c) arc CE < arc DF (d) None of these
(a) 130° (b) 110°
(c) 65° (d) 50°
8. Let C be the mid-point of an arc AB of a circle such that 183. mAB =° If the region bounded by the arc ACB and line segment AB is denoted by S, then the centre O of the circle lies
(a) in the interior of S (b) in the exterior of S (c) on the segment AB (d) on AB and bisects AB
9. In a circle, the major arc is 3 times the minor arc. The corresponding central angles and the degree measures of two arcs are
(a) 90° and 270° (b) 90° and 90°
(c) 270° and 90° (d) 60° and 210°
10. Two circles intersect at two points C and F. Through C, two line segments ACD and BCE are drawn to intersect the circle at A, D, B and E, respectively (see figure). If ∠AFB = 48°, find ∠DFE.
Kajal made two rangoli in front of her house. She started one with a circle of diameter 50 cm and made triangles as shown in figure. In another circle, she drew one isosceles triangle as shown in figure.
(a) 48° (b) 60°
(c) 80° (d) 120°
Assertion-Reason
Based Questions
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): The length of a chord which is at a distance of 5 cm from the center of a circle of radius 10 cm is 17.32 cm.
Reason (R): The perpendicular from the center of a circle to a chord bisects the chord.
2. Assertion (A): Two diameters of a circle intersect each other at right angles. Then, the quadrilateral formed by joining their end-point is a rectangle.
Reason (R): Angle in a semi-circle is a right angle.
3. Assertion (A): If AOB is a diameter of a circle and C is a point on the circle, then AC 2 + BC 2 = AB 2
Reason (R): Angle in a semi-circle is a right angle.
4. Assertion (A): A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm.
Reason (R): Through three collinear points a circle can be drawn.
Case Study Based Questions
1. Read the following and answer any four questions from (a) to (d).
Rangoli hold a significant role in everyday life of a Hindu household. It has different names based on the state and culture. It represents happiness, positivity and liveliness of a household.
(a) While making her first rangoli, Kajal divided the circle in figure (i) into 8 equal parts. What is the angle subtended by each arc at the center?
(b) In her figure (ii), Kajal drew a chord AB of length 40 cm. What is its distance from the center of the circle?
(c) While working on figure (iii), Kajal made PQ as the diameter of the circle. What is the measure of ∠PRQ in this case?
(d) In figure (iii), Kajal designed triangle PQR as an isosceles triangle. What is the measure of ∠RPQ ?
2. Two new roads, Road E and Road F were constructed between Society 4 and 1 and Society 1 and 2, respectively. Society 1
Society 4 Road D Road A RoadF RoadE Road B RoadC P
Society 2 Society 3
Based on the above information, answer the following questions:
(a) What would be the measure of the sum of angles formed by the straight roads at Society 1 and Society 3?
(b) Krish says, ‘The distance to go from Society 4 to Society 2 using Road D will be longer than the distance using Road E ’. Is Krish correct? Justify your answer with examples.
(c) If the length of Road B is equal to Road D, what can you say about the arcs they subtend?
(d) Priya said, ‘Minor arc corresponding to Road B is congruent to minor arc corresponding to Road D. Do you agree with Priya? Give reason to support your answer.’
Answers
Multiple Choice Questions
1. (d) ΔADP ∼ΔCBP
∠APD =∠CPB (Vertically opposite angle)
∠ADP =∠CBP (Angle subtends on the same segment)
∴ AA similarity
ΔADP ∼ΔCBP
2. (d) 90°
∠ACD =∠ABD (Angle in the same segment are equal)
⇒∠ACD = y
Consider the ΔACM in which
∠ACM + x + 90° = 180°
y + x + 90° = 180°
x + y = 90°
3. (b) 200°
O is the center, CD = DE = EF = FG, and
∠COD = 40°. Find reflex ∠COG.
Since all chords are equal, so all central angles are equal. There are 4 equal parts from C to G:
∠COG = 4 × 40° = 160°
Reflex ∠COG = 360° - 160° = 200°
4. (a) arc EC = arc DF A
∠A = 90° (i) AO = BO (ii)
From equation (i) and (ii), we get AMBO is a square.
∴∠EMD = 90° and ∠CHF = 90°
Area (arc ECM) = 360 φ× 2πr (φ = 90°)
Area (arc DMF) = 360 φ× 2πr (φ = 90°)
arc EC = arc DF
5. (b) 5 : 6
∠QPS +∠QRS = 180°
2x + 4x = 180°
x = 30°
∠PSR +∠PQR = 180°
2y + 3y = 180°
y = 36°
6. (a) 32°
∠CAD = 58° (given)
∠CBD =∠CAD (Angles in the same segment/)
⇒∠CBD = 58°
(Angle DCB is the angle in a semicircle)
∠DCB = 90°
And angles in a triangle sum to 180°.
Thus, ∠CDB = (180 - 90 - 58)°
∠CDB = 32°
7. (a) 130°
We have, ∠ACB = 1 2 ∠AOB = 65°
(Angle subtended at centre)
∴∠DCB = 180° -∠ACB = 180° - 65° = 115°
Now, reflex ∠BO'D = 2∠BCD
⇒ 360° - x = 2 × 115°
⇒ x = 130°
8. (b) in the exterior of S
Arc AB = 183°, C is the midpoint. Region S is bounded by arc ACB and chord AB.
Since arc is >180°, midpoint lies opposite to center. So center O lies in exterior of region S
9. (c) 104°
∠ADC and ∠ABC are opposite angles in a cyclic quadrilateral so they add up to 180°.
∠ABC + 128 = 180
∠ABC = 52°
∠AOC is at the centre and ∠ABC is at the circumference, so ∠AOC is twice the size of ∠ABC
Angle AOC = 2 × 52 = 104°
10. (a) 48°
We know that angles in the same segment of a circle are equal.
305
366 x y ==
Thus, 5 : 6
So, we get ∠AFB =∠ACB (i) and ∠EFD =∠ECD (ii)
Also, ∠ACB =∠ECD (iii)
( Vertically opposite angles)
Using (i) and (iii) we get
∠AFB =∠ECD (iv)
Using (iv) and (ii) we get
∠AFB =∠EFD (v)
∠AFB = 48° (Given)
Thus, from (v) we get
∠AFB =∠EFD = 48°
Assertion-Reason Based Questions
1. (a) Both A and R are true, and R is the correct explanation of A.
Let PQ be a chord of a circle with centre O and radius 10 cm. Draw OR ⊥ PQ
cm 10 cm
Now, OP = 10 cm and OR ≈ 5 cm
In right triangle ORP, we get: OP 2 = PR 2 + OR 2
⇒ PR 2 = OP 2 - OR 2
⇒ PR 2 = 102 - 52 = 75
⇒ PR = 75 = 8.66 cm
Since the perpendicular from the centre to a chord bisects the chord.
Therefore, PQ = 2 × PR = 2 × 8.66 = 17.32 cm
So, Assertion is correct. Also, Reason is correct and it is the correct explanation of Assertion.
2. (b) Both A and R are true, but R is not the correct explanation of A.
Let AB and CD be two perpendiculars diameters of a circle with centre O.
∠AOC =∠BOC (each equal to 90°) and
OC = OC (common)
ΔAOC ≅ΔBOC (by SAS congruence rule)
⇒ AC = BC (by Corresponding parts of congruent triangles)
Also, ∠ACB = 90° (Angle in a semi-circle is a right angle)
Similarly, we get, BC = BD and ∠CBD = 90°
DB = DA and ∠BDA = 90°
Hence, ACBD is a square.
It implies that ACBD is a rectangle.
Also, Reason is correct, but it is not necessary that Reason is the correct explanation.
3. (a) Both A and R are true, and R is the correct explanation of A.
Assertion is true. (Theorem: The angle in a semicircle is a right angle).
Reason, we find that
∠ACB = 90°. Applying Pythagoras theorem in ΔACB, we obtain
AB 2 = AC 2 + BC 2
Clearly, Reason is a correct explanation for assertion.
4. (c) A is true, but R is false.
Assertion is true, because a circle of radius 3 cm can be drawn which has AB as its diameter.
Reason is not true because a circle through two points cannot pass through a point which is collinear to these two points.
Case Study Based Questions
1. (a) Angle formed by each arc will be 360 45 8 ° =°
Kajal calculated the angle as 45°
(b) OA = 25 cm and AB = 40 cm
AM = 20 cm
(As altitude divides base of an isosceles triangle in two equal parts)
In ΔOAM, by Pythagoras
theorem OA2 = AM 2 + OM 2
⇒ (25)2 - (20)2
(OM)2 ⇒ 625 - 400
OM 2 = 225
⇒ OM = 15 cm
In Δ's AOC and BOC, we have
OA = OB (O is the mid-point of AB)
Hence, distance of chord from centre is 15 cm. AMB O 40 cm 25 cm 25cm
(c) Since angle in a semicircle is equal to 90°.
∴∠PRQ = 90°
(d) Since PQR is an isosceles triangle,
∴ PR = RQ
⇒∠RPQ =∠RQP = x (Let)
Also, ∠PRQ = 90°
Hence, x + x + 90° = 180°
⇒ 2x + 90° = 180°
⇒ x = 45°
∠RPQ = 45°
2. (a) The sum of opposite angles in a cyclic quadrilateral is equal to 180°.
∠Society 1 +∠Society 3 = 180°
Additional Practice Questions
1. In the given figure, O is the centre of the circle and XOY is a diameter. If XZ is any other chord of the circle, then which of the following is correct?
(b) In a right triangle, the sum of legs is longest for an isosceles triangle when hypotenuse remains same. Therefore, length of Road B + Road D is greater than Road E + Road F. Yes, Krish is correct.
(c) Since Road B and Road D are equal chords, their corresponding minor arcs are also equal.
Conclusion: Minor arc corresponding to Road B is congruent to the minor arc corresponding to Road D
(d) Yes Priya is correct. Road B and Road D are equal and corresponding to them, minor segments are congruent.
4. In the diagram, A, B, C and D are points on the circumference of a circle, centre O.
(a) XY > XZ
(b) XZ < OZ
(c) OX + OZ < XZ
(d) XZ + ZY < XY
2. In a circle, the major arc is 3 times the minor arc. The corresponding central angles and the degree measures of two arcs are
(a) 90° and 270°
(b) 90° and 90°
(c) 270° and 90°
(d) 60° and 210°
3. Use the info given in the figure. What is the size of ∠ACD?
5. What is the value of x?
(a) 62° (b) 95° (c) 85° (d) 118°
6. The diagram shows points D, E, F and G on the circumference of a circle. EG is a diameter. DEFG is a kite. E
What is the value of x ?
7. The diagram shows a circle, centre O. A, B and C are points on the circumference of the circle. Angle AOC is 140°.
11. In figure, O is the center of the circle and ∠BOC = 120°, find ∠CDE. O A
What is the size of ∠ABC ? (a) 40° (b) 110° (c) 120° (d) 140°
8. What is the size of ∠AOD ?
12. In figure, AB is the diameter of the circle with center O. If ∠DAB = 70° and ∠DBC = 30°, determine ∠ABD.
30°
13. In the given figure, O is the center of the circle. Find the value of x O C
(a) 34° (b) 68° (c) 292° (d) 146°
9. In figure, find ∠CAB +∠CBA.
10. In the given figure, determine a, b and c
14. A friendly cricket match is being organized between two teams. The proceeds of this match will be given for the aid to the ‘Charitable Hospital’ for economically weak children. The field is circular with a ring of uniform width which as shown in the figure for spectators. If O is the center of the field and four poles are fixed at points A, B, C and D lying in a straight line. Prove that AB = CD = 1 2 (AD - BC). O
15. Prove that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
16. Two equal chords PQ and RS of a circle with center O, when produced meet at a point M as shown in the figure. Prove that QM = SM
20. In the given figure, O is the centre of a circle and PO bisects ∠APD. Prove that AB = CD.
17. In the given figure if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to
If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, then prove that the chords are equal.
21. In the following figure, equal chords AB and CD of a circle with centre O, cut at right angles at E. If M and N are the mid-points of AB and CD respectively, then prove that OMEN is a square.
18. In the given figure, O is the center of a circle and ∠ADC = 130°. If ∠BAC = x°, then find the value of x.
22. In the following figure, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC, OP ⊥ AB and OQ ⊥ AC, then prove that PB = QC.
19. In the figure, A, B and C are three points on a circle with center O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, Find ∠ADC
23. Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.
Challenge
1. BC is a chord with centre O. A is a point on an arc BC as shown in the given figure. Prove that:
(a) ∠BAC +∠OBC = 90°, if A is the point on the major arc.
(b) ∠BAC -∠OBC = 90°, if A is the point on the minor arc.
Answers
Additional Practice Questions
1. (a) XY > XZ
2. (a) 90° and 270°
3. (b) 54°
4. (c) 250°
5. (d) 118°
6. (c) 25°
7. (b) 110°
8. (b) 68°
9. 90°
2. Prove that the sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to 6 right angles.
3. In the given figure, the points M, R, N, P, S and Q are concyclic. Find ∠PQR +∠OPR +∠NMS +∠OSN, if O is the centre of the circle. O Q M R N P S
10. a = 105°, b = 13°, c = 62°
11. 60°
12. 20°
13. 140°
17. 2 cm
18. 40°
19. 45°
Challenge Yourself
3. 180°
Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 60 mins
Scan me for Solutions
Max. Marks: 40
Multiple Choice Questions [3×1=3Marks]
1. In the given figure, ABCD is a cyclic quadrilateral in which BC is diameter, then find ∠BAC.
(a) 90° (b) 100° (c) 80° (d) 110°
2. In figure, PQ and MN are equal chords of a circle with center O and diameter 20 cm. What is the distance between the two chords?
6
3. In figure, if O is the center of the circle, then the value of x is
(a) 30° (b) 40° (c) 50° (d) 60°
Assertion-Reason Based Questions [2×1=2Marks]
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
4. Assertion (A): In the given figure, ∠PQO = 50° and ∠PRO = 45°. Then the measure of ∠QOR = 190°.
Reason (R): Angle subtended by an arc of a circle at the center of the circle is twice the angle subtended by that arc on the remaining part of circle.
5. Assertion (A): In the given figure, O is the center of the circle and AB is the diameter with ∠APC = 145°, then ∠BQC = 125°.
Reason (R): In a cyclic quadrilateral, opposite angles are supplementary.
Very Short Answer Questions [2 × 2=4Marks]
6. Prove that the perpendicular from the center of a circle to a chord bisects the chord.
7. In the given figure, ∠ACP = 40° and ∠BPD = 120°, then ∠CBD = ___________ .
Short Answer Questions [4 × 3=12Marks]
8. In the given figure, O is the center of the circle and chord AC and BD intersects at P such that ∠APB = 120° and ∠PBC = 15°, find the value of ∠ADP
9. A chord of length 10 cm is at a distance of 12 cm from the center of a circle. Find the radius of the circle.
10. In the given figure, AB and CD are two chords of a circle with center O such that MP = NP. If OM ⊥ AB and ON ⊥ DC, show that AB = CD
11. If O is the circumcentre of a ΔABC and OD ⊥ BC, then prove that ∠BOD =∠BAC.
Long Answer Questions
[3 × 5=15Marks]
12. Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
13. In figure, PQRS is a cyclic quadrilateral, find measure of each of its angles.
14. A circular park of radius 15 cm is situated in a colony. Three boys Aman, Rahul and Daksh are sitting at equal distance on its boundary each having a toy telephone in his hand to talk each other. Find the length of the string of each phone.
Case Study Based Questions
[1 × 4=4Marks]
15. Government of India is working regularly for the upliftment of handicapped persons by making them self dependent. For this, three STD booths to be operated by handicapped persons were installed at A, B and C as shown in the figure. These three booths are equidistant from each other as shown in the figure.
(a) What type of triangle is formed by the positions of booths at points A, B, and C ?
(b) What is the measure of ∠ABC formed at booth B ?
(c) If the distance between booths A and B is 6 cm, what is the sum of the distances BC + CA ?
(d) What is the measure of ∠BOC, the angle formed at the center of the circle connecting the booths?
10 Heron’s Formula
Chapter at a Glance
Area: The area of a shape is the amount of surface it covers. It is measured in square units such as square meters (m2) or square centimeters (cm2).
● For a triangle with sides a, b, c :
Semi-perimeter () 2 abc s ++ =
Area ()()() ssasbsc =−−− (Heron’s formula)
● For a triangle with base (b) and altitude (h):
Area 1 2 bh=××
● For an isosceles triangle, with base a and equal sides b :
Area 22 4 4 aba=−
● For an equilateral triangle with side a:
Altitude 3 a a = Area 2 3 4 a =
NCERT Zone
NCERT Exercise 10.1
1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Sol. Each side of the triangle = a
Perimeter of the triangle = 3a \ 3 2 a s =
\ Area of the signal board (triangle)
()()() ssasbsc =−−−
()()() ssasasa =−−−() abc∴==
()() sassa=−−
Hence, area of the signal board is
Now, perimeter = 180 cm
Each side of the triangle 180 cm 60 cm 3 ==
Area of the triangle
2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of `5000 per m2 per year. A company hired one of its walls for 3 months.
How much rent did it pay?
Sol. Here, we first find the area of the triangular side walls. a = 122 m, b = 120 m and c = 22 m 12212022 m 2 132 m s ++ ∴= = Area of the triangular side wall ()()() ssasbsc =−−− ()()()
2 1321012110 m 1320 m =××× = Rent of 1 m2 of the wall for 1 year = `5000
\ Rent of 1 m2 of the wall for 1 month = ` 5000 12
\ Rent of the complete wall (1320 m2) for 3 months
= ` 5000 13203 12 ××= `16,50,000
3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
Sol. Here, a = 15 m, b = 11 m and c = 6 m
Area of the triangle ()()() ssasbsc =−−−
()()() 2 1616151611166 m =−−− 2 161510 m =×××
2 202 m =
Hence, the area painted in colour 2 202 m =
4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Sol. Here, a = 18 cm, b = 10 cm and c = ?
Perimeter of the triangle = 42 cm
⇒ a + b + c = 42
⇒ 18 + 10 + c = 42
⇒ c = 42 - 28 = 14
Now, 42 22 abc s ++ == cm = 21 cm
Area of the triangle ()()() ssasbsc =−−−
()()() 2 21211821102114 cm =−−−
2 213117 cm =××× 2 733117 cm =×××× 2 7311 cm =×
2 2111 cm =
5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Sol. Let the side of the triangle be 12x cm, 17x cm and 25x cm. Perimeter of the triangle = 540 cm
12x + 17x + 25x = 540
54x = 540
⇒ x = 10
The actual sides are: a = 12x = 12 × 10 = 120 cm
b = 17x = 17 × 10 = 170 cm
c = 25x = 25 × 10 = 250 cm
Semi-perimeter: 2 abc s ++ = 120170250 2 ++ = 540 270 cm 2 ==
Area of the triangle using Heron’s formula: Area ()()() ssasbsc =−−−
()()() 270270120270170270250 =−−−
27015010020=×××
100271520=××
= 100 × 9 × 5 × 2
= 9000 cm2
6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Sol. Here, a = b = 12 cm
Also, a + b + c = 30
⇒ 12 + 12 + c = 30
⇒ c = 30 - 24 = 6
Multiple Choice Questions
1. The sides of a triangle are 56 cm, 60 cm, and 52 cm long. Then the area of the triangle is
(a) 1322 cm2
(b) 1311 cm2
(c) 1344 cm2
(d) 1392 cm2
(NCERTExemplar)
2. An isosceles right triangle has an area of 8 cm2. The length of its hypotenuse is
(a) 32 cm
(b) 16 cm
(c) 48 cm
(d) 24 cm
3. The perimeter of an equilateral triangle is 60 m. Then the area of triangle is
(a) 2 103 m
(b) 2 153 m
(c) 2 203 m
(d) 2 1003 m
(NCERTExemplar)
4. The length of the hypotenuse of an isosceles right triangle with an area of 72 cm2 is (a) 12 cm
(b) 122 cm
(c) 24 cm
(d) 12.5 cm
(NCERTExemplar)
5. A triangular colorful scenery is made on a wall with sides 50 cm, 50 cm, and 80 cm. A golden thread is to hang from the vertex so as to just reach the side 80 cm. How much length of the golden thread is required?
(a) 40 cm (b) 80 cm
(c) 50 cm
(d) 30 cm
6. An umbrella is made by stitching 8 triangular pieces of cloth of two different colors. Each piece measures 20 cm, 50 cm, 50 cm.
\ Area of the triange ()()() ssasbsc =−−−
How much cloth of each color is required?
(a) 2 8006 cm
(b) 2 10006 cm
(c) 2 16006 cm
(d) 2 5006 cm
7. A student is given three sticks of lengths 12 cm, 6 cm, and 4 cm. He is asked to make a triangle and find the area of the triangle formed. The area of the triangle is:
(a) 36 cm2
(b) 12 cm2
(c) 24 cm2
(d) Providing area not possible
8. The base of a right triangle is 16 cm, and the hypotenuse is 34 cm. Then the area of the triangle is:
(a) 30 cm2
(b) 240 cm2
(c) 2400 cm2
(d) 272 cm2
9. The area of an equilateral triangle is 2 163 cm Then half of the perimeter of the triangle is:
(a) 12 cm
(b) 6 cm
(c) 10 cm
(d) 8 cm
10. The cost of leveling the ground in the form of a triangle having sides 51 m, 37 m, 20 m at the rate of `3 per m2 is:
(a) `306
(b) `918
(c) `725
(d) `900
Answers
1. (c)
2. (a)
3. (d)
4. (b)
5. (d)
Constructed Response Questions
Very Short Answer Questions
1. If the side of an equilateral triangle is x units, then find the area of the triangle.
Sol. Given side of equilateral triangle = x As we know, Area of equilateral 2 3 4 a ∆=
Are of equilateral 2 3 unit 4 x ∆=
2. Find the length of each side of an equilateral triangle having an area of 2 93 cm .
Sol. Given, area of the equilateral triangle 2 93 cm = Area 3 4 =× (side)2
2 3
4 ×= (side)2 = 9 × 4 = 36 side 366 cm ==
Hence, the side of the equilateral triangle is 6 cm.
3. Find the area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm.
Sol. Area of an isosceles triangle is: 22 4 4 aba
4. The semi-perimeter of a triangle is 132 cm, and the product of the differences of the semi-perimeter and its respective sides (in cm) is 100 times the semi-perimeter. Find the area of the triangle.
Sol. Given: s = 132 cm (s - a)(s - b)(s - c) = 100 × 132 = 13200
6. (a)
7. (d)
8. (b)
9. (a)
10. (b)
Area of ()()() ssasbsc ∆=−−− 13213200=× = 132 × 10 = 1320 cm2
5. If the area of an equilateral triangle is 2 163 cm , then find the perimeter of the triangle.
Sol. Area of an equilateral triangle is 2 3 4 a = 2 2 3 163 4 64 648 cm a a a = = ⇒==
Thus, the perimeter of the equilateral triangle is: 3a = 3 × 8 = 24 cm
6. An isosceles right triangle has an area of 8 cm2. Find the length of its hypotenuse.
Sol. Area of an isosceles right triangle is 2 2 a = 8
⇒ a = 16 = 4 cm
Hypotenuse 22442 cm a =×=×=
7. If the base of a right-angled triangle is 15 cm and its hypotenuse is 25 cm, then find its area.
Sol. In right-angled ΔABC, Hypotenuse2 = Base2 + Height2
252 = 152 + Height2
Height2 = 625 - 225 = 400
Height 40020 cm ==
Area = 1 2 × Base × Height = 1 2 × 15 × 20 2 300 150 cm 2 ==
The area of the triangle is 150 cm2.
8. The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle. (NCERTExemplar) a a
Sol. Let ABC be an isosceles triangle with perimeter 32 cm. We have, ratio of equal side to its base is 3 : 2.
Let sides of triangle be AB = AC = 3x, BC = 2x
Q Perimeter of a triangle = 32 m
3x + 3x + 2x = 32
8x = 32
⇒ x = 4
AB = AC = 3 × 4 = 12 cm and BC = 2x = 2 × 4 = 8 cm
The sides of a triangle are a = 12 cm, b = 12 cm and c = 8 cm.
ABCssasbsc ∆=−−−
2 abc s ++ = 1212832 16 cm 22 s ++ === ()()() ()()() 2 2 1616121612168 16448 4422 cm 322 cm
=×× = Hence, the area of an isosceles triangle is 2 322 cm
9. The longest side of a right-angled triangle is 125 m, and one of the remaining two sides is 100 m. Find its area using Heron’s formula.
Sol. Using the Pythagorean theorem to find the third side:
Third side ()() 12510022=− 1562510000 5625 75 m =− = = 10075125300 150 cm 22 s ++ ===
Area of ()()()
Short Answer Questions
1. In a rectangular field of dimensions 60 m × 50 m, a triangular park is constructed. If the dimensions of the park are 50 m, 45 m, and 35 m, find the area of the remaining field.
Sol. Area of rectangular field = length × breadth
= 60 × 50 = 3000 m2
Now, a = 50 m, b = 45 m, and c = 35 m
504535130 65 m 222 abc s ++++ ====
Area of triangle ()()() ssasbsc =−−− ()()() 65655065456535 65152030 5532132 15026 =−−− =××× =××××× = = 764.85 m2
Remaining area = Area of rectangle - Area of triangle = 3000 - 764.85 = 2235.15 m2
The area of the remaining field is 2235.15 m2.
2. Find the area of an isosceles triangle whose one side is 10 cm greater than each of its equal sides and the perimeter is 100 cm.
Sol. Let the equal sides of the isosceles triangle be x cm. Thus, the length of the greater side = (x + 10) cm.
Given that the perimeter of the isosceles triangle is 100 cm: x + x + (x + 10) = 100
⇒ 3x = 100 - 10 = 90 90 30 cm 3 x ⇒==
Base = 10 + x = 10 + 30 = 40 cm a = 30, b = 30, c = 40 303040 50 22 abc s ++++ ===
Area of triangle ()()() 50503050305040=−−− 2 50202010 20105 2005 cm =××× =× =
The area of the isosceles triangle is 2 2005 cm .
3. Find the area of the shaded region in the given figure. (All measurements are in cm)
Perimeter of ΔABC = AB + BC + AC = 13 + 14 + 15 = 42 cm
42 21 cm 2 s ==
Area of ΔABC ()()() ssasbsc =−−−
= Area of shaded portion = Area of ΔABC - Area of ΔADB = 84 - 30 = 54 cm2
4. The lengths of the sides of a triangle are 7 cm, 13 cm, and 12 cm. Find the length of the perpendicular from the opposite vertex to the side whose length is 12 cm.
7 cm 12 cm 13 cm
Sol. Let, a = 7 cm, b = 13 cm, c = 12 cm 7131232 16 cm 222 abc s ++++ ====
Heron’s formula, the area of ΔABC is:
Area of ΔABC ()()() ssasbsc =−−−
5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Sol. Let the sides of the triangle in cm be a = 12x, b = 17x, and c = 25x, where x is the constant ratio.
Given that the Perimeter of the triangle is 540 cm, 12x + 17x + 25x = 540
54x = 540
⇒ x = 10 cm
Then, Sides are a = 12(10) = 120 cm, b = 17(10) = 170 cm,
c = 25(10) = 250 cm
Perimeter of the triangle540 270 cm 22 s ===
Heron’s formula, the area of the triangle is:
Area ()()() ssasbsc =−−−
32105 cm 9000 cm s
27015010020 cm
310531010052 cm
Area of triangle = 9000 cm2
Long Answer Questions
1. The perimeter of a right-angled triangle is 12 cm, and its hypotenuse is 5 cm. Find the other two sides and calculate its area. Verify the result using Heron’s formula.
Sol. Let ΔABC be the right-angled triangle.
AB = y cm, BC = x cm, and AC = 5 cm
Perimeter x + y + 5 = 12
⇒ x + y = 7 (1)
Pythagoras Theorem: x2 + y2 = 52 x2 + y2 = 25 (2)
Squaring Equation (1)(x + y)2 = 72
x2 + y2 + 2xy = 49
Using equation (2) we get:
25 + 2xy = 49
⇒ 2xy = 24 xy = 12 (3)
Using (1) and (3) we get:
x(7 - x) = 12
7x - x2 = 12
⇒ x2 - 7x + 12 = 0
x2 - 4x - 3x + 12 = 0
(x - 3)(x - 4) = 0
Thus, x = 3 cm, y = 4 cm (or vice versa).
Area 2 11 346 cm 22 xy =××=××=
Verifying: 34512 6 222 abc s ++++ ====
Area ()()() ssasbsc =−−− ()()() 2
2. A triangular park has sides 60 m, 40 m, and 26 m. A gardener has to put a fence all around its boundary and also plant grass inside. Find the area in which grass will be planted. Also, calculate the cost of fencing it with barbed wire at the rate of `30 per meter, leaving a spare 2 m wide for a gate on one side.
Sol. Let, a = 60 cm, b = 40 cm, c = 26 cm.
Semi-perimeter, 604026126 63 22 s ++ ===
Area of triangular park ()()() ssasbsc =−−− ()()() 2 63636063406326 6332337 401 m =−−− =××× =
Total perimeter = 60 + 40 + 26 = 126 m
Since 2 meters is left for a gate, the total fencing length required is: 126 - 2 = 124 m Cost = 124 × 30 = `3720
3. Two identical circles with same inside design as shown in the figure are to be made at the entrance. The identical triangular leaves are to be painted red and the remaining are to be painted green. Find the total area to be painted red.
Sol. For one identical triangular leaf, let a = 28 cm, b = 15 cm and c = 41 cm
Also, 2 abc s ++ = 281541 2 84 42 cm 2 ++ =
Area of one triangular leaf ()()() ssasbsc =−−− ()()() 2 42422842154241 4214271 3141439 914 126 cm
There are 6 leaves in a circle.
So, total number of leaves in 2 circles = 2 × 6 = 12 \ Area of 12 leaves = (12 × 126) cm2 = 1512 cm2
Hence, total area to be painted red = 1512 cm2
Competency Based Questions
Multiple Choice Questions
1. In a family with two sons, a father has a field in the form of a right-angled triangle with sides 18 m and 40 m. He wants to give independent charge to his sons, so he divided the field in the ratio 2 : 1 : 1. The bigger part he kept for himself and divided the remaining equally among the sons. Find the total area distributed to the sons.
(a) 360 m2 (b) 90 m2
(c) 180 m2 (d) 200 m2
2. The area of an isosceles triangle having base x cm and one side y cm is:
(a)
3. A gardener has to put double fence all round a triangular path with sides 120 m, 80 m and 60 m. In the middle of each sides, there is a gate of width 10 m. Find the length of the wire needed for the fence.
(a) 250 m (b) 490 m
(c) 230 m (d) 460 m
4. If each side of an equilateral triangle is tripled then what is the percentage increase in the area of the triangle?
(a) 300% (b) 600%
(c) 800% (d) 900%
5. A triangular plot has sides in the ratio 3 : 4 : 5 and its perimeter is 72 meters. The owner sold 2 rd 3 of the plot at `1200 per m². How much money did he earn?
(a) `1,62,000 (b) `1,72,800
(c) `2,16,000 (d) `1,80,000
6. In a triangle, the average of any two sides is 6 cm more than half of the third side. The area of the triangle is
(a) 2 723 cm (b) 2 643 cm (c) 2 483 cm (d) 2 363 cm
7. Each side of a triangle is multiplied with the sum of the squares of the other two sides. If the sum of all such possible result is 6 times the product of the sides, then the triangle must be (a) equilateral (b) isosceles (c) scalene (d) right angled
Assertion-Reason Based Questions
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): If the area of an equilateral triangle is 2 813 cm , then semi perimeter of triangle is 20 cm.
Reason (R): Semi-perimeter of a triangle is 2 abc s ++ = where a, b, c are sides of triangle.
2. Assertion (A): Using Heron’s formula, area of an equilateral triangle with side 16 cm is 2 643 cm .
Reason (R): Heron’s formula ()()() ssasbsc =−−− where, 2 abc s ++ = and a, b, c are sides a triangle.
3. Assertion (A): The area of an isosceles triangle each of whose equal side is 13 cm and whose base is 24 cm is 60 cm2.
Reason (R): The area of an isosceles triangle having base a and each equal side b is 22 4. 4 bab
Answers
Multiple Choice Questions
1. (c) 180 m2
Farmer divides the field in the ration 2 : 1 : 1, \ Area of right angular field = 1 2 × base × altitude
= 1 2 × 18 × 40 = 360 m2
Case Study Based Questions
1. Ajay bought some land for carrying out his wholesale business as shown in the figure below. He plans to divide this land into 3 parts for warehouse, inventory, and canteen.
Now using the given information, answer the following questions.
(a) Find the area of Inventory?
(b) Find area allotted for canteen?
(c) Find cost of whole land at the rate of `500 per m2 ?
(d) What fraction of total land is used for the Warehouse?
2. Green cleaning refers to cleaning methods and products that are environmentally friendly ingredients and procedures to preserve human health and environmental quality. To preserve the environment, Ramesh made a slide in the park of his society. One of the side walls of the slide had a message: ‘‘Take Short Walk in Park down a Happy Trail’’ (see figure). 6 m
15 m 11m
Take Short Walk in Park down a Happy Trail
If the sides of the wall are 15 m, 11 m, and 6 m, then answer the questions by looking at the figure.
21.41 =
(a) Find the semi-perimeter of the triangle.
(b) Find the area (in m2) of the wall.
(c) Find the cost of painting the side wall, if the price is `8 per m2.
(d) What is the height of the triangle with base 15 m?
Area distributed to each son = 1 4 (360) = 90 m2
Total area distributed to sons = 2(90) = 180 m2
2. (b) 2 22 cm 24 xx y
Given: Base x cm and Side y cm
Height h using Pythagoras theorem: 2 2 22 24 xxhyy
=−=−
Area = 1 2 × base × height = 1 2 × x × 2 2 4 x y
Area 2 22 cm 24 xx y =−
3. (d) 460 m
a = 120 m,
b = 80 m,
c = 60 m
Length of wire needed
= 2[perimeter of path - 3 (length of gate)]
= 2[120 + 80 + 60 - 3(10)]
= 2[230]
= 460 m
4. (c) 800%
Suppose the original side of the equilateral triangle = a.
Original Area: 2 1 3 4 Aa = (1)
New side = 3a
New Area: () 2 2 2 33 39 44 Aaa ==× (2)
From (1) and (2) we get:
A2 = 9A1
= (8) × 100 = 800%
5. (b) `1,72,800
Since ratio = 3 : 4 : 5, let sides be 3x, 4x and 5x
Thus, 3x + 4x + 5x = 12x = 72 ⇒ x = 6
So, sides are 18, 24, 30 18243072 36 22 s ++ ===
Sold area 2 2 216144 m 3 =×=
Money earned = Sold area × Rate = 144 × 1200 = 1,72,800 rupees
6. (d) 2 363 cm
Let the sides of the triangle be a, b, and c.
As per the question: 6,6 and 6 222222 abcbcacab +++ =+=+=+
a + b = c + 12, b + c = a + 12 and c + a = b + 12
a + b - c = 12 (i)
b + c - a = 12 (ii)
c + a - b = 12 (iii)
Adding (i), (ii) and (iii) we get: (a + b - c) + (b + c - a) + (
⇒ a + b + c = 36
⇒ s = 36 2 = 18
c = a + b - 12 from (i) we get
a + b + (a + b - 12) = 36
⇒ a + b = 24
c = 24 - 12 = 12 cm
Using (ii)
b + 12 - a = 12
⇒ b = a
a + b = 24
⇒ 2a = 24
⇒ a = 12, b = 12
Area ()()() ssasbsc =−−− 2 18666 363 cm =××× =
7. (a) Equilateral
Let the sides of the triangle be a, b, and c.
As per question:
a(b2 + c2) + b(c2 + a2) + c(a2 + b2) = 6abc
⇒ a(b2 + c2 - 2bc) + b(c2 + a2 - 2ac) +
c(a2 + b2 - 2ab) = 0
⇒ a(b - c)2 + b(c - a)2 + c(a - b)2 = 0
⇒ a(b - c) = 0, b(c - a) = 0, c(a - b) = 0
⇒ b = c, c = a, a = b
⇒ a = b = c
So, triangle is equilateral.
Assertion-Reason Based Questions
1. (d) A is false, but R is true.
In case of A:
Area of equilateral triangle 2 3 4 a = 2 3 813 4 a = ⇒ 81 × 4 = a2
⇒ a = 18 cm 181818 27 cm 2 s ++ ⇒== \ A is false.
In case of R: R is correct
2. (a) Both A and R are true, and R is the correct explanation of A.
In case of A: Side = 16 cm 16161648 24 cm 22 s ++ ===
Area ()()() ssasbsc =−−−
()()() 2 24241624162416 24888 643 cm =−−− =××× = A is true.
R is correct, also R is correct explanation of A.
3. (c) A is true, but R is false.
In case of R: R is not true, because the area of an isosceles triangle having base a and each equal side b is 22 4. 4 aba
In case of A: Substituting a = 24 and b = 13, we get
Area 22 24 41324 4 =×− 2 6676576 610 60 cm =− =× = A is correct.
Case Study Based Questions
1. (a) 65516 8 m 22 s ++ ===
Area ()()() ssasbsc =−−− ()()() 2 8868585 8233 12 m =−−− =××× =
(b) 54312 6 m 22 s ++ ===
Area ()()() 6656463 =−−− 2 6123 6 m =××× = (c) 85518 9 22 s ++ ===
Area of warehouse ()()() 9989595 =−−− 2 9144 12 m =××× =
Area of land = Area of canteen + Area of inventory + Area of warehouse
Total area = 12 + 6 + 12 = 30 m2
Cost = 30 × `500 = `15000
(d) Warehouse area = 12 m2
Total land = 30 m2
Fraction = 122 305 =
Fraction used for warehouse 2 5 =
2. (a) 1511632 16 m 22 s ++ ===
(b) Area ()()() ssasbsc =−−− ()()() 2 1616151611166 161510 1650 800 28.2 m =−−− =××× =× = =
(c) Cost = Area × Rate
Cost = 28.2 × 8 = 225.60 rupees
(d) Let’s use area = 1 2 × base × height
We already found area = 28.2 m² and base = 15 m 1 1528.2 2 28.22 15 56.4 3.76 m 15 h h ××= × ⇒= =≈
Height ≈ 3.76 m
Additional Practice Questions
1. The area of an equilateral triangle with side 23 cm is (a) 33 cm2
(b) 23 cm2 (c) 43 cm2
(d) 3 cm2
(NCERTExemplar)
2. The sides of a triangle are 20 cm, 37 cm, and 51 cm long. Then the area of the triangle is: (a) 306 cm2 (b) 612 cm2 (c) 102 cm2 (d) 153 cm2
3. The angles of a triangle are in the ratio 5 : 3 : 7, then the triangle is (a) Acute angled (b) Obtuse angled (c) Right triangle (d) Isosceles triangle
4. The length of each side of an equilateral triangle having an area of 93 cm2 is (a) 8 cm (b) 36 cm (c) 4 cm (d) 6 cm
5. The edges of a triangular board are 12 cm, 17 cm and 25 cm. The cost of painting it one of the surface at the rate of 50 paisa per cm2 is (a) `22.50 (b) `45 (c) `55 (d) `90
6. The sides of a triangle are 13 cm, 14 cm, and 15 cm. If the side lengths are increased by 20%, what is the percentage increase in the area of the triangle? (a) 20% (b) 44% (c) 40% (d) 60%
7. In the triangle ABC, AD is a median. If the area of ΔADC is 15 cm2, then what is the area of ΔABC?
8. Find the sides of an isosceles right triangle of hypotenuse 52 cm.
9. The area of a triangle is 48 cm2. If its base is 12 cm, find its altitude.
10. The perimeter of a triangular field is 450 m and its sides are in the ratio 13 : 12 : 5. Find the area of the triangle.
11. Find the area of ΔABC in which AB = 4 cm, BC = 5 cm and ∠A = 90°.
12. If a kite is to be made in the shape of an isosceles triangle with base 8 cm and each of the equal sides 6 cm, then what is the area of paper required to make the kite?
13. In a scalene triangle, one side exceeds the other two sides by 4 cm and 5 cm respectively. If the perimeter of the triangle is 36 cm, then what is the area of the triangle?
14. Find the area of the quadrilateral ABCD, in which AB = 7 cm, BC = 6 cm, CD = 12 cm, DA = 15 cm and AC = 9 cm.
15. Calculate the area of the shaded region in given figure.
122m
120 m
16. A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Find the area of the parallelogram. (NCERTExemplar)
17. A floral design on a floor is made up of 16 tiles, which are triangular. The sides of the triangle are 9 cm, 28 cm, and 35 cm. Find the cost of polishing the tiles at the rate of 50 paise per cm2.
18. A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown in given figure. The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.
19. The teacher of class IX asked the student to form groups and make 8-models on any of the topics related to Mathematics, as part of their project based learning. Varun and his group prepared the model of a pyramid. Once it was completed they measured the sides of the four triangles and found their length as shown in the figure.
(a) What is the perimeter of ΔAOC ?
(b) What is the value of s, the semi-perimeter of ΔBOC ?
(c) What is the area of ΔABC ?
(d) What is the area of ΔAOB ?
Challenge Yourself Scan me
1. If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides?
2. In the given figure, ABC is a triangle in which CDEFG is a pentagon. Triangles ADE and BFG are equilateral triangles each with side 2 cm and EF = 2 cm. Find the area of the pentagon.
3. The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards) on all its three sides, then the total area of the re-entrant hexagon thus obtained is.
4. Three circles of radius a, b, and c touch each other externally. The area of the triangle formed by joining their centres is:
(a) () abcabc ++
(b) () abcabbcca ++++
(c) ab + bc + ca
(d) None of these
5. In the figure given alongside, ABCDEF is a regular hexagon and ∠AOF = 90°. FO is parallel to ED. What is the ratio of the area of triangle AOF to that of the hexagon ABCDEF ?
(a) () 2 32 q + (b)
2 231 4 q + (c) () 2 431 4 q (d) () 2 532 4 q
(Hint: Add area if all 4 triangles)
Answers
Additional Practice Questions
1. (a) 2 33 cm 2. (a) 306 cm2
3. (a) Acute angled 4. (d) 6 cm
5. (b) `45
`705.60
18. 840 cm2 of paper of each type
(b) 44%
1 12
1 6
1 24
1 18
(Hint: split the figure in 4 triangles and a rectangle)
19. (a) 82 cm (b) 39 cm (c) 386.3 cm2 (d) 300 cm2
and
(a) 1 12
SELF-ASSESSMENT
Time: 60 mins
Multiple Choice Questions
Scan me for Solutions
Max. Marks: 40
[3×1=3Marks]
1. The perimeter of an equilateral triangle is 48 m. Its area is: (a) 2 643 m (b) 2 1443 m (c) 2 163 m (d) 2 2563 m
2. If the base and height of a triangle are doubled, then its area (a) becomes 2 times (b) becomes 4 times (c) is halved (d) does not change
3. The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at `0.09 per cm2 is (a) `2.00 (b) `2.16 (c) `2.48 (d) `3.00
Assertion-Reasoning Based Questions
[2×1=2Marks]
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
4. Assertion (A): The area of an equilateral triangle the length of whose altitude is 6 cm, is 2 123 cm .
Reason (R): The area of an equilateral triangle with altitude p is 2 . 3 p
5. Assertion (A): If the side of a rhombus is 10 cm and one diagonal is 16 cm, the area of the rhombus is 96 cm2. Reason (R): The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm respectively. The area of the parallelogram is 30 cm2.
Very Short Answer Questions [2 × 2=4Marks]
6. Find the area of the triangle given in the figure. 24 cm 26 cm
7. Find the perimeter of an equilateral triangle, if its area is 2 643 cm.
Short Answer Questions [4 × 3=12Marks]
8. The perimeter of an isosceles triangle is 42 cm and its base is 3 2
times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.
9. Find the area of the triangular field of sides 25 m, 60 m and 65 m. Also, find the cost of laying the grass in the triangular field at the rate of `8 per m2.
10. Find the area of the blades of the magnetic compass shown in given figure () Take 113.32. = 1 cm 5 cm
11. Find the area of a triangle whose perimeter is 180 cm and its two sides are 80 cm and 18 cm. Calculate the altitude of triangle corresponding to its shortest side.
Long Answer Questions
12. Find the area of a trapezium whose parallel sides 25 cm, 13 cm and other sides are 15 cm and 15 cm.
[3 × 5=15Marks]
13. OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O. If the radius of the circle is 10 cm, find the area.
14. In figure ΔABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a parallelogram DBCE is constructed. Find the height DF of the parallelogram.
Case Study Based Questions
[1 × 4=4Marks]
15. To beautify parks in a city, city municipal corporation decided to make triangular flower beds in parks as shown in figure. The dimensions of a triangular flower bed are 75 m × 80 m × 85 m. Based on this information answer the following questions:
(a) If each triangular flower bed is to be fenced with two parallel wires one below the other, what is the total length of wire required?
(b) What is the area of a flower bed?
(c) If each triangular bed is an equilateral triangle of side 60 m, what is its area?
(d) If each triangular bed is in the form of an isosceles triangle with base 60 m and equal sides of length 40 m each, what is the area of one flower bed?
11 Surface Areas and Volumes
Chapter at a Glance
Cone
A cone is a solid shape that has a circular base and a curved surface that tapers smoothly to a point called the vertex.
Examples: ice cream cone, birthday cap
Base: The circular, flat surface of the cone.
Height: The perpendicular distance from the vertex to the center of the base (denoted as h).
Slant Height: The diagonal distance from the vertex to any point on the circular edge of the base (denoted as l).
Radius: The distance from the center of the base to its boundary (denoted as r).
Vertex: The point where the curved surface of the cone meets at the top, is called the vertex.
For a cone having height h, radius r, and slant height l:
● Slant Height of Cone (l ): 22lrh =+
● Curved Surface Area of Cone: =π CSA rl
● Total Surface Area of Cone: π()=+ TSA rrl
● Volume of Cone: 2 1 3 Vrh =π
Sphere
A sphere is a perfectly round three-dimensional shape. It has no edges or vertices. For a sphere with a radius r :
● Surface Area of Sphere: 2 4 rπ
● Volume of Sphere: 3 4 3 rπ
Hemisphere
A hemisphere is exactly half of a sphere. It has one curved surface and one flat circular base. Examples: bowl, dome.
For a sphere with a radius r :
● Curved Surface Area of Hemisphere 2 2 r=π
● Total Surface Area of Hemisphere 2 3 r=π
●
NCERT Zone
NCERT Exercise 11.1
1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area
Sol. Here, r = 10.5 2 cm 5.25 = cm, 10 l = cm.
Curved surface area of the cone = rlπ 22 22 5.2510 cm165 cm 7 =××=
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Sol. Here, l = 21 m, 24 m12 m 2 r ==
Total surface area of the cone =() rlrπ+
() 2 22 12 2112 m 7 =××+
22 22 1233 m1244.57m 7 =××=
3. Curved surface area of a cone is 308 2 cm and its slant height is 14 cm. Find
(a) radius of the base and (b) total surface area of the cone.
Sol. Here, 14 l = cm, curved surface area 2 308 cm, ? r ==
(a) Curved surface area of the cone = rlπ 22 30814 7 r ⇒=××
308 7 22 2 r ⇒== ×
Hence, base radius of the cone = 7 cm.
(b) Total surface area of the cone =() rlrπ+
() 222 22 7147cm2221 cm462 cm 7 =×+=×=
4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(a) slant height of the tent.
(b) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is `70.
Sol. Here, 10 h = m, 24 r = m
(a) We have, 222 lhr =+
()() 102422=+
100576676=+=
67626 l ⇒== m
(b) Curved surface area of the tent = rlπ 22 2426 7 =×× m2
Cost of canvas = `70 22 2426 7 ××× = `1,37,280
5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for Stitching margins and wastage in cutting is approximately 20 cm. (use 3.14) π=
Sol. Here h = 6 m, 8 m r =
We have, 222 lhr =+
366410010 m =+==
Curved surface area of the tent = rlπ 2 3.14610 m =××
∴ Required length of tarpaulin
Surface area of cone Extra length
Width =+ () 3.14610 m20 cm 3 ×× =+ 62.8 m0.2 m63 m =+=
6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface at the rate of `210 per 100 m2.
Sol. Here, 25 l = m, 14 m7 2 r == m
Curved surface area of the tomb = rlπ = 22 22 725 m550 m 7 ××=
Cost of white washing 100 2 m 210 = `
∴ cost of white washing 2 210 550 m5501155 100 =×=``
7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Sol. Here, 7 cm r = , 24 h = cm
We have, 22lhr =+ =()2 2 247 + 5764962525 cm =+==
Total curved surface area of 1 cap
= rl π= 22 22 725 cm550 cm 7 ××=
Area of sheet required to make 10 such caps = 10 22 550 cm5500 cm ×=
8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones
is to be painted and the cost of painting is `12 per m2, what will be the cost of painting all these cones?
(Use π= 3.14 and take 1.04 = 1.02)
Sol. Here, r = 40 2 cm = 20 cm = 0.20 m, h = 1 m ()2 22210.2lhr=+=+
1.041.02 m ==
∴ Curved surface area of 1 cone = rlπ
Curved surface area of 50 cones
= 50 × 3.14 × 0.2 × 1.02 m2 = 32.028 m2
Cost of painting an area of 1 m2 = `12
∴ Cost of painting an area of 32.028 m2 = `12 × 32.028 = `384.34
NCERT Exercise 11.2
1. Find the surface area of a sphere of radius: (a) 10.5 cm (b) 5.6 cm (c) 14 cm
Sol. (a) r = 10.5 cm
Surface area of the sphere 2 4 r=π 22
410.510.5 7 =××× cm2 = 1386 cm2
(b) r = 5.6 cm
Surface area of sphere 2 4 r=π 2 22 45.6 × 5. cm 7 6 =××= 394.24 cm2 (c) r = 14 cm
Surface area of the sphere 2 4 r=π 22 41414 7 =×××= 2464 cm2
2. Find the surface area of sphere of a diameter: (a) 14 cm (b) 21 cm (c) 3.5 m
Sol. (a) 14 2 r = cm = 7 cm
Surface area of the sphere = 2 4 rπ 22 22 47 cm 7 =×× = 88 × 7 cm2 = 616 cm2
(b) 21 2 r = cm = 10.5 cm
Surface area of the sphere 2 4 r=π
()2 2 22 410.5 cm 7 =×× = 1386 cm2
(c) 3.5 m1.75m 2 r ==
Surface area of the sphere 2 4 r=π
()2 2 22 41.75 cm 7 =×× = 38.5 m2
3. Find the total surface area of a hemisphere of radius 10 cm. (Use π= 3.14)
Sol. r = 10 cm
Total surface area of the hemisphere 2 3 r=π
()2 2 33.1410 cm =×× 2 33.14100 cm =××= 942 cm2
4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Sol. When r = 7 cm,
Surface area of the balloon 2 4 r=π 477 π =××× cm2
When r = 14 cm,
Surface area of the balloon 2 4 r=π 41414 π =××× cm2
Required ratio of the surface areas of the balloon 4771 414144 π π ××× == ××× = 1:4
5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of `16 per 100 cm2.
Sol. Here, 10.5 2 r = cm 5.25 = cm
Inner surface area of the bowl 2 2 r=π
()2 2 22 25.25cm 7 =×× = 173.25 cm2
Cost of tin plating 100 cm2 = `16
∴ Cost of tin plating 173.25 cm2 = ` 16 100 × 173.25 = `27.72
6. Find the radius of a sphere whose surface area is 154 cm2
Sol. Surface area of the sphere 2 4 r=π
⇒ 2 22 1544 7 r =××
⇒ 2 154777 4224 r ×× == ×
⇒ r = 7 3.5 2 =
Hence, radius of the sphere 3.5 = cm 10.5 cm
7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Sol. Let diameter of the earth = 2r
Then radius of the earth = r
∴ Diameter of the moon 2 42 rr ==
∴ Radius of the moon = 4 r
Now, surface area of the moon 2 4
8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Sol. Inner radius of the bowl (r) = 5 cm
Thickness of the steel = 0.25 cm
∴ Outer radius of the bowl (R) = (5 + 0.25) cm = 5.25 cm
Outer curved surface area of the bowl = 2πR2
9. A right circular cylinder just encloses a sphere of radius r (see figure). Find (a) surface area of the sphere, (b) cur ved surface area of the cylinder, (c) ratio of the areas obtained in (a) and (b).
Sol. Here, radius of the sphere = r radius of the cylinder = r and, height of the cylinder = 2r
(a) Surface area of the sphere 2 4 r=π
(b) Curved surface area of the cylinder 2 rh=π 2 224rrr ππ ××=
(c) Required ratio = 2 2 41 1 4 r r π π =
NCERT Exercise 11.3
1. Find the volume of the right circular cone with (a) Radius 6 cm, height 7 cm
(b) Radius 3.5 cm, height 12 cm
Sol. (a) Here, r = 6 cm, h = 7 cm
Volume of the cone 2 1 3 rh=π 5 cm 0.25 cm
33 122 667 cm264 cm
37 =××××=
(b) Here, r = 3.5 cm, h = 12 cm
Volume of the cone 2 1 3 rh=π
33 122 3.53.512 cm154 cm 37 =××××=
2. Find the capacity in litres of a conical vessel with
(a) Radius 7 cm, slant height 25 cm
(b) Height 12 cm, slant height 13 cm
Sol. (a) Here, 7 cm, 25 cm rl== ∴ 22 6254957624 cm hlr=−=−==
33 122 22554 5512 cm cm 37 7 ××× =××××= 2255411 litres litres 7100035 ××× == ×
3. The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use 3.14) π=
Sol. Here, 15 h = cm, volume 1570 = cm3
Volume of the cone 2 1 3 rh=π 2 1 15703.1415 3 r ⇒=××× 2 1570 3 100 3.14 15 r × ⇒== × 10 r ⇒=
Hence, radius of the base = 10 cm.
4. If the volume of a right circular cone of height 9 cm is 48 3 cm, π find the diameter of its base.
Sol. Here, 3 9 cm, volume48 cm h ==π
Volume of the cone 2 1 3 rh=π 2 1 48 9 3 r ππ ⇒=×× 9 cm ?
2 48 3 16 9 r × ⇒== ×
π 4 r ⇒=
Hence, base diameter of the cone = 24 cm8 cm. ×=
5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Sol. Here, === 3.5 m1.75 m, 12m 2 rh and 1m3 = 1 kl
Capacity of the pit = 2 1 3 rh π = 3 122 1.751.7512 m 37 ××××
3 38.5 m38.5 kl. ==
6. The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find
(a) Height of the cone
(b) Slant height of the cone
(c) Curved surface area of the cone.
Sol. Here, 28 cm14 cm, 2 r == Volume = 9856 cm3
(a) Volume of the cone = 2 1 3 rhπ 122 98561414 37 h ⇒=×××× 985637 48 cm 221414 h ×× ⇒== ××
(b) Slant height ()()22 22 4814 lhr=+=+
2304196250050=+==
Hence, slant height of the cone = 50 cm
(c) Curved surface area of the cone = rlπ 22 22 1450 cm2200 cm 7 =××=
7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Sol. The solid formed is a cone, whose height 12 h = cm, base radius r = 5 cm.
∴ Volume of the cone = 2 1 3 rhπ 3 1 5512 cm 3 =×××× π = 100π cm3
8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in questions 7 and 8.
Sol. Here radius r of the cone = 12 cm and height h of the cone = 5 cm.
∴ Volume of the cone 2 1 3 rh=π
12 × 12 × 5 = 240π cm3
Hence, required ratio = 100 5 5:12 240 12 π π ==
9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Sol. Here, 10.5 m5.25 m, 3 m 2 rh ===
Volume of the heap = 2 1 3 rhπ
= 33 122 5.255.253 m86.625 m 37 ××××=
Now, ()2 22235.25lhr=+=+
927.56256.05 m =+=
Curved surface area of the cone = rlπ 22 22 5.256.05 m99.825 m 7 =××=
Hence, 99.825 m2 of canvas is needed.
NCERT Exercise 11.4
1. Find the volume of sphere whose radius is (a) 7 cm (b) 0.63 m
Sol. (a) r = 7 cm
Volume of the sphere = 3 4 3 rπ = 3 422 777 cm 37 ×××× 3 1 1437 cm 3 =
(b) Here, 0.63 r = m
Volume of the sphere = 3 4 3 rπ =()3 3 422 0.63 m 37 ×× 1.05 = 3 m (approx.)
2. Find the amount of water displaced by a solid spherical ball of diameter
(a) 28 cm (b) 0.21 m
Sol. (a) Here, 28 2 r = cm 7 = cm
Volume of water displaced by the spherical ball
(b) Here, 0.21 m0.105 m 2 r ==
Volume of water displaced by spherical ball = 3 4 3 rπ
3 422 0.1050.1050.105 m 37 ×××× = 0.004851 m3
3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3 ?
Sol. 4.2 cm2.1 cm 2 r == Volume of the ball 3 4 3 r=π
3 422 2.12.12.1 cm
×××× = 38.808 cm3
Density of the metal = 8.9 g/cm3
∴ Mass of the ball = 8.9 × 38.808 g = 345.39 g (approx.)
4. The diameter of the moon is approximately one fourth of the diameter of earth. What fraction of the volume of the earth is the volume of the moon?
Sol. Let the diameter of the earth be 2r. The radius of the earth = r
diameter of the moon
Radius of the moon 4
5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Hence, the hemispherical bowl can hold 303 1000 litres = 0.303 litres of milk.
6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of iron used to make the tank.
Sol. Here, the inner radius of the tank () 1 r = m
Thickness of the iron sheet = 1 cm = 0.01 m
∴External radius of the tank ()() 1 0.01m1.01m R =+=
Volume of the iron used to make the tank = Volume of Outer hemisphere – Volume of inner hemisphere () 33 2 3 Rr=− π () () 3 33 222 1.011m 37 =××− 3 222 0.030301 m 37 =×× = 0.06348 m3
7. Find the volume of a sphere whose surface area is 154 cm2
Sol. Here, 2 4154 r π= ⇒ 22 22 4154 7 rr=××= ⇒ 2 154 7 49 4224 r × == × ⇒ 7 3.5 cm 2 r ==
∴Volume of the sphere 3 4 3 r=π = 33 422 3.53.53. 1 5 cm cm 3 7 7 79.6 ××××=
8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of `498.96. If the cost of white-washing is `2.00 per square metre, find the (a) inside surface area of the dome, (b) Volume of the air inside the dome.
Sol. (a) Inner surface of the dome = 2 Total cost
Cost of white washing per m 22 498.96 m249.48 m 2 ==
(b) Let the radius of the dome be r m. then, 2 2249.48 r π= 2 22 2249.48 7 r ⇒××=
2 249.487 39.69 222 r × ⇒== × 6.3 cm r = ∴Volume of the air inside the dome 3
3 r=π = 33 222 6.36.36.3 m523.9 m 37 ××××=
9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S '. Find the (a) radius r ' of the new sphere. (b) ratio of S and S '
(b) Surface area (S ) of the sphere with radius 2 4 rr =π
Surface area (S') of the sphere with radius 2 4 rr = ′′ π ()2 2 4336rr ππ== 2 2 41 9 36 Sr Sr π π ∴= ′ == 1:9
10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3 ) is needed to fill this capsule?
3.5 mm
Sol. (a) Volume of a sphere of radius 3
3 rr =π
Volume of 27 such spheres = 33 4 2736 3 rr ππ×=
Multiple Choice Questions
1. The radius of a cone is 3 cm and vertical height is 4 cm. Then, the area of the curved surface is (a) 47.14 cm² (b) 42.15 cm² (c) 41.74 cm² (d) 44.17 cm²
2. The diameter of a cone whose height and slant height are respectively 12 cm and 20 cm is (a) 28 cm (b) 32 cm (c) 36 cm (d) 42 cm
3. The radius and slant height of a cone are in the ratio of 4:7. If its curved surface area is 792 cm², then its radius is (a) 12 cm (b) 3 cm (c) 24 cm (d) 6 cm
4. The total surface area of a cone whose radius is 2 r and slant height 2l is (a) () 2 rlrπ+ (b) 4 rlrπ
(c)
Sol. Here, r = 3.5 mm1.75mm 2 = Volume of the capsule 3 4 3 r=π = 3 422 1.751.751.75 mm 37 ×××× = 22.46 mm3
Hence, 22.46 mm3 of medicine is needed to fill the capsule.
5. If the volume of a right circular cone of height 9 cm is 3 27cm π , then the diameter of its base is (a) 12 cm (b) 9 cm (c) 6 cm (d) 5.4 cm
6. The ratio of surface area and volume of the sphere of unit radius is (a) 3:4 (b) 4:3 (c) 3:1 (d) 1:3
7. If the volume and the surface area of a sphere are numerically the same, then its radius is (a) 1 unit (b) 2 units (c) 3 units (d) 4 units
8. The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratios of the surface areas of the balloon in the two cases is (a) 1:4 (b) 1:3
(c) 2:3 (d) 2:1
9. Volume of a hemisphere is 19404 cm3. The total surface area is:
(a) 2772 cm2 (b) 4158 cm2
(c) 5544 cm2 (d) 1346 cm2
10. A spherical lead ball of radius 8 cm is melted and small lead balls of radius 16 mm are made. The total number of possible small lead balls is:
(a) 250 (b) 125
(c) 400 (d) 1250
11. The capacity in litres of a conical vessel with radius 7 cm and slant height 25 cm is:
(a) 2.464 L (b) 3.396 L
(c) 1.232 L (d) 0.2464 L
12. The area of sheet required to cover a hemispherical bowl of radius 20 cm is:
(a) 800π cm2 (b) 1200π cm2
(c) 1600π cm2 (d) 800π cm2
13. The surface area of a spherical ball is 5544 cm2. Its diameter is:
(a) 21 cm (b) 21 cm 2
(c) 42 cm (d) 63 cm
14. If the radius of a sphere is doubled, then its surface area will be:
(a) 2 times (b) 3 times (c) 4 times (d) 1.5 times
15. A hemispherical bowl is made of steel, 0.25 cm thick. If the inner radius of the bowl is 3.75 cm, then the outer surface area of the bowl is:
Answers
1. (a) 47.14 cm² 2. (b) 32 cm
3. (a) 12 cm 4. (b) 4 rlrπ
5. (c) 6 cm 6. (c) 3:1
7. (c) 3 units 8. (a) 1:4
(a) 100.57
16. The volume of a sphere whose surface area is 616 cm2 is: (a) 1179.67 cm3 (b) 1473.33 cm3 (c) 1437.33 cm3 (d) 1237.33 cm3
17. If the radius of a sphere is 2r, then its volume will be: (a) 3 4 3 rπ (b) 3 4 r
18. In a right circular cone, the cross section made by a plane parallel to the base is a (a) Sphere (b) Hemisphere
(c) Circle (d) Semicircle
19. The ratio of the radii of two spheres whose volumes are in the ratio 64:27 is (a) it is 8:3. (b) it is 16:9. (c) it is 10:7. (d) it is 4:3.
20. The curved surface area of a cone whose radius is one-third of its height is:
1. Find the radius of a sphere whose surface area is 616 cm².
Sol. 2 Surface Area4 r=π 2 4616 r π= 2 616 154 4 r π== 2 1547 22 49 22 7 r × === π 497cm r ⇒==
The radius of the sphere is 7cm.
2. A cone and a hemisphere stand on equal bases and have the same height. Find the ratio of their volumes.
Sol. Volume of the Cone
of the Hemisphere
= 1:2
The ratio of the volumes of the cone and the hemisphere is 1:2
3. The curved surface area of a conical vessel is 10 times its slant height. Find the diameter of the vessel.
Sol. Let the slant height of the conical vessel be l units. CSA rl=π 10l = 10 r π = Diameter, D = 2r 1020 2 D ππ
4. Cheery took a spherical orange and wrapped a thread around its middle boundary. She measured the length of the thread to be 22 cm. What is the diameter of the orange?
Sol. The circumference of the spherical orange is equal to the length of the thread: 222cm r π=
The diameter D of the orange is twice the radius: 223.57cmDr
The diameter of the orange is 7 cm.
5. Find the volume of a hemispherical bowl with a radius of 9 cm and express the volume in litres.
Sol. Given 9 cm r =
litres: 3 1 1cmlitres 1000 = 1527.43
Volume in litres 100 lit 0 res =
Volume in litres = 1.527 litres
6. A semi-circular sheet of metal of diameter 28 cm is bent into an open conical cup. Find the depth and capacity of cup.
Sol. When sheet is bent into conical cup, the radius of the sheet becomes the slant height of the cup and the circumference of the sheet becomes the circumference of the base of the cone.
∴ l = Slant height of the conical cup = 14 cm
Let r cm be the radius and h cm the height (depth) of the conical cup. Then,
Circumference of the base of the conical cup = Circumference of the sheet
214 r ⇒=× ππ
⇒ r = 7 cm
Now, 222 lrh =+ 222214773 cm hlr ⇒=−=−= () 71.732cm12.12 cm =×=
∴ Depth of the cup 12.12 cm =
∴ Capacity of the cup = Volume of the cup
cm
7. The radius of a planet is one-fourth the radius of Earth. What fraction of Earth’s volume is the volume of this planet?
Sol. Earth’s radius = R Planet’s radius = 4 R Volume of a sphere 3
2. Find the volume of a sphere whose surface area is 154 cm².
Sol. Let the radius of sphere be r. Therefore, surface area 2
From (i) we get 1 64 PE VV =×
Volume of the planet is 1 64 of Earth’s volume.
Short Answer Questions
1. Find the curved surface area and volume of a cone with the given height is 16 cm and base radius is 12 cm. (Use π= 3.14)
Sol. l = slant height
3. Two hemispherical domes are to be painted as shown in the given figure. If the circumferences of the bases of the domes are 17.6 cm and 70.4 cm respectively, then find the cost of painting at the rate of `10 per cm².
Sol. Let the radii of the bases of the two domes be r and R 2 Cr =π For the first dome: 1 1 17.617.67 2.8cm 22 2 44 2 7 C r × ==== × π For the second dome: 2 70.470.47 11.2cm 22 2 44 2 7 C R × ==== × π The surface area of a hemisphere is given by: 2 2 Ar =π For the first dome:
For the second dome: 22 22 22 22(11.2) 7 Ar π ==××
6. How many litres of milk can a hemispherical bowl of diameter 10 cm hold?
(Use π= 3.14)
Sol. Radius of hemispherical bowl: 10 5cm. 2 r ==
Therefore, the cost of painting the two hemispherical domes is `8377.6.
4. Determine how many full bags of wheat, each occupying 1.96 m³, can be emptied into a conical tent with given dimensions.
Radius of the conical tent: 8.4m r =
Height of the conical tent: 3.5m h =
Volume of each bag of wheat: 3 1.96m
Sol. 2 1 3 Vrh =π 2
Number of bags132 =
5. A shot put is a solid metallic sphere with a radius of 4.9 cm.
(a) Find the volume of the shot put.
(b) Find the mass of the shot put if its density is 7.8 g/cm3
Amount of milk that the hemispherical bowl can hold:
Convert cm3 to litres 3 (1litre1000cm = )
== 261. Volume in litres litres. 1000 67 0.262
7. The surface area of a sphere of radius 5 cm is five times the curved surface area of a cone of radius 4 cm. Find the height and volume of the cone.
Sol. 222 1 Surface area of sphere44(5)100cm r ===πππ 2 2 Curved surface area of cone4cm rll==ππ () 100545cm ll =⇒=ππ 222 2 lrh =+ 222 22 542516 93cm.hhhh =+⇒=+⇒=⇒=
Volume of the Cone 2 2 1 3 Vrh =π ()()() 2 11 (4)316316 33 V ===πππ 3 22352 16 50.29cm 77 V =×==
8. The total cost of making a spherical ball is `33,957 at the rate of `7 per cubic metre. What will be the radius of this ball? 22 Use 7 = π
Sol. Let the radius of the ball be r metres.
Volume of a Sphere 3 4 3 Vr =π
The cost per cubic metre is `7
Total Cost = Volume × Cost per cubic metre 33,957 = Volume × 7 3 33,957 Volume 4851 m 7 == 10 cm
48513714553 1158.83 42212.56 ×× ==≈ × 3 1158.8310.5 m r ⇒=≈
The radius of the spherical ball is approximately 10.5 metres.
Long Answer Questions
1. A conical tent has the area of its base as 154 cm² and its curved surface area as 550 cm². Find the volume of the tent.
Sol. Area of the base of a cone is: 2 154 r π= 22 22 1547 154 49 7 22 rr × =⇒==
497cm. r ⇒==
Curved surface area of a cone is: CSA rl=π 22 5507 7 l =×× 550 55022 25cm. 22 ll =⇒==
222 lrh =+
252 = 72 + h2 ⇒ 625 = 49 + h2
⇒ h2 = 625 49 = 576 ⇒ h 57624cm.==
Volume of a cone is: 2 1 3 Vrh =π 2 122122 7244924 3737 V =×××=××× 1241 224922168 373 V =×××=×× 3 3696 1232cm. 3 V ==
2. A hemispherical dome, open at base is made of sheet of fiber. If the diameter of hemispherical dome is 80 cm and 13 170 of sheet actually used was wasted in making the dome, then find the cost of dome at the rate of `35 per 100 cm².
Sol. Curved surface area of a hemisphere 2 2 r=π r = 40
23.14(40)2 CSA =××=10,048 cm². 13 170 of the sheet is wasted, 13157
Usable sheet1 170170 =−= 80 cm
157 Sheet usedCSA 170 =×
Assume that the area of fibre sheet used is equal to x.
Actual sheet used is approximately 10,880 cm².
The cost is given as `35 per 100 cm²,
Actual Sheet Used Total Cost 35
=× ` 1088035 3808 100 × ==``
The total cost of making the hemispherical dome is approximately `3,808.
3. If h, c, V are respectively the height, the curved surface and the volume of a cone, prove that: 3222 390 VhChV π−+= .
Sol. Let r and l denote respectively the radius of the base and slant height of the cone. 222 1
Then, , and . 3 lrhVrhCrl ππ=+==
393222 VhChV∴−+ π
23 1 3 3 rhh=×× ππ 2 22 2 1 ()9 3 rlhrh ππ
2242222242 rhrlhrhπππ =−+
22422222242222 rhrhrhrhlrh πππ =−++=+
224242224242 0 rhrhrhrh ππππ =−−+=
4. The solid spheres made of the same metal have weights 5920 gm and 740 gm respectively. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.
(NCERTExemplar)
Sol. It is given that the radius of the smaller sphere is 5 cm2.5 cm 2 = .
Let r be the radius of the larger sphere.
Let 1V and 2 V be the volumes of the larger and the smaller spheres respectively.
Then, 3 3 12 445 and 332 VrVππ
Let gm x /cm3 be the density of the metal. Then, 12 5920 and 740 VxVx==
33 8 8125 125 rr ⇒=⇒=
⇒ r = 5
Hence, the radius of the larger sphere is 5 cm.
5. A cone of height 24 cm has a curved surface area 550 cm2. Find its volume.
Sol. Height of the cone () 24 cm h =
Let r cm be the radius of the base and cm l be the slant height of the cone. Then, 2222224576lrhrr =+=+=+
Squaring both the sides we get ()22 57630625 rr += 222576306250rrr+−=
Let, 2 rx = 2 576306250xx+−= 2 62549306250xxx ⇒+−−= ()()625496250xxx ⇒+−+= r 24 cm l
Competency Based Questions
Multiple Choice Questions
1. A semicircular thin sheet of a metal of diameter 28 cm is bent and an open conical cup is made. The capacity of the cup is
()()625490xx ⇒+−= 6250,490xx ⇒+=−= () 625 rejected, 49 xx ⇒=−= 2 4949rr ∴=⇒= ⇒ r = 7 cm 112222
Volume of the cone 724 337 rh∴π==××× 3 22 Volume of the cone 49241232cm 21 =××=
6. The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Sol. Let the original diameter of the sphere be 2x.
Then, original radius of the sphere = x
Original curved surface area 2 4 x=π
Decreased diameter of the sphere 225 x =−% of 3 22 22 x xxx =−=
Decreased radius of the sphere 3 4 x = ∴ Decreased curved surface area 2 2 39 4 44 xxππ
Decrease in area 222 97 4 44 xxx πππ=−=
Hence, percentage decrease in area: Decrease 100 Original ×
2. A hemispherical bowl is 176 cm round the brim. Supposing it to be half full, how many persons may be served from it in hemispherical glasses 4 cm in diameter at the top?
(a) 1372 (b) 1272 (c) 1172 (d) 1472
3. The curved surface area of one cone is twice that of the other while the slant height of the later is twice that of the former. The ratio of their radii is (a) 2:1 (b) 4:1 (c) 8:1 (d) 1:1
4. The slant height of the largest possible cone that can be inserted in a hemisphere of volume 3 144 cm π , is
(a) 92 cm (b) 122 cm (c) 72 cm (d) 62 cm
5. The diagram shows the cross section of six identical marbles touching each other on a horizontal surface.
PQ
If the volume of a marble is 3 9 cm 2 π , calculate the length of PQ , in cm.
(a) 9 (b) 27 (c) 18 (d) 22
6. A sphere and a right circular cone of same radius have equal volumes. By what percentage does the height of the cone exceed its diameter?
(a) 50% (b) 100% (c) 150% (d) 200%
7. A largest sphere of radius 3.5 cm is carved out from a cubical solid. The difference between their surface areas is
(a) 2 224 cm
(b) 2 140 cm
(c) 2 176 cm
(d) 2 80.5 cm
8. A sphere and a cone have equal bases. If their heights are also equal, the ratio of their curved surface will be:
(a) 1:3 (b) 4:5 (c) 4:1 (d) 5:1
9. A spherical iron shell with external diameter 21 cm weighs 5 22775 21 grams. Find the thickness of the shell if the metal weighs 10 gms per cm3.
(a) 3 cm (b) 1 cm (c) 2 cm (d) 2.5 cm
10. The diagram shows a composite solid consisting of a cone and a hemisphere. Calculate the volume of the whole solid. ( 3.142 π= )
(a) 374.94 mm³
(b) 356.88 mm³
(c) 392.10 mm³
(d) 341.56 mm³
Assertion-Reason Based Questions
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): A cylinder and a right circular cone have the same base and same height.
If the volume of the cone is 25 cubic units, then the volume of the cylinder is 75 cubic units.
Reason (R): When a cylinder and a right circular cone have the same base and same height, then the volume of the cylinder is three times the volume of the cone.
2. Assertion (A): If the surface areas of two spheres are in the ratio 9:25, then their radii are in the ratio 3:5.
Reason (R): If surface areas of two spheres are in the ratio S 1: S 2, then their radii are in the ratio 12 : SS
Case Study Based Questions
1. Arnavi is fond of playing with clay. It develops coordination and motor skills and builds imagination. She made a big ball with all the clay available with her.
(a) What is the volume of the clay she had if the radius of the ball formed is 9 cm?
(b) Arnavi can reshape the clay into small balls, each of radius 1.5 cm. Find the number of balls formed?
(c) Find the surface area of each of the small balls?
2. A vendor sells ice cream in a cone-shaped wafer with a hemispherical scoop of ice cream on top. The radius of the cone and scoop is 3.5 cm, and the height of the cone is 10 cm.
Answer these question based on above information:
(a) What is the slant height of the cone used in the above ice cream?
(b) What is the total surface area of the ice cream cone?
(c) What is the total volume of ice cream, including the cone and the scoop?
(d) If the cost of 1 cm³ of ice cream is `0.50, what is the total cost of one filled ice cream (Round to nearest whole number)?
Answers
Multiple Choice Questions
1. (c) 10783 3 S QR R P S Q P 14 cm 14 cm
The radius () r of the semicircle 14 cm = Also, for the conical cup (slant height) 14 cm lr==
Let R be the radius of the base of the conical cup. Then,
Circumference of the base of the cone = Circumference of the semicircle 14 22 7 cm 22 RrRrRrππ ⇒=⇒=⇒===
Let h be the vertical height of the cone. Then, l 2 = R 2 + h2 ⇒ h2 = l 2 R 2 = (14)2 72 = 196 49 = 147 14773 cm h ⇒== 2 1 Capacity of the cup 3 Rh∴=π 122 7773 37 =×××× cm3 3 1078 3 cm 3 =
2. (a) 1372
Given, perimeter of the circular base of the cup = 176 cm. If r is the radius of the hemispherical bowl, then 2176 r π= 1767 28 cm 222 r × ⇒== ×
Total number of persons who can be served 1 Volume of hemispherical bowl 2 Volume of one hemispherical glass × = () () 3 3 12 28 2744 23 1372 2 2 2
3. (b)
Base
It is given that ()1211221121 2222 SSrlrlrlrl ππ =⇒=⇒= [] 21 2 (given) ll = 112212 24 rlrlrr ⇒=⇒=⇒ ππ r1:r2 = 4:1
4. (d) 62 cm
Let r be the radius of the hemisphere. It is given that Volume of hemisphere 3 144 cm =π The base of the cone should be same as the circular base of the hemisphere. Therefore, radius of the base of the cone is r and its height is also r 22 Slant height 262 cm rrr ∴=+==
5. (c) 18 9 Volume of each marble 2 π = 3 49 32 r π π= 3393273 2482 rrr π π =×⇒=⇒= cm
Diameter of each marble 3 23 cm 2 =×=
∴=×=×= 6diameter of each marble 6318cm PQ
6. (b) 100%
Let r be the radius of the cone and sphere. Let h be the height of the cone. Then, according to the question, 4132 4 33 rrhhrππ=⇒=
7. (b) 2 140 cm
s ∴= Length of an edge of the cubical solid = 2r = 7 cm. Surface area of cube: 22 2 667649294cm s =×=×=
We have, r = radius of the sphere 3.5 cm =
Surface area of sphere = 22 22 44(3.5) 7 r π=×× 2 221078 412.25154cm 77 ××==
Difference = 2 294154140cm −=
Curved
8. (b) 4:5
Let radius of cone and hemisphere. r = () Then, height of cone 2 hr = () 222 Slant height of cone 55lhrrr ∴=+==
∴Required ratio 2 4:rrl=ππ = 4πr2:π× r × r = 5 4:5. =
9. (c) 2 cm
Let the internal radius of the shell be r cm.
∴ Internal volume of the shell 33 4 cm 3 r=π
External radius of the shell 21 cm 2 = .
External volume of the shell
Weight of 3 1cm of metal 10 g = .
Volume of the metal in the shell 3 51 22775cm 2110 =×
=×=
∴ Internal vol. of the shell 3 4782854043 4851 cm 2121 =−=
Thickness of shell 10.5
10. (a) 374.94 mm³
Volume of cone 2 11 33 rh ==× π 3.142
Assertion-Reason Based Questions
1. (a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Let r be the radius of the common base and h be the height of the cylinder as well as that of cone. Then,
the cylinder , Vrh =π
Clearly, 12 3 VV = So, reason is true.
Replacing 2 25 V = in 12 3 VV = , we obtain 1 75 V = cubic units.
So, assertion is true. Clearly, reason is a correct explanation for assertion.
2. (a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Let 12 , rr be the radii of two spheres. Then,
So, Reason is true.
Replacing 1S by 9 and 2 S by 25 in (i), we obtain 1 2 3 5 r r = . So, Assertion is also true.
Also, Reason is a correct explanation for Assertion.
Case Study Based Questions
1.
Additional Practice Questions
1. If a cone is cut into two parts by a horizontal plane passing through the mid-point of its axis, the ratio of the volumes of upper and lower part is (a) 1:2
(b) 2:1
(c) 1:7
(d) 1:8
2. The surface area of a sphere is 2 616 cm . Its radius is (a) 7 cm (b) 14 cm (c) 7 cm 2 (d) 21 cm
3. The volume of a hemisphere is 3 19404 cm . The total surface area of the hemisphere is (a) 2 16632 cm (b) 2 3696 cm (c) 2 4158 cm (d) 2 8316 cm
4. If , hS and V denote respectively the height, curved surface area and volume of a right circular cone, then 393222 VhShV π−+ is equal to (a) 8 (b) 0 (c) 4π (d) 2 32π
5. The slant height of a cone is increased by 10%. If the radius remains the same, the curved surface area is increased by (a) 10% (b) 12.1%
(c) 20%
(d) 21%
6. A hemispherical container with radius 6 cm contains 325 ml of milk. Calculate the volume of milk that is needed to fill the container completely. () 3.142 π=
7. If the number of square centimetres in the surface area of a sphere is equal to the number of cubic cm in its volume. Find the diameter of the sphere?
8. Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water.
9. How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? 22 Take 7 π
10. A hemispherical bowl has its external diameter equal to 10 cm and its thickness is 1 cm. What is the whole surface area of the bowl?
11. A corn cob (see figure), shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length as 20 cm If each 1 cm2 of the surface of the cob carries an average of four grains, find how many grains you would find on the entire cob?
12. How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter.
13. Two cones have their heights in the ratio 1:3 and the radii of their base in the ratio 5:1. Show that their volumes are in the ratio 3:1.
14. A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?
15. A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of `2 per square metre, if the radius of the base is 14 metres.
16. If the radius of a sphere is increased by 10%. Prove that the volume will be increased by 33.1% approximately.
(NCERTExemplar)
17. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use 3.14 π= ).
h = 8 m
r = 6 m
18. The volume of the two spheres are in the ratio 64:27. Find the difference of their surface areas, if the sum of their radii is 7 cm.
19. A conical tent is 9 m high and the radius of its base is 12 m.
(a) What is the cost of the canvas required to make it, if a square metre canvas costs `10?
(b) How many persons can be accommodated in the tent, if each person requires 2 square metre on the ground and 3 15 m of space to breathe in?
20. Earth’s moon is the only place beyond the earth where humans have step foot. Moon influences human societies and cultures since time immemorial.
Challenge Yourself
1. A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm5 cm2 cm ×× . Assuming 22 7 π= . What is the percentage of wood wasted in the process?
2. The radius and height of a right solid circular cone () ABC are respectively 6 cm and 27 cm . A coaxial cone (DEF) of radius 3 cm and height 7 cm is cut out of the cone as shown in the given figure. What is the whole surface area of the solid thus formed?
Earth Moon
Answer the following questions on the basis of following information about the moon and the Earth:
“The diameter of the moon is approximately onefourth of the diameter of the earth.”
(a) If ‘d ’ be the diameter of the earth then what is the radius of moon in terms of d?
(b) What fraction of the volume of the earth is the volume of the moon?
(c) What is the ratio of the surface areas of the earth and the moon?
OR
(d) Write the expression that represents the volume of the spherical shell, where R and r are the radii of the outer and the inner spheres respectively as shown in the figure. r R r R
4. The length and width of a swimming pool are 50 m and 15 m respectively. If the depth of the swimming pool at one end is 10 m and at the other end 20 m, then find the volume of water in the swimming pool? Scan me
3. A cone of height 3 cm is inscribed in a sphere of radius 2 cm. Find the ratio of the volume of the cone to that of the sphere.
Answers
Additional Practice Questions
1. (c) 1:7
3. (c) 2 4158 cm
5. (a) 10%
7. 6 cm
9. 110 m
11. 531
13. Hence, proved.
15. `2068
17. 63 m
19. (a) `5657
(b) 90
2. (a) 7 cm
4. (b) 0
6. 127.45 ml
8. 38.808 ml
10. 286 cm2
12. 2541
14. 126
16. Hence, proved.
18. 88 cm2
20. (a) 8 d
(b) volume of the moon is 1 64 of the volume of the earth.
(c) (b) 16:1
(d) 4433 33 VRr =−ππ
Challenge Yourself
1. 2 92 3 %
2. 2 87 cm π
3. 9:32
4. 3 11250 m
SELF-ASSESSMENT
Time: 60 mins
Multiple Choice Questions
Scan me for Solutions
Max. Marks: 40
[3×1=3Marks]
1. If the radius and height of a right circular cone are r and h respectively, the slant height of the cone is (a) () 1 22 3 hr + (b) 1 ()3 hr + (c) () 1 22 2 hr + (d) 22hr +
2. A solid metal sphere is cut through the centre in two equal parts. If the radius of the sphere is 14 cm, then the total surface area of each part is
3. The radius of a hemispherical balloon increases from 4 cm to 8 cm as air is being pumped into it. The ratio of surface areas of the balloon in the two cases is (a) 2:3 (b) 1:3 (c) 2:1 (d) 1:4
Assertion-Reason Based Questions
[2×1=2Marks]
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R) Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
4. Assertion (A): If the volumes of two spheres are in the ratio 27:8 then their surface areas are in the ratio 3:2. Reason (R): 32 4 Volume of sphere , Surface area of a sphere 4 3 rr =ππ =
5. Assertion (A): If the height of cone is 24 cm and diameter of base is 14 cm, then the slant height of cone is 25 cm. Reason (R): If r be radius and h be the slant height of cone then the slant height () 22hr=+
Very Short Answer Questions
[2×2=4Marks]
6. A shot put is a metallic sphere of radius 4 cm If the density of the metal is 10 g per cm3, find the mass of the shot put.
7. A glass hemispherical bowl is of radius 9 cm How many litres of water it can hold?
Short Answer Questions
8. The radius and slant height of a cone are in the ratio of 3:5. If its curved surface area is 2 2310 cm , find its radius, height and slant height. h r l
[4×3=12Marks]
9. Monica has a piece of canvas whose area is 2 551 m . She uses it to have a conical tent made, with a base radius of 7 m Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 2 1 m , find the volume of the tent that can be made with it.
10. A storage tank consists of a circular cylinder, with a hemisphere adjoined on both ends. If the external diameter of the cylinder be 1.4 m and its length be 5 m, what will be the cost of painting it on the outside at the rate of `10 per square metre?
11. A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of `10 per 2 m .
Long Answer Questions
[3×5=15Marks]
12. The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted and recast into a solid cone of height 2 2 cm 3 , find the diameter of the cone.
13. There are 50 students in a blind school. Mr and Mrs Khatri wished to serve them milk. They have two options for serving the milk.
Option A: A hemispherical bowl with radius 10.5 cm made up of eco-friendly material.
Option B: A hemispherical bowl with radius 7 cm made up of plastic.
(a) How many litres of milk is required if option A is taken?
(b) How many litres of milk is required if option B is taken?
14. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Case Study Based Questions
15. Traffic cone are used outdoor during road work in various situation. Such as traffic redirection, advance warning and hazards of the prevention traffic.
A traffic cone has the radius 2.1 cm and height 20 cm Answer the following question based on the above data.
(a) What is the slant height of the traffic cone?
(b) What will be the total surface area of the traffic cone?
(c) Find the price of the painting the curved surface of 20 such traffic cone. If the price of painting is `8/m2?
(d) What will be volume of each traffic cone?
[1×4=4Marks]
Canvas
12 Statistics
Chapter at a Glance
1. Statistics is the study of the collection, organization, analysis, interpretation and presentation of data.
2. Data refers to numerical facts or qualitative information collected for a specific purpose. It can be primary (collected first-hand) or secondary (already available and used by someone else).
3. Graphical Representation of Data: A picture is better than a thousand words, graphical representation makes data easier to understand.
Common methods include:
● Bar Graphs: Used to represent categorical data using rectangular bars.
● Histograms: Similar to bar graphs, but used for data with continuous class intervals.
● Frequency Polygon: Constructed by plotting midpoints of class intervals and joining them with line segments.
NCERT Zone
NCERT Exercise 12.1
1. A survey conducted by an organisation for the cause of illness and death among the women between the ages 15−44 (in years) worldwide, found the following figures (in %):
(a) Represent the information given above graphically.
(b) Which condition is the major cause of women’s ill health and death worldwide?
(c) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (b) above being the major cause.
Sol. (a) By representing causes on x-axis and female fatality rate on y-axis and choosing an appropriate scale (1 unit = 5% for y-axis), the graph of the given above information can be drawn as follows.
All the rectangle bars are of the same width and have equal spacing between them.
(b) Major cause of women’s ill health and death worldwide is reproductive health conditions, as 31.8% of all women are affected by it.
(c) The factors are as follows.
(i) Lack of medical facilities
(ii) Lack of correct knowledge of treatment
2. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.
Scheduled Caste (SC)
Tribe (ST)
(a) Represent the information above by a bar graph.
(b) In the classroom discuss what conclusions can be arrived at from the graph.
Sol. (a) By representing section (variable) on x-axis and number of girls per thousand boys on y-axis, the graph of the information given above can be drawn by choosing an appropriate scale (1 unit = 100 girls for y-axis) Number of Girls per Thousand boys
Number of Girls per Thousand Boys in Different Sections of Indian Society
Political Party A B C D E F
Seats Won 75 55 37 29 10 37
(a) Draw a bar graph to represent the polling results.
(b) Which political party won the maximum number of seats?
Sol. (a) By taking polling results on x-axis and seats won as y-axis and choosing an appropriate scale (1 unit = 10 seats for y-axis), the required graph of the above information can be drawn as follows.
Seats Won by Political Parties in State Assembly Elections
Here, the rectangle bars are of the same length and have equal spacing in between them.
(b) Political party ‘A’ won maximum 75 number of seats.
4. The length of 40 leaves of a plant are measured correct to one millimeter, and the obtained data is represented in the following table:
Here, all the rectangle bars are of the same length and have equal spacing in between them.
(b) It can be observed that maximum number of girls per thousand boys (i.e., 970) is for ST and minimum number of girls per thousand boys (i.e., 910) is for urban.
Also, the number of girls per thousand boys is greater in rural areas than that in urban areas, backward districts than that in non-backward districts, SC and ST than that in non SC/ST.
3. Given below are the seats won by different political parties in the polling outcome of a state assembly elections:
(a) Draw a histogram to represent the given data.
(b) Is there any other suitable graphical representation for the same data?
(c) Is it correct to conclude that the maximum numbers of leaves are 153 mm long? Why?
Sol. (a) It can be observed that the length of leaves is represented in a discontinuous class interval having a difference of 1 in between them. Therefore, 1 0.5 2 = has to be added to each upper class limit and also have to subtract 0.5 from the lower class limits so as to make the class intervals continuous.
Taking the length of leaves on x-axis and the number of leaves on y-axis, the histogram of this information can be drawn as above. Here, 1 unit on y-axis represents 2 leaves.
(b) Other suitable graphical representation of this data is frequency polygon.
Here, 1 unit on y-axis represents 10 lamps.
(b) It can be concluded that the number of neon lamps having their lifetime more than 700 is the sum of the number of neon lamps having their lifetime as 700 800, 800 900, and 900 1000. Hence, the number of neon lamps having their lifetime more than 700 hours is (74 + 62 + 48) =184
6. The following table gives the distribution of students of two sections according to the mark obtained by them:
(c) No, as maximum number of leaves (i.e., 12) has their length in between 144.5 mm and 153.5 mm. It is not necessary that all have their lengths as 153 mm.
5. The following table gives the life times of neon lamps:
Sol.
(a) Represent the given information with the help of a histogram.
(b) How many lamps have a lifetime of more than 700 hours?
(a) We can take life time (in hours) of neon lamps on x-axis and the number of lamps on y-axis, the histogram of the given information can be drawn as follows.
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Sol. We can find the class marks of the given class intervals by using the following formula. Class mark = 2 Upper
Taking class marks on x-axis and frequency on y-axis and choosing a scale of 1 unit = 3 for y-axis, the frequency polygon can be drawn as follows.
0510152025303540455055 −5 Marks
From the graph, it can be observed that the performance of students of section ‘A’ is better than the students of section ‘B’ in terms of good marks.
7. The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:
We can take class marks on x-axis and runs scored on y-axis, a frequency polygon can be drawn as follows:
8. A random survey of the number of children of various age groups playing in park was found as follows:
Represent the data of both the teams on the same graph by frequency polygons.
[Hint: First make the class intervals continuous.]
Sol. It can be observed that the class intervals of the given data are not continuous. There is a gap of 1 in between each interval.
As class mark of each interval can be found by using the following formula. Class mark = 2 Upper class limit Lower class limit +
Therefore, 0.5 has to be added to the upper class limits and 0.5 has to be subtracted from the lower class limits.
Continuous data with class mark of each class interval can be represented as follows.
Draw a histogram to represent the data above.
Sol. It can be observed that the data has class intervals of varying width.
The proportion of children per 1 year interval can be calculated as follows.
Taking the age of children on x-axis and proportion of children per 1 year interval on y-axis, the histogram can be drawn as follows.
(b) Write the class interval in which the maximum number of surname lie.
Sol. (a) Here, it can be observed that the data has class intervals of varying width. The proportion of the number of surnames per 2 letters interval can be calculated as follows.
9. 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:
By taking the number of letters on x -axis and the proportion of the number of surnames per 2 letters interval on y -axis and choosing an appropriate scale (1 unit = 4 students for y -axis), the histogram can be constructed as follows.
Histogram of Number of Letters in Surnames
(a) Draw a histogram to depict the given information.
Multiple Choice Questions
1. A bar graph is drawn to the scale 1 cm k = units, then a bar of length cm k represents
(a) 1 unit (b) k units
(c) 2k units (d) 2 k units
(b) The class inter val in which the maximum number of surnames lies is 6 8 as it has 44 surnames in it i.e., the maximum for this data.
2. In figure, bar graph represents sales of two wheelers and four wheelers in a mega city from 2013 to 2016. In which year the difference between the sales of two wheelers and four wheelers is less?
(a) 2013 (b) 2014 (c) 2015 (d) 2016
Two wheelers Four wheelers
3. In a bar graph, the height of a bar is 5 cm and it represents 40 units. The height of the bar representing 56 units is
(a) 11.2 cm (b) 5.6 cm (c) 7 cm (d) 8 cm
4. The width of each of five continuous classes in a frequency distribution is 5 and lower class-limit of the lowest class is 10. The upper class-limit of the highest class is:
(a) 15 (b) 25 (c) 35 (d) 40
5. The class mark is 45. What could be the class interval if the class width is 10?
(a) 35 45 (b) 45 55
(c) 40 50 (d) 38 48
6. If x be the mid-point and 1 be the upper class limit of a class in a continuous frequency distribution. What is the lower limit of the class?
Answers
1. (d) 2 k units.
2. (b) 2014
3. (c) 7 cm
4. (c) 35
5. (c) 40 50
Constructed Response Questions
Very Short Answer Questions
1. A histogram is to be constructed for the following table:
(a) x 1 (b) 3x + 8
(c) 2x + 2 (d) 2x 1
7. In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is:
(a) 6 (b) 7 (c) 8 (d) 12
8. In a bar graph, the height of a bar is proportional to the
(a) width of the bar
(b) range of the data
(c) value of the component
(d) number of observations in the data
9. Consider the following frequency distribution:
Class interval 5 10 10 15 15 25 25 45 45 75
Frequency 6 12 10 8 15
To draw a histogram to represent the above frequency distribution the adjusted frequency for the class 25 45 is (a) 6 (b) 5 (c) 3 (d) 2 (NCERTExemplar)
10. A frequency polygon is constructed by plotting frequency of the class interval and the (a) upper limit of the class
(b) lower limit of the class
(c) mid value of the class
(d) any values of the class
6. d) 2x 1
7. (b) 7
8. (c) value of the component
9. (d) 2
10. (c) mid value of the class
2. If the class marks in frequency distribution are 19.5, 26.5, 33.5, 40.5, then find the class corresponding to the class mark 33.5
Sol. The class size of the distribution = 40.5 33.5 = 7
The required class of the class mark 33.5 is
33.5 – – 33.5
, i.e., 30 37.
3. The following table shows the number of students in Class IX during the academic years 1996 1997 to 2000 2001:
No. of Students 50 75 125 150 200
Represent the above data by a bar graph.
Sol. To represent the above data by a bar graph, we first draw a horizontal and a vertical line. Since five values of the numerical data are given. So, we mark five points on the horizontal line at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the numerical values of the data as shown in figure.
Sol. Total number of students = 500 + 350 + 400 + 750 + 500 + 100 + 400 = 3000.
(a) The given bar graph represents the number of students wearing shoes of different numbers out of a total of 3000 selected students.
(b) The students wear shoes bearing numbers 4, 5, 6, 7, 8, 9 and 10.
(c) The number of students wearing shoe No. 6 is 400.
(d) Shoe number 7 is worn by the maximum number of students. These students are 750 in number.
5. The number of family members in 10 families are given below: 6, 4, 2, 5, 4, 5, 3, 2, 3, 4
Represent the data in the form of frequency distribution table.
Sol.
4. Read the bar graph shown in given figure and answer the following questions:
(a) What is the information given by the bar graph?
(b) What are the different numbers of the shoeworn by the students?
(c) What is the number of students wearing shoe No. 6?
(d) Which shoe number is worn by the maximum number of students? Also give its number.
6. Read the bar graph. Find the percentage of excess expenditure on wheat than pulses and ghee taken together.
Sol. Expenditure on pulses and ghee = 10%+ 20%= 30%
Expenditure on wheat = 35%
Excess expenditure on wheat = 35% 30%= 5%
7. Study the following graph and answer the question given below 1952
Voters Turn-out (in percentage)
Y X
(a) In which years is the highest and lowest ever voters turn-out (in %)?
(b) For which two years, the numeric difference in voters turn-out (in %) was nearly equal to 10%?
Sol. From the given graph, we have
(a) The highest voter turnout was in 1984 (≈ 64%), and the lowest was in 1962 (≈ 54%).
(b) The difference in voter turnout between 1962 and 1984 is nearly 10%.
Short Answer Questions
1. Convert the given frequency distribution into a continuous grouped frequency distribution:
The difference = 154 153 = 1
Half the difference 1 10.5 2 =×=
So, the new class interval formed from 150 153 is (150 0.5) (153 + 0.5), i.e., 149.5 153.5. Continuous classes formed are:
Therefore, 153.5 is included in the class interval 153.5 157.5 and 157.5 in 157.5 161.5.
2. Represent the following frequency distribution by means of a histogram.
3. A family with a monthly income of `20,000 had planned the following expenditure per month under various heads.
In which intervals would 153.5 and
be included?
(NCERTExemplar)
Sol. Consider the classes 150 153 and 154 157.
The lower limit of 154 157 = 154
The upper limit of 150 153 = 153
Draw a bar graph for the data above.
4. In 2019, there are thousands number of visitors come to Agra from June to December.
Sol. (a) Class size is 5.
(b) First class interval is 10 15.
Lower limit = 10, Upper limit = 15
(c) Class mark of 25 30 2530 27.5 2 + ==
(d) 30 35 has maximum frequency, i.e., 48.
6. For the following data, draw a histogram and a frequency polygon:
Read the graph and answer the questions:
(a) Which month recorded the highest number of visitors?
(b) How many people visited the Agra in 2019?
(c) In which month did equal number of visitors come?
Sol. From the graph we have,
(a) The highest number of visitors come in month of December.
(b) People visited Agra from June to December in 2019
() 2030204035254510002,15,000 =++++++×=
(c) In the month of June and August equal number of visitors came.
5. The table given below shows the number of persons from various age groups who participated in a campaign for promoting “USE OF CLEAN FUEL”.
(a) Determine the class size.
(b) Write the class limits of first class interval.
(c) What is the class mark of fourth class interval?
(d) Which class inter val has maximum frequency?
Sol. In figure, histogram and frequency polygon are drawn on the same scale.
To get the frequency diagram we need to connect the midpoints of class intervals.
Long Answer Questions
1. Construct a histogram from the following distribution of total marks obtained by 65 students of IX class in the final examination.
(mid-points)
Sol. Since the difference between the second and first mid-point is 16015010 −= .
105. 2 hh ∴=⇒=
So, lower and upper limits of the first class are 1505 and 1505 + i.e. 145 and 155 respectively.
∴ First class interval is 145 155.
Using the same procedure, we get the classes of other mid-points as under:
The histogram of the above frequency distribution is given in figure.
2. Draw a histogram to represent the following grouped frequency.
Also draw frequency polygon.
Sol. Prepare a table,
3. Draw a histogram to represent the
Prepare a table,
4. A random survey of number of children of various age group playing in a park was found as follows:
Draw the histogram of above data.
Sol. Prepare a table,
According to the table, prepare histogram
the frequency polygon,
5. Draw a frequency polygon for the following distribution:
6. The marks obtained (out of 100) by a class of 80 students are given below:
Construct a histogram to represent the data above.
Determine between which two intervals is the drop in frequency density the highest?
Sol. Prepare the following table,
Prepare the table,
Prepare the histogram,
Competency Based Questions
Multiple Choice Questions
1. A bar graph is drawn to the scale of 1 cm x = units. If the length of a bar representing a quantity of 702 units is 3.6 cm, then x = (a) 165 (b) 175 (c) 185 (d) 195
2. The histogram shows information about the weights, in grams, of some plums.
Farm A
A sample of 50 eggs is taken from Farm B.
The histogram shows information about the masses of the eggs from Farm B.
B
The estimate for the proportion of these plums with a weight of less than 100 grams is (a) 5 81 (b) 5 76 (c) 71 81 (d) 76 81
3. A sample of 50 eggs is taken from Farm A.
The table shows information about the masses of the eggs from Farm A.
For medium eggs, 53 g < mass 63 g <
The Farm A sample has more medium eggs than the Farm B sample.
Using the table and the histogram, estimate how many more.
(a) 27 (b) 18 (c) 9 (d) 36
4. The histogram shows information about the ages of all the passengers travelling on a plane. No one on the plane is older than 80 years.
A passenger on the plane is picked at random. Work out an estimate for the probability that this person is older than 55 years. (a) 11 62 (b) 8 62 (c) 16 62 (d) 23 62
5. A bar graph shows the monthly electricity usage (in units) of a family over 6 months.
6. The total number of students who took between 30 and 70 seconds is to be divided between two categories: Students who took 30 50 seconds, Students who took 50 70 seconds
What is the ratio of students in the interval 30 50 seconds to those in 50 70 seconds?
Give your answer in the simplest form.
(a) 5:17 (b) 5:7 (c) 9:17 (d) 4:7
7. Calculate the percentage of students who took more than 70 seconds to complete the puzzle out of the total number of students shown in the histogram. Give your answer correct to 2 decimal places.
(a) 5.88% (b) 10% (c) 6.45% (d) 1.35%
8. If 6 students took between 10 and 30 seconds, Express the number of students who took less than 30 seconds as a fraction of the total, in simplest form.
(a) 5 17 (b) 4 17 (c) 4 14 (d) 5 14
9. The histogram shows some information about the ages of the 134 members of a sports club.
If the cost per unit is `5, in which month did the family spend `200 more than in February?
(a) January (b) April (c) March (d) June
Use the below graph to answer Question 6 to 8.
A group of students were asked to complete a puzzle. The histogram shows the distribution of the times taken.
20% of the members of the sports club who are over 50 years of age are female.
Work out an estimate for the number of female members who are over 50 years of age.
(a) 5 (b) 6 (c) 7 (d) 8
Assertion-Reason Based Questions
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): The graphical representation of the frequency distribution
given in figure.
Reason (R): In a histogram, the areas of the rectangles are proportional to the frequencies.
2. Assertion (A): A bar graph is a pictorial representation of the numerical data by a number of rectangles of uniform width erected horizontally or vertically with equal spacing between them.
Reason (R): In order to draw the histogram when mid-points of class intervals are given, it is assumed that the frequency corresponding to the variate value a (say) is spread over the interval 2 h a to 2 h a + , where h is the jump from one value to other.
Case Study Based Questions
1. Child labour refers to any work or activity that deprives children of their childhood. It is a violation of children’s
rights. This can harm them mentally or physically. It also exposes them to hazardous situations or stop them from going to school. Naman collected data on number of child labours (in million) in different countries, which is shown in the bar graph below.
Read the following passage and answer any four questions of the following:
(a) Which countr y has the highest child labour?
(b) Which countr y has the lowest child labour?
(c) Name the countries having more than 4 million child labour.
(d) What is the total number of child laborers in all the countries shown?
2. India’s national animal, the tiger, has long been under government monitoring due to a sharp decline in its population. However, a report by the Ministry of Environment, Forests, and Climate Change states that tiger population has started to rise again at an annual growth rate of 6% between 2006 and 2018.
Pranav has found a report on status of tigers in Indiain last 100 years. He has some queries as follow:
(a) In which year, number of tigers were minimum?
(b) What does this bar graph show?
Answers
Multiple Choice Questions
1. (d) 195
We have, 1 cm units 3.6 cm3.6 units xx =⇒=
∴ 3.6x = 702 702 3.6 x = 195 x =
2. (d) 76 81
Use Frequency = Frequency density × Class width to work out the frequency of each bar on the histogram.
The last bar will need splitting at a weight of 100 grams. 1 2 3 4 Y
13161824576 ++++=
Total: 76581 +=
One mark for each
Now write the proportion as a fraction.
The proportion of plums with a weight less than 100 grams is 76 81
3. (c) 9
Number of “medium” eggs from Farm A:
81927 += medium eggs
Frequency of the 50 55 bar on the Farm B: 51.05 eggs ×=
(c) How much decrease in number of tigers was there between 1900 and 1960?
(d) What was the total decrease in the number of tigers between 1900 and 2007?
Frequency of 53 55: 50.42 eggs ×=
Frequency of the 55 65 bar on the Farm B: 102.020 eggs ×=
Frequency of 5563: 200.816 eggs ×=
Total number of “medium” eggs at Farm B: 21618 += eggs
How many more medium eggs Farm B has compared to Farm A: 2718 9 −=
Farm A has 9 more medium eggs than Farm B
4. (a) 11 62
No scale is given so this question can be done by looking at the relative areas (or, more simply, by counting squares) Count the number of small squares in total (in all six bars)
20 + 90 + 160 + 150 + 120 + 80 = 620 small squares in total
Count the number of small squares for ages greater than 55
30 + 80 = 110 small squares
Estimate the probability that the person is over 55 (by dividing 110 by 620) 11011
62062 =
5. (a) January
Cost in February =180 × `5 = `900
Find month where cost = `900 + `200 = `1100 1100 220 units 5 = ` ` January = 220 units
6. (b) 5:7
Students (30 50 sec) = 50
Students (50 70 sec) = 70 5
Ratio50:70 7 === 5:7
7. (a) 5.88%
Students > 70 sec = 10
Total students = 170 10
Percentage1005.88% 170
Students (10 30 sec) = 40
Total from full table = 170
Fraction: 404 17017 =
9. (c) 7
From the histogram, the intervals over 50 years are:
50 60 from the 40 60 bar (FD ≈ 1.4)
60 80 (frequency density ≈ 0.7)
80 100 (frequency density ≈ 0.4)
FrequencyFrequency DensityClass Width =×
Frequency = 1.4 × 10 = 14
Frequency = 0.7 × 20 = 14
Frequency = 0.4 × 20 = 8
1414836 members ++=
Calculate 20% of 36
20 367.27
100 ×=≈
Estimated number of female members over 50 years of age = 7
Assertion-Reason
Based Questions
1. (c) A is true, but R is false.
Additional Practice Questions
1. The histogram given below shows the marks obtained by students in a online exam conducted during the remedial classes in the school: Using the histogram the number of students obtained marks less than 40 is
The histogram shown in figure is not a correct representation of the data because the classes are not of the uniform width. Height of the rectangle representing class 60100 should be 40 2512.5 20 ×= not 25. So, Assertion is not true.
Reason is true.
2. (b) Both A and R are true, but R is not the correct explanation of A.
Case Study Based Questions
1. (a) India
(b) United States
(c) India and Bangladesh
(d) Add all the values:
= 41.5 million = 41.5 million
2. (a) In 2007 number of tigers were minimum.
(b) This bar graph shows decline in number of tigers. (c) Number of tigers in 1900 = 40,000 and Number of tigers in 1960 = 18000 Therefore, Decrease in number of tigers = 22000 (d) Tigers in 1900 = 40,000 Tigers in 2007 = 1411
Decrease = 40,000 − 1411
2. The following data shows the weight of 30 apples. Taking class size of 20 and taking 60 80 as first class interval weighing more than 180 grams.
The number of apples weighing more than 180 grams is
(a) 1 (b) 2
(c) 3 (d) 4
3. If the class mark of the frequency distributions are 9.5, 16.5, 23.5, 30.5 then the class interval corresponding to the class mark 16.5 is
(a) 13 20 (b) 12 17
(c) 14 20 (d) 15 17
4. Below is a histogram showing how long people can hold their breath.
8. The following table gives the life time of 400 neon lamps:
There were 54 people who could hold it for at least 1 minute.
Work out how many could hold their breath for between 20 and 40 seconds.
5. A grouped frequency distribution table with classes of equal sizes using 63 72 (72 included) as one of the classes is constructed for the following data:
6. The value of π up to 50 decimal places is given below: 3.14159265358979323846264338327950288 419716939937510
(a) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(b) What are the most and the least frequently occurring digits?
7. Draw histogram for the following data:
(a) Represent the given information with help of histogram.
(b) How many lamps have life time of more than 700 hours?
9. Draw a frequency polygon for the following data:
10. Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them:
Represent the marks of the students of both the sections on the same graph by two frequency polygons. What do you observe?
11. The marks by 750 students in an examination are given in the form of a frequency distribution table.
Read this data in the form of a histogram and construct a frequency polygon.
12. Hydroponics is an innovative farming technique that lays the foundation for future agricultural careers.
Class IX students explored hydroponics and discovered its benefits by growing 35 plants without soil in short span of time. The table below shows the
Challenge Yourself
1. The histogram shows information about the ages of the members of a football supporters club.
recorded lengths of the leaves of these plants grown in plastic containers:
There are 20 members aged between 25 and 30 One member of the club is chosen at random. What is the probability that this member is more than 30 years old?
2. The histogram gives information about the weights of some fish.
Read and answer the related questions.
(a) How many plants grew in length less than 12 cm?
(b) What is the class mark of the class interval 9.8 10.2?
(c) What is the class width of each inter val?
(d) What does the data suggest about plant growth using hydroponics?
The number of fish with a weight between 400 g and 450 g is 7 more than the number of fish with a weight between 250 g and 300 g.
Calculate the total number of fish represented by the histogram.
3. The histogram gives information about the heights, in metres, of the trees in a park. The histogram is incomplete.
20% of the trees in the park have a height between 10 metres and 12.5 metres.
None of the trees in the park have a height greater than 25 metres.
Complete the histogram. Scan me
4. The histogram shows information about the speed of cars as they pass a checkpoint.
The scale on the frequency density axis is missing. The histogram shows information about 480 cars.
How many cars does the first bar represent?
Answers
Additional Practice Questions
1. (c) 50
2. (b) 2
3. (a) 13 20
4. 114 people
5. 10 classes
6. (a)
(b) Most frequently occurring digit: 3 and 9 (appears 6 times)
Least frequently occurring digit: 0 (appears 1 time)
(b) Lamps having life time for
9. Prepare the table,
2.5 16 (2.5, 16)
40 (7.5, 40) 12.5 32 (12.5, 32)
17.5 24 (17.5, 24)
22.5 8 (22.5, 8)
Prepare the frequency polygon according to the above data,
24)
8)
From the frequency polygon of two sections we observe that maximum marks of 67.5 is scored by 40 students of section B.
12. (a) 35 plants (b) 10 cm
(c) 0.4 cm
(d) The majority of plants (28 out of 35) have leaf lengths greater than 10.2 cm. This suggests that hydroponics is highly effective in promoting healthy plant growth in a short time, as most plants reached significant lengths without using soil.
Yourself 1. 61 118 2. 68 3. Bar on height 3.2 4. 300 cars
X Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 60 mins
Multiple Choice Questions
Scan me for Solutions
Max. Marks: 40
[3 × 1 = 3Marks]
1. Which one of the following is not the graphical representation of statistical data? (a) Bar graph (b) Histogram (c) Frequency polygon (d) Cumulative frequency distribution
2. In a histogram the class intervals or the groups are taken along (a) y -axis (b) x -axis (c) both of x -axis and y -axis (d) in between x and y axis
3. The mid-value of a class interval is 42. If the class size is 10, then the upper and lower limits of the class are (a) 38 and 47 (b) 37 and 47 (c) 37.5 and 47.5 (d) 47.5 and 37.5
Assertion-Reason Based Questions
[2 × 1 = 2Marks]
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
4. Assertion (A): In a histogram, the area of each rectangle is proportional to the class size of the corresponding interval.
Reason (R): To draw the histogram of a continuous frequency distribution with unequal class intervals, the frequencies of classes are adjusted by using the formula:
Minimum class size
Adjusted frequency of a class
Frequency of the class Class size =×
5. Assertion (A): A frequency polygon can not be drawn with the help of a Histogram.
Reason (R): A frequency histogram is a type of bar graph that shows the frequency or number of times an outcome occurs.
Very Short Answer Questions
6. Below is a histogram showing the times taken to complete a quiz.
[2 × 2 = 4Marks]
44 people took between 0 and 1.5 minutes. Work out how many people took between 3 and 4 minutes.
7. The following bar graph shows the number of vehicles passing through a road crossing in Delhi in different time intervals on a particular day. Read the bar graph and answer the following questions:
(a) What does the bar graph represent?
(b) When is the hourly traffic maximum?
Short Answer Questions
8. The expenditure of a family on different heads in a month is given below:
Expenditure (in `) 4000
Draw a bar graph to represent the data above.
9. The following table gives the frequencies of most commonly used letters a, e, i, o, r, t, u form a page of a book:
Represent the information above by a bar graph.
10. Expenditure on Education of a country during a five year period (2002 2006), in crores of rupees, is given below:
Represent the information above by a bar graph.
11. Draw histogram for following frequency distribution:
Long Answer Questions
[3x5 = 15Marks]
12. The marks scored by 750 students in an examination are given in the form of a frequency distribution table.
Represent this data in the form of a histogram and construct a frequency polygon.
13. Draw a histogram for the following data:
14. The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:
Represent the data of both the teams on the same graph paper by frequency polygons.
15. Corona Virus disease (COVID-19) is an infectious disease caused by a newly discovered variant of the coronavirus. The data below shows the number of confirmed and recovery cases of the COVID-19 pandemic in four Indian States/Union Territories as of 9th December 2020, 9:05 a.m. IST.
Study the table and its bar graph given below to answer the following questions.
(a) Which State/UT shows maximum number of confirmed cases?
(b) Which two States/State and UT together exceed the confirmed number of cases than MH.
(c) What is the total number of recovery cases in lakh in 4 States/UT?
(d) Which state has the lowest recovery rate compared to its confirmed cases?
Self-Assessment Answers
Chapter 1
1. (d) 0
2. (a) 0
3. (b) 0.378
4. (a) Both A and R are true, and R is the correct explanation of A.
5. (a) Both A and R are true, and R is the correct explanation of A.
6. 7 9 7. 0.142857
8. 2 3 9. 3 115
85280 +
166 99
64
15. (a) 8 (b) 1 (c) 128
Chapter 2
1. (b) Linear Polynomial
2. (b) x 2 - 2x - 80
3. (b) (x - a)
4. (a) Both A and R are true, and R is the correct explanation of A.
5. (a) Both A and R are true, and R is the correct explanation of A.
6. 2029
7. 1 4
8. 0
9. 3
10. (3m - 11n)2
13. (x - 1) (x - 2) (x + 1) 14. ±76
15. (a) (x + 5) (b) 3 (c) k = 1
(d) The possible expressions are (2x + 3) and (2x - 1) or (2x - 1) and (2x + 3).
Chapter 3
1. (b) Second quadrant
2. (c) Origin
3. (c) On x-axis
4. (a) Both A and R are true, and R is the correct explanation of A.
5. (c) A is true, but R is false.
6. (a) Quadrant IV (b) 2 units
7. (a) (8, 0) or (-8, 0) (b) 4, 0, 4, -3
8. (a) (-6, -2) (b) -5
9. (a) b =-5 (b) (-5, -3)
10. (a) k can be any negative number (b) (-7, 4)
11. The coordinates are:
A(-3, 0)
B(3, 0)
C(3, 3)
D(-3, 3)
12. P(y, x) = (3, -3) → lies in IV quadrant
Q(2, x) = (2, -3) → lies in IV quadrant
R(x2 , y - 1) = [(-3)2, (3 - 1)] = (9, 2)
→ lies in I quadrant
S(2x, -3y) = [2(-3), -3(3)] = (-6, -9)
→ lies in III quadrant
13. (a) Reflection in the x-axis: P = (2, 5)
(b) Reflection in the y-axis: P = (-2, -5)
(c) Reflection in the origin: P = (-2, 5)
(d) Shifting 3 units right and 4 units up: Pshifted = (2 + 3, -5 + 4) = (5, -1)
14. (a) a = 4, b =-11
(b) Distance from the x-axis 11 units
Distance from the y-axis 4 unit
(c) IV Quadrant
15. (a) (-4, -2)
(b) III quadrant
(c) 4.47
(d) (8, 8)
Chapter 4
1. (d) () 2, 1.5
2. (b) Vertical Line
3. (a) 12 cm
4. (a) Both A and R are true, and R is the correct explanation of A.
5. (d) Assertion is false, but Reason is true.
6. 195 , 1111xy==
7. 4 x =
8. 1 m =
9. 1 6 t =
10. 1 4, 3
11. () 0,3 , () 5,1 Infinite more
12. 11 , 46 ab==
13. Area = 20 sq. units
14. First Girl = `45, Second girl = `31
15. (a) 36210 xy+= , 59330 xy+=
(b) cost of one notebook = `30, cost of one pen = `20
(c) `470
Chapter 5
1. (a) Thales
2. (b) Deductive reasoning
3. (b) a definition.
4. (a) Both A and R are true, and R is the correct explanation of A.
5. (b) Both A and R are true, but R is not the correct explanation of A.
7. x = 2a, Second thing y = 2a then, x = y
11. (a) Infinite, if they are collinear.
(b) Only one, if they are non-collinear
13. (a) False.
(b) False.
(c) True.
(d) True.
(e) True.
15. (a) (i) All of these (b) (ii) Things which are halves of the same thing are equal to one another.
(c) (iii) Collinear Points
(d) (iv) C is an interior point of AB
Chapter 6
1. (b) Exactly 270°
2. (a) One
3. (b) Adjacent Angles
4. (a) Both A and R are true, and R is the correct explanation of A.
5. (a) Both A and R are true, and R is the correct explanation of A
6. 105º
7. a = 10°
8. ∠BOC = 86°
9. ∠ABC = 146°
12. 125,125, 35 xyz =°=°=°
13. (b) 30°
15. (a) 145° (b) 35° (c) 130°
(d) These angles do not form a straight line; they are part of different lines intersecting at point O.
Chapter 7
1. (c) 6 cm
2. (b) 50°
3. (a) ∠ABD
4. (a) Both A and R are true, and R is the correct explanation of A.
5. (d) A is false, but R is true.
6. No, because angle must be included angle
7. 30°
15. (a) ΔADC
(b) SAS congruency
(c) BC = CD
(d) Isosceles Triangle
Chapter 8
1. (c) 10 cm
2. (c) trapezium
3. (b) diagonals of ABCD are perpendicular
4. (a) Both A and R are true, and R is the correct explanation of A.
5. (a) Both A and R are true, and R is the correct explanation of A.
6. 16 cm
7. 120°
10. 140 x =°
15. (a) Parallelogram
(b) Rhombus
(c) ∠C = 135° and ∠D = 135°.
(d) Arjun should understand that using quadrilateral frames like PQRS with equal opposite sides and angles improves structural stability. These shapes distribute external forces like wind and vibrations evenly throughout the structure, reducing stress on any single point. This makes the Eiffel Tower and similar structures highly resistant to collapse and deformation.
Chapter 9
1. (a) 90°
2. (c) 12 cm
3. (c) 50°
4. (a) Both A and R are true, and R is the correct explanation of A.
5. (a) Both A and R are true, and R is the correct explanation of A.
7. 20°
8. 105°
9. 13 cm
12. (a) Chord BD (b) 50 m
(c) 2400 m2
13. ∠P = 150°, ∠Q = 45°, ∠R = 30°, ∠S = 135°
14. 153 m
15. (a) ABC is an equilateral triangle (b) 60°
(c) 12 cm
(d) 120°
Chapter 10
1. (a) 2 643 m
2. (b) becomes 4 times
3. (b) `2.16
4. (a) Both A and R are true, and R is the correct explanation of A.
5. (c) A is true, but R is false.
6. 120 cm2
7. 48 cm
8. 12 cm, 12 cm, 18 cm, 2 277 cm,37 cm
9. 750 m2, `6000
10. 4.98 cm2
11. Area = 720 cm2, Altitude = 80 cm
12. 2 5721 cm
13. 2 503 cm
14. 3 cm
15. (a) 480 m
(b) 2 60021 m
(c) 2 9003 m
(d) 2 3007 m
Chapter 11
1. (c) () 1 22 2 hr +
2. (d) 2 1848 cm
3. (d) 1:4
4. (d) A is false, but R is true.
5. (a) Both A and R are true, and R is the correct explanation of A.
6. 2.68 kg
7. 1.527 l
8. Radius = 21 cm, Slant height = 35 cm and Height = 28 cm
9. 1232 m3
10. `281.60
11. `382.80
12. 143 cm
13. (a) 121.275 litres (b) 35.934 litres
14. Volume = 86.625 m3, Canvas = 99.825 m2
15. (a) 20.109 cm
(b) 0.014586 m2
(c) `2.34
(d) 92.4 cm3
Chapter 12
1. (d) Cumulative frequency distribution
2. (b) x -axis
3. (b) 37 and 47
4. (d) Assertion is false, but Reason is true.
5. (d) Assertion is false, but Reason is true.
6. 22 people
7. (a) The bar graph represents the number of vehicle passing through a particular crossing of Delhi in different time intervals on a particular day. (b) The maximum traffic is between 9 to 10 hours.
This section consists of 20 questions of 1 mark each�
1� In the figure, O is the centre of the circle. If 20 ABC ∠= , then AOC∠ is equal to: (a) 60 (b) 10o (c) 40 (d) 20 2� 2,1xy==− is a solution of the linear equation
(a) 20 xy+= (b) 20xy+= (c) 24xy+= (d) 25 xy+=
3� The value of 1.999… in the form of p q , where p and q are integers and q ≠ 0. (a) 19 10 (b) 1999 1000 (c) 2 (d) 1 9
4� () 5,7 P be a point on the graph. Draw the PMy ⊥ -axis. The coordinates of M are
(a) () 0,7
(c) () 7,0
(b) () 0,0
(d) () 7,5
5� If the measures of the sides of a triangle is given by a, b and c, then we calculate the semi-perimeter by the formula
(a) 2 abc ++ (b) 3 abc ++
(c) 4 abc ++
(d) () 2 abc ++
6� In a figure, if ∠= ,110OPRSOPQ ‖ and 130 QRS ∠= , then PQR∠ is equal to
(a) 40 (b) 50 (c) 70 (d) 60
7� If 3 x is a factor of 2 15 xax , then a =
(a) 5 (b) 2 (c) 5 (d) 3
8� The value of ()()() 333 222222 333 ()()() abbcca abbcca −+−+− −+−+− is
(a) ()()() 3 abbcca
(b) ()()() abbcca +++
(c) ()()()()()() 3 abbccaabbcca +++−−−
(d) ()()() 2 abbcca
9� If a solid sphere of radius r is melted and cast into the shape of a solid cone of height r , then the radius of the base of the cone is
(a) 3r (b) r (c) 2r (d) 4r
10� If ++=+= 2 2 2 11 24, then aa aa
(a) 12 (b) 13 (c) 14 (d) 14
11� Simplify: (22)2
(a) 642 + (b) 643
(c) 622 (d) 642
12� The value of 151535 ÷ is
(a) 53 (b) 35 (c) 3 (d) 5
13� The value of 1 16 4 4 256 81 x y
is
(c) 4
14� ABCD is a Rhombus such that 40 ACB ∠= , then ADB∠ is (a) 100 (b) 40 (c) 60 (d) 50
15� To draw a histogram to represent the following frequency distribution:
The adjusted frequency for the class 25–45 is (a) 6 (b) 5 (c) 2 (d) 3
16� In the adjoining figure, BCAC = . If 115 ACD ∠= , the A∠ is
(a) 50 (b) 65 (c) 57.5 (d) 70
17� An irrational number between 1 7 and 2 7 is (a) 12
18� Let ABCD be a parallelogram. The diagonals bisect each other at E. If = 5 cm DE and = 7 cm AE , then find BD (a) 14 cm (b) 10 cm (c) 7 cm (d) 5 cm
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
19� Assertion (A): For all values of
3 ,, 2 kk is a solution of the linear equation 230 x += .
Reason (R): The linear equation += 0 axb can be expressed as a linear equation in two variables as axyb ++ 0 = .
20� Assertion (A): If the diagonals of a parallelogram ABCD are equal, then 90 ABC ∠= .
Reason (R): If the diagonals of a parallelogram are equal, it becomes a rectangle.
SECTION B
This section has 5 very short answer type questions of 2 marks each�
21� The radii of two cones are in the ratio 2:1 and their volumes are equal. What is the ratio of their heights? OR
A hollow spherical shell is made of a metal of density 4.5 g per 3 cm . If its internal and external radii are 8 cm and 9 cm respectively, find the weight of the shell.
22� Read the bar graph shown in given figure and answer the following questions:
(a) What is the information given by the bar graph?
(b) What are the different numbers of the shoe-worn by the students?
(c) What is the number of students wearing shoe No. 6?
(d) Which shoe number is worn by the maximum number of students? Also give its number.
23� BE and CF are two equal altitudes of a triangle ABC .Using RHS congruence rule, prove that the triangle ABC is isosceles.
24� Express 0.47 in the form p q , where p and q are integers and 0 q ≠
25� In figure = , ACXDC is the mid-point of AB and D is the mid-point of XY. Using a Euclid’s axiom, show that AB = XY.
This section has 6 short answer type questions of 3 marks each�
26� The marks scored by 750 students in an examination are given in the form of a frequency distribution table.
No� of Students 16 45 156 284 172 59 18
Represent this data in the form of a histogram and construct a frequency polygon. OR
Read the bar graph given in Figure and answer the following questions:
(a) What information is given by the bar graph?
(b) In which years the areas under the sugarcane crop were the maximum and the minimum?
(c) State whether true or false:
The area under the sugarcane crop in the year 1982-83 is three times that of the year 1950–51.
27� A city has two main roads which cross each other at the centre of the City. These two roads are along the North–South direction and East–West direction. All the other street of the city run parallel to these roads and are 200 m apart. There are about 5 streets in each direction. Using 1 cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines.
There are many cross streets in your model. A particular cross street is made by two streets, one running in north south direction and another in the East–West direction, which cross street is refer to in the following manner: If the 2nd Street running in the north south direction and 5th in the east west direction meet at some crossing, then we will call this cross street (2, 5). Using this convention, find:
(a) How many cross streets can be referred to as (4, 3).
(b) How many cross streets can be referred to as (3, 4)
28� If the length of a rectangle is decreased by 3 units and breadth increased by 4 units, then the area will increase by 9 sq. units. Represent this situation as a linear equation in two variables. Also, find the breadth when the length is 10 units.
29� If 5 11 11 3 211 xy + =+ , find the values of x and y.
30� It is given that 325 abc += , then find the value of 278125333 abc +− if 0 abc =
31� In figure X and Y are respectively the mid-points of the opposite sides AD and BC of a parallelogram ABCD . Also, BX and DY intersect AC at P and Q, respectively. Show that == APPQQC .
SECTION D
This section has 4 long answer questions of 5 marks each�
32� Using factor theorem, factorize the polynomial: 326310xxx−++
33� A hemispherical dome, open at base is made of sheet of fiber. If the diameter of hemispherical dome is 80 cm and 13 170 of sheet actually used was wasted in making the dome, then find the cost of dome at the rate of `35 per 100 cm².
34� ABC is a right triangle with ABAC = . Bisector of ∠A meets BC at D. Prove that 2 BCAD = .
35� In each of the figures given below, ABCD ‖ . Find the value of x .
OR
In the given figure, ∠=∠=∠=∠= ,,,,90ABCDEFDBGxEDHyAEBzEAB ‖‖ and 65 BEF ∠= Find the values of , xy and z .
SECTION E
This section has 3 case-study based questions of 4 marks each�
36� Read the following text carefully and answer the questions that follow: Rohan draws a circle of radius 10 cm with the help of a compass and scale. He also draws two chords, AB and CD in such a way that the perpendicular distance from the center to AB and CD are 6 cm and 8 cm respectively. Now, he has some doubts that are given below.
(a) Show that the perpendicular drawn from the Centre of a circle to a chord bisects the chord.
(b) What is the length of CD ?
(c) What is the length of AB ? OR
How many circles can be drawn from given three non collinear points?
37� A shopkeeper sells two varieties of rice: Type A and Type B. The cost incurred for the farmer to grow Type A and Type B rice were `12 per kg and `10 per kg respectively. And the selling price of Type A rice and Type B rice were `18 per kg and `14 per kg respectively. Now if the farmer has sold ‘x ’ kg of Type A rice and ‘y’ kg of Type B rice, then answer the following questions.
(a) The total cost price for the farmer is given by
(b) The total selling price for the farmer is given by
(c) The total profit earned by the farmer is given by
38� Read the text carefully and answer the questions:
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. 25 such spheres are used for this purpose and are to be painted silver. Each support is a cylinder and is to be painted black.
(a) What will be the total surface area of the spheres all around the wall?
(b) Find the cost of orange paint required if this paint costs 20 paise per 2 cm .
(c) (i) How much orange paint in liters is required for painting the supports if the paint required is 3 ml per 2 cm ? Or
(ii) What will be the volume of total spheres all around the wall?
Sample Question Paper 1
SOLUTIONS
1� (c) 40
Angle made by a chord at the centre is twice the angle made by it on any point of the circumference.
So, 222040AOCABC ∠=∠=×=
2� (b) 20xy+= () 221220 +−=−=
3� (c) 2
let 1.999 x =…
Multiply both sides by 10:
1019.999 x =…
Now subtract the first equation from the second: 1019.9991.999 xx−=…−…
918 x = 18 2 9 x ==
So, the value of 1.999... in the form of p q is: 2
4� (a) () 0,7
Here, PM Perpendicular to y-axis.
So point M lies on the y-axis, and for any point on y -axis always the value of 0 x =
So Co-ordinate of ()=−0,7 M
5� (a) 2 abc ++
Semi-perimeter is given by 2 abc ++ .
6� (d) 60
Produce OP to intersect RQ at point N
Now, OP‖RS and transversal RN intersects them at N and R respectively.
Apply power on numerator and denominator separately
= 1 4 4 1 16 4 (81) (256) y x
Thus, 4 3 4 y x = .
14� (d) 50
Explanation: In Rhombus, digonals bisect each other right angle. By using angle sum property in any of the four triangles formed by intersection of diagonals, we get 50 CBD ∠= and CBDADC∠=∠ (alternate angles).
So, 50 ADC ∠= .
15� (c) 2
Adjusted frequency frequency of the class 5 width of the class
Therefore, Adjusted frequency of 2545 8 52 20 =×=
16� (c) 57.5
As BCAC = , therefore triangle ABC is an isoscelestriangle.
Given ∠=∠=−= 115,18011565ACDACB (Linear Pair)
As ACBC = , therefore AB∠=∠ .
As sum of all the three angles of atriangle is 180 .
Therefore, 180 ABACB ∠+∠+∠=
57.5 AB ∠=∠=
17� (a) 12 77 ×
An irrational number between a and b is given by ab
So, an irrational number between 1 7 and 2 7 is 12 77 × .
18� (b) 10 cm
= 5 cm DE and = 7 cm AE
∴==×=22510 cm BDDE (∴ E is the mid-point of ) BD
19� (c) A is true but R is false.
3 , 2 k is a solution of 230 x +=
3 23330 2
3 , 2 k is the solution of 230 x += for all values of k .
Also 0 axb+= can be expressed as a linear equation in two variables as 00axyb+⋅+=
20� (a) Both A and R are true and R is the correct explanation of A.
In a parallelogram, if diagonals are equal, then it is a rectangle. In a rectangle, all angles are 90°. So, if diagonals of a parallelogram are equal, it must be a rectangle, and hence ∠ABC = 90°.
Both A and R are true and R is the correct explanation of A.
21� Radii of two cones are in the ratio of 2 :1 =
Let 12 , rr be the radii of two cones and 12 , hh be their respective heights.
Then, 1 2 2 1 r r =
Now, Volume of first cone
Volume of the second cone 2 11 2 22 1 3 1 3 rh rh π π = 2 2 1111 2 22 22 rhrh rhrh
2 11 22 4 2 1 hh hh
Their volumes are equal 1 2 4 1 h h ∴= 1 2 1 4 h h ⇒=
∴ Their ratio is 1:4 = OR
Internal radius of the hollow spherical shell, = 8 cm r
External radius of the hollow spherical shell, = 9 cm R
Weight of the shell = volume of the shell × density per cubic cm
2728 4.54092 g4.092 kg 3 =×≈=
Therefore, Weight of the shell 4.092 kg = .
22� Total number of students
= 500 + 350 + 400 + 750 + 500 + 100 + 400 = 3000.
(a) The given bar graph represents the number of students wearing shoes of different numbers out of a total of 3000 selected students.
(b) The students wear shoes bearing numbers 4, 5, 6, 7, 8, 9 and 10.
(c) The number of students wearing shoe No. 6 is 400.
(d) Shoe number 7 is worn by the maximum number of students. These students are 750 in number.
23� BE and CF are altitudes of a ΔABC.
∴∠BEC =∠CFB = 90°
Now, in right triangles BEC and CFB, we have Hypotenuse BC = Hypotenuse BC (Common)
Side BE = Side CF (Given)
∴ΔBEC ≅ΔCFB (By RHS congruence)
∴∠BCE =∠CBF (Corresponding parts of congruent triangles)
Now, in ΔABC, ∠B =∠C
∴AB = AC (Sides opposite to equal angles are equal)
Hence, ΔABC is an isosceles triangle. Hence, proved.
24� x = 0.47 = 0.4777 {as 7 is repeating}
We multiply both the sides by 10.
10x = 4.7
10x = 4.3 + 0.47
10x = 4.3 + x
10x – x = 4.3 4.343 9 990 x ==
25� In the above figure, we have
AB = AC + BC = AC + AC = 2AC (1) (Since, C is the mid-point of AB )
XY = XD + DY = XD + XD = 2XD (2) (Since, D is the mid-point of XY )
Also, AC = XD (Given) (3)
From (1), (2)and(3), we get
AB = XY, According to Euclid, things which are double of the same things are equal to one another.
26� In Figure, a histogram and a frequency table of the above frequency distribution are drawn on the same scale.
Number of Students
To construct a frequency polygon without using the histogram of a given frequency distribution, we use the following algorithm.
STEP-I: Obtain the frequency distribution.
STEP-II: Compute the mid-points of class intervals i.e. class marks.
STEP-III: Represent class marks on X-axis on a suitable scale.
STEP-IV: Represent frequencies on Y-axis on a suitable scale.
STEP-V: Plot the points, where x denotes class mark and f corresponding frequency.
STEP-VI: Join the points plotted in step V by line segments.
STEP-VII: Take two class intervals of zero frequency, one at the beginning and the other at the end. Obtain their mid-points. These classes are known as imagined classes.
STEP-VIII: Complete the frequency polygon by joining the mid-points of first and last class intervals to the mid-points of the imagined classes adjacent to them. OR
(a) It gives the information about the areas (in lakh hectors) under sugarcane crop during different years in India.
(b) The areas under the sugarcane crops were the maximum and minimum in 1982–83 and 1950–51 respectively.
(c) The area under sugarcane crop in the year 1982–83 = 34 lakh hectares.
The area under sugarcane crop in the year 1950–51 = 17 lakh hectares.
Clearly, the area under sugarcane crop in the year 1982–83 is not 3 times that of the year 1950–51.
So, the given statement is false.
27� Let us draw two perpendicular lines as the two main roads of the city that cross each other at the center. Let us mark them as North–South and East–West. As given in the question, let us take the scale as 1cm 200 m. = Draw five streets that are parallel to both the main roads (which intersect), to get the given below figure. Street plan is as shown in the figure:
Substitute in equation we get: () 4103210 y −−= 403210 y ⇒−−= 19 193 6.33 units 3 yy ⇒=⇒=≈
So, XBYD is a parallelogram (A pair of opposite sides equal and parallel) i.e., PXQD ‖
Therefore, APPQ = (1)
(From ΔAQD where X is mid-point of AD)
Similarly, from Δ= , CPBCQPQ (2)
Thus, APPQCQ == [From (1) and (2)]
32� Let, () 326310fxxxx=−++
The constant term in ()fx is 10
The factors of 10 are ±±±±1,2,5,10
Let, 10 x +=
1 x ⇒=−
Substitute the value of x in ()fx
()() 32 1(1)6(1)3110 f −=−−−+−+ 16310=−−−+
0 =
Similarly, () 2 x and () 5 x are other factors of ()fx
Since, ()fx is a polynomial having a degree 3, it cannot have more than three linear factors.
()()()() 125fxkxxx ∴=+−−
Substitute 0 x = on both sides
()()() 326310125 xxxkxxx ⇒−++=+−−
()()() 00010125 k ⇒−++=−−
()1010 k ⇒=
1 k ⇒=
Substitute 1 k = in ()()()() 125fxkxxx =+−−
()()()()() 1125fxxxx =+−−
so, ()()() 326310125 xxxxxx −++=+−−
This is the required factorisation of ()fx
33�
80 cm
Curved surface area of a hemisphere 2 2 r=π
40 r =
23.14(40)2
CSA =××=10,048 cm². 13 170 of the sheet is wasted, 13157
Usable sheet1 170170 =−= 157 Sheet usedCSA 170 =×
Assume that the area of fibre sheet used is equal to x. 157 170 10048 10048 10880 170 157 xxx ⇒=⇒=×⇒= Actual sheet used is approximately 10,880 cm². The cost is given as ` 35 per 100 cm², =× Actual Sheet Used Total Cost 35 100 × == 1088035 3808 100
The total cost of making the hemispherical dome is approximately ` 3,808.
34� B AC D 3 2 4 1
In right ΔABC, we have ABAC = CB⇒∠=∠ (1) (Angles opposite to equal sides are equal) 90 A ∴∠=° (2) In a triangle, the sum of all three angles is 180°.
180 ABC ∴∠+∠+∠=° 90 180 BB ⇒°+∠+∠=° [using (1) and (2)] 21809090 B ⇒∠=°−°=° ° ⇒∠==° 90 45 2 B 45 BC ⇒∠=∠=° (3) Since AD is the bisector of ∠A = 90°
∴∠=∠=×∠=×°=° 11 9045 22 BADCADA (4)
From (3) and (4), we have BADBABD∠=∠=∠ and CADCACD∠=∠=∠ BDAD⇒= and CDAD = Now, BCBDDCADAD =+=+ 2 BCAD⇒= Hence, proved.
35� A
C E B D O 55° 25° x°
Draw EO||AB||CD
Then, EOBEODx ∠+∠= Now, EOAB ‖ and BO is the transversal.
180 EOBABO ∴∠+∠= (Consecutive Interior Angles)
55180 EOB ⇒∠+=
125 EOB ⇒∠= Again, EOCD ‖ and DO is the transversal.
180 EODCDO ∴∠+∠= (Consecutive Interior Angles)
25180 EOD ⇒∠+=
155 EOD ⇒∠= Therefore, xEOBEOD =∠+∠ (125155) x =+ 280 x =
OR
EFCD ‖ and ED is the transversal.
180 FEDEDH ∴∠+∠= (co-interior angles)
65180 y ⇒+=
() 18065115 y ⇒=−=
Now CHAG ‖ and DB is the transversal 115 xy ∴== (corresponding angles)
Now, ABG is a straight line.
180 ABEEBG ∴∠+∠= (sum of linear pair of angles is 180 )
180 ABEx ⇒∠+=
115180 ABE ⇒∠+=
() 18011565 ABE ⇒∠=−=
We know that the sum of the angles of a triangle is 180 . From ΔEAB , we get
180 EABABEBEA ∠+∠+∠= 9065180 z ⇒++=
() 18015525 z ⇒=−=
115,115xy ∴== and 25 z =
36� (a) In Δ AOP and ΔBOP () ∠=∠ = = ∆≅∆ = (Given) (Common) (radius of circle)
APOBPO OPOP AOOB AOPBOP APBPCPCT
(b) In right ∆COQ
222 COOQCQ =+ 108222 CQ ⇒=+ 2 1006436 CQ ⇒=−= 6 CQ ⇒= CD 2CQ = ⇒= 12 cm CD
(c) In right ∆AOB
222 AOOPAP =+ 106222 AP ⇒=+
2 1003664 AP ⇒=−= 8 AP ⇒= 2 ABAP = ⇒= 16 cm AB OR
There is one and only one circle passing through three given non-collinear points.
37� (a) The cost incurred for the farmer to grow Type A rice and Type B rice were `12 per kg and `10 per kg. Hence, total cost = `() 1210xy +
(b) The selling price of Type A rice and Type B rice were `18 per kg and `14 per kg.
Hence, total selling price = `() 1814xy +
(c) Cost price = `() 1210xy +
Selling price = `() 1814xy +
Total profit =()()() +−+=+ 18141210 64 xyxyxy ` OR
This section consists of 20 questions of 1 mark each�
1� ABCD is a trapezium in which ABDC ‖ . M and N are the mid-points of AD and BC respectively. If 12 AB = cm,14 cm MN = , then CD = (a) 10 cm (b) 14 cm (c) 12 cm (d) 16 cm
2� In figure, for which value of x is 12ll ‖ ? (a) 43 (b) 37 (c) 45 (d) 47
3� In the given figure, P and Q are centers of two circles intersecting at B and C . ACD is a straight line. Then, the measure of BQD∠ is (a) 130
(b) 150
(c)
(d)
4� The value of 33 22 (0.013)(0.007) (0.013)0.0130.007(0.007) + −×+ , is (a) 0.0091 (b) 0.02
0.006 (d) 0.00185
5� The length of the sides of a triangle are 5 cm,7 cm and 8 cm. Area of the triangle is:
6� Abscissa of a point is negative in (a) quadrant IV only (b) quadrant II and III (c) quadrant I and IV (d) quadrant I only
7� In a bar graph if 1 cm represents 30 km, then the length of bar needed to represent 75 km is (a) 3.5 cm
8� A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. The volume of the solid is
9� Each equal side of an isosceles triangle is 13 cm and its base is 24 cm Area of the triangle is:
10� In a trapezium , ABCDE and F be the midpoints of the diagonals AC and BD respectively. Then, EF = ?
11� If 32 32 x = + and 32 32 y + = , then 22xxyy++= (a) 102 (b) 101 (c) 99 (d) 98
12� The value of 2 53 is
13� Express ‘x ’ in terms of ‘y ’ in the equation 2350 xy−−= .
14� In the figure, O is the centre of the circle. If 25 OPQ ∠= and 20 ORQ ∠= , then the measures of POR∠ and PQR∠ are respectively:
15� The value of pqqrrpxxx is equal to (a) xpqr (b) 0 (c) x (d) 1
16� In the given figure, the value of x which makes POQ a straight line is:
2x - 30 (a) 40 (b) 30 (c) 35 (d) 25
17� If 52 7 2 3816561 3 3 x x ×× = , then x = (a) 1 3 (b) 3 (c) -3 (d) 1 3
18� In the given figure, AOB is a diameter of a circle and CDAB ‖ . If 30 BAD ∠= , then CAD ∠= ?
(a) 45 (b) 60 (c) 50 (d) 30
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
19� Assertion (A): 0.271 is a terminating decimal and we can express this number as 271 1000 which is of the form p q , where p and q are integers and 0 q ≠ .
Reason (R): A terminating or non-terminating decimal expansion can be expressed as rational number.
20� Assertion (A): A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm.
Reason (R): Through three collinear points a circle can be drawn.
SECTION B
This section has 5 very short answer type questions of 2 marks each�
21� If 322 x =+ , find the value of 2 2 1 x x + . OR
Prove that: 1111 1 37755331 +++= ++++ .
22� Seema has a 10 m10 m × kitchen garden attached to her kitchen. She divides it into a 1010 × grid and wants to grow some vegetables and herbs used in the kitchen. She puts some soil and manure in that and sows a green chilly plant at A, a coriander plant at B and a tomato plant at C. Her friend Kusum visited the garden and praised the plants grown there. She pointed out that they seem to be in a straight line. See the below diagram carefully and answer the following questions:
(a) Write the coordinates of the points A, B, and C, taking the 1010 × grid as coordinate axes.
(b) By distance formula or some other formula, check whether the points are collinear.
23� Insert five rational numbers between 2 3 and 3 4 . OR
Evaluate: 1 1 0 5 (32)(7)(64)2 +−+
24� The surface areas of two spheres are in the ratio 1:4 . Find the ratio of their volumes. OR
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
25� Look at the figure. Show that length AH > sum of lengths of AB + BC + CD.
SECTION C
This section has 6 short answer type questions of 3 marks each�
26� If both ( 2 x ) and 1 2 x are factors of 2 5 pxxr ++ , Show that pr = .
27� Find four solutions for the following equation: 1250 xy+=
28� Locate 10 on the number line.
29� In the given figure, ABC and DBC are two triangles on the same base BC such that ABAC = and DBDC = . Prove that ABDACD∠∠ =
OR
In given figure, AD and BE are respectively altitudes of a triangle ABC such that AEBD = . Prove that ADBE = .
30� In figure, if ABDE ‖ and BDFG ‖ such that 125 FGH ∠= and 55 B ∠= , find x and y.
31� The weight distribution (in kg) of 100 people is given below.
Construct a histogram for the above distribution.
OR
The length of 40 leaves of a plant are measured correct to one millimeter, and the data obtained is represented in the following table:
(a) Draw a histogram to represent the given data.
(b) Is there any other suitable graphical representation for the same data?
(c) Is it correct to conclude that maximum leaves are 153 mm long? Why?
SECTION D
This section has 4 long answer questions of 5 marks each.
32. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BADABE∠=∠ and EPADPB∠=∠ (See the given figure). Show that
(a) DAPEBP∆≅∆
(b) ADBE = OR
Line l is the bisector of A∠ and B is any point on l. BP and BQ are perpendiculars from B to the arms of A∠ (see the given figure).
Show that:
(a) APBAQB∆≅∆
(b) BPBQ = or B is equidistant from the arms of A∠ .
33. The polynomial () 4322337pxxxxaxa =−+−+− when divided by 1 x + leave remainder 19. Find the remainder when ()px is divided by 2 x +
34. A conical tent is 10 m high and the radius of its base is 24 m. Find the (a) slant height of the tent.
(b) cost of the canvas required to make the tent, if the cost of 2 1 m canvas is `70.
35. In the given figure, ,40,35ABCDABOCDO ∠=∠= ‖ . Find the value of the reflex BOD∠ and hence the value of x.
In the given figure, ABCD ‖ . Prove that 180 pqr+−= .
SECTION E
This section has 3 case-study based questions of 4 marks each.
36. Read the text carefully and answer the questions:
There is a Diwali celebration in the a school in New Delhi. Girls are asked to prepare Rangoli in a triangular shape. They made a rangoli in the shape of triangle ABC. Dimensions of ABC∆ are 26 cm, 28 cm 25 cm.
(a) In figure R and Q are mid-points of AB and AC respectively. Find the length of RQ.
(b) Find the length of Garland which is to be placed along the side of ∆QPR.
(c) (i) , RP and Q are the mid-points of , ABBC , and AC respectively. Then find the relation between area of PQR∆ and area of ABC∆ .
(ii) ,, RPQ are the mid-points of corresponding sides ,, ABBCCA in ABC∆ , then name the figure so obtained BPQR.
37. Read the following text carefully and answer the questions that follow:
In Agra in a grinding mill, there were installed 5 types of mills. These mills used steel balls of radius 5 mm, 7 mm, 10 mm, 14 mm and 16 mm respectively. All the balls were in the spherical shape.
For repairing purpose mills need 10 balls of 7 mm radius and 20 balls of 3.5 mm radius. The workshop was having 3 20000 mm steel.
This 3 20000 mm steel was melted and 10 balls of 7 mm radius and 20 balls of 3.5 mm radius were made and the remaining steel was stored for future use.
(a) What was the volume of one ball of 3.5 mm radius?
(b) What was the surface area of one ball of 3.5 mm radius?
(c) (i) What was the volume of 10 balls of radius 7 mm ?
(ii) How much steel was kept for future use?
38. Raj, studying in class 9, goes to his school every day by walking. There are two routes from his house to the school. One of the routes connects his house to the school directly, while the other takes a sharp left-turn around a garden. Both routes are shown in the graph below.
(a) Find the distance from Raj’s house to the garden.
(b) Find the distance from garden to Raj’s school.
(c) (i) Find the shortest distance from Raj’s house to his school.
(ii) If Raj decides to travel by Auto rickshaw at the fare of ` 10/km , then how much will it cost to travel from his house to the garden?
Sample Question Paper 2
SOLUTION
1� (d) 16 cm
Given, ABCD is a trapezium ABDC ‖
M, N are mid points of AD & BC 12 cm,14 cm ABMN==
ABMNCD ‖‖ [M, N are mid points of &]ADBC
Now, ACD is straight line, so, ∠ACB +∠DCB = 180°.
18075105 DCB ∠=−=
Now, angle subtended by arc BD on centre is twice of 2105210 DCB ∠=×=
Now, 360210150 BQD ∠=−=
4� (b) 0.02
Assume 0.013 a = and 0.007 b = . Then the given expression can be rewritten as 33 22 ab aabb + −+
Recall the formula for sum of two cubes () () 3322 ababaabb +=+−+
MPCD = and 1 2 NPAB =
MPNP = By mid point theorem, 1 2
() 1 2 MNABCD∴=+
() 1 1412 2 CD ⇒=+
281216 cm CD ⇒=−=
2� (d) 47
Let if 12ll ‖ and AB is tranverse to it
Then, PBA∠ should be equal to BAS∠ (Alternate angles)
So if 12ll ‖ , then 70 BAS ∠= 783543 BAC∠ ⇒=−= (i)
Now, in ABC∆
C180xBAC ∠∠ °++=° 9043180 x ° ⇒°+°+=° 180904347 x ⇒°=°−°−°=° 47 x ⇒°=°
So, if 47 x °=° then 12ll ‖
3� (b) 150 C 150° B AP D Q
150 APB ∠= , so, 75 ACB ∠= {Angle subtended by an arc at centre is twice the angle subtended at any point on circumference}
Using the above formula, the expression becomes () () 22 22 abaabb aabb +−+ −+
Note that both a and b are positive. So, neither 33ab + nor any factor of it can be zero.
Therefore we can cancel the term () 22aabb −+ from both numerator and denominator. Then the expression becomes () () 22 22 abaabb ab aabb +−+ =+ −+
0.0130.007=+ 0.02 =
5� (b) 2 103 cm 578 10 cm 2 s ++ ==
Area of triangle ()()() ssasbsc =−−− ()()() 10105107108 =−−− 10532=××× 103 = cm2
6� (b) quadrant II and III
The abscissa (x-axis) is -ve in 2nd and 3rd quadrant only because, Sign of point in 2nd quadrant is () , −+ , and in 3rd quadrant, it is () , .
7� (b) 2.5 cm
1 cm30 km = So, for 75 km
75 2.5 cm 30 =
1 cm 1 cm
Radii of cone 1 cm r ==
Radius of hemisphere () 1 cm1 cm rh ===
Height of cone () 1 1 cm hh==
Volume of solid = Volume of cone + Volume of a hemisphere () 121232 2 333 rhrrhr πππ =+=+ () 2 1 (1)121 3 π =××+× 3 1 3cm 3 ππ=××=
9� (c) 2 60 cm 131324 25 cm 2 s ++ ==
Area of triangle ()()() ssasbsc =−−− ()()() 25251325132524=−−− 2512121=×××
60sq.cm =
10� (c) () 1 2 ABCD
Construction: Join CF and extent it to cut AB at point M
Firstly, in triangle MFB and triangle DFC
DFFB = (As F is the mid-point of DB)
DFCMFB∠=∠ (Vertically opposite angle)
DFCFBM∠=∠ (Alternate interior angle)
∴ By ASA congruence rule
MFBDFC∆≅∆
Now, in triangle CAM
E and F are the mid-points of AC and CM respectively () 1 2 EFAM∴= () 1 2
EFABMB =− () 1 2 EFABCD =−
11� (c) 99
Given 32 32 x = + and 32 32 y + = , Consider, 32 32 x = + 3232 3232 =× +− 2 22 (32) (3)(2) = + () () 22 (3)(2)232 32 +− = 3226 1 +− = 526=−
Hence, 526 x =− 22 (526) x ⇒=− 2524206=+− 49206=− i.e 2 49206 x =− (i) Again consider 32 32 y + = 3232 3232 ++ =× −+ 2 22 (32) (3)(2) + = () () 22 (3)(2)232 32 ++ = 3226 1 ++ = 526=+
Hence, 526 y =+ 22 (526) y ⇒=+ 2524206=++ 49206=+ i.e. 2 49206 y =+ (ii) Then, 22xxyy ++ = 3232 49206 49206 3232 −+ −+×++ +− [from (i) and (ii)] = 98 + 1 = 99
12� (c) 53 +
Explanation: 2 53
multiplying nunominator and denominator by 53 + , we get () () () 253 5353 + −+ ()253 53 53 + ==+
0.271 is a terminating decimal and we can express this number as 271 1000 which is of the form p q , where p and q are integers and q ≠ 0. This statement is correct. 0.271 is a terminating decimal and 271 1000 is a rational number (in the form p q ).
A terminating or non-terminating decimal expansion can be expressed as a rational number.
This statement is incorrect.
Only terminating or non-terminating recurring decimals can be expressed as rational numbers.
Non-terminating non-recurring decimals (like π or 2 ) are irrational numbers, not rational.
20� (c) A is true, but R is false.
Assertion is true, because a circle of radius 3 cm can be drawn which has AB as its diameter.
Reason is not true because a circle through two points cannot pass through a point which is collinear to these two points.
21� Given, 322 x =+ () 11 322 x ∴= +
() () 322 1 322322 =× +−
22 322 (3)(22) =
37755331 2 −+−+−+− = 2 2 == 1 = RHS
22� (a) A(2, 2) B(5, 4) C(7, 6) (b) 22 (52)(22) AB =−+− 94=+ 13 = 22 (75)(64) BC =−+−
44=+ 22 = 22 (72)(62) AC =−+−
2516=+ 41 = 1322 ABBC+=+ 41 AC = ABBCAC∴+≠ ,, ABC ∴ are not collinear 23� 2248 33412 −×− −== × 3339 44312 × == × So, the five rational numbers between 2 3 and 3 4 are 8765 ,,, 12121212 and 4 12 OR 1 1 5 (32)(7)(64)2 +−+° ()[] 1 5 5 21881 a ++×°= 1 5 5 218 × =++ 218=++ = 11
24� Suppose that the radii of the spheres are r and R . We have: 2 2 41 4 4 r R π π = 11 42 r R ⇒==
Therefore, The ratio of the volumes of the spheres is 1:8.
Inner radius of hemispherical tank (r ) = 1 m = 100 cm
Thickness of sheet 1 cm =
∴ Outer radius of hemispherical tank (R) = 100 + 1 = 101 cm
222 (101)(100)cm
[] 2 44 10303011000000cm 21 = 2 0.06348 m =
25� From the given figure, we have [, ABBCCDADABSC ++= and CD are the parts of ] AD
Here, AD is also the parts of AH . By Euclid’s axiom, the whole is greater than the part. i.e., AHAD > .
Therefore, length AH > sum of lengths of ABBCCD ++ .
26� Suppose, () 2 5 pxpxxr =++
As () 2 x is a factor of ()px
20 p ∴= () 2 (2)520 pr ⇒++= 4100 pr ⇒++= (1)
Again, 1 2 x
is factor of ()px 1 0 2 p
Now, 2 111 5 222 ppr
15
42pr=++
∴=⇒++=
115 0 0 242 ppr
From equation (1), we have 410 pr+=−
From equation (2), we have 1040pr++=
410pr ⇒+=−
44 prpr∴+=+ [∴ Each 10]=−
33prpr ∴=⇒=
Hence, proved.
27� 1250 xy+=
512yx⇒=− 12 5 yx⇒=
Put 0 x = , then () 12 00 5 y ==
Put 5 x = , then () 12 512 5 y ==−
Put 10 x = , then () 12 1024 5 y ==−
Put 15 x = , then () 12 1536 5 y ==−
()()() 0,0,5,12,10,24∴−− and () 15,36 are the four solutions of the equation 1250 xy+=
28� We can write 10 as 10913122 =+=+
Draw 3 OA = units, on the number line
Draw 1 BA = unit, perpendicular to OA
Join OB
Figure:
(2)
Now, by Pythagoras theorem, 222 OBABOA =+ 2221310 OB =+=
10 OB ⇒=
Taking O as centre and OB as a radius, draw an arc which intersects the number line at point C. Clearly, OC corresponds to 10 on the number line.
29� In ∆ABC, AB = AC (Given)
∴∠ABC =∠ACB (angles opposite to equal sides are equal)
Similarly in ∆DBC, DB = DC (Given) (1)
∴∠DBC =∠DCB (2)
Adding (1) and (2)
∠ABC +∠DBC =∠ACB +∠DCB or ∠ABD =∠ACD OR
In ∆DPDB and ∆DPEA,
∠PDB =∠PEA (Each 90°)
∠BPD =∠APE (Vertically opposite angles)
AE = BD (Given)
∴∆DPDB ≅∆DPEA (By AAS property)
∴ PA = PB (1)
(Corresponding part of congruent triangle)
PD = PE (2)
(Corresponding part of congruent triangle)
PA + PD = PB + PE
⇒ AD = BE (By adding (1) and (2))
30� Here, as AB || DE and BD is the transversal, so according to the property, “alternate interior angles are equal”, we get
DB∠∠ =
55 D ∠= (1)
Similarly, as BD || FG and DF is the transversal DF∠∠ =
55 F ∠= (Using 1)
Further, EGH is a straight line. So, using the property angles forming a linear pair are supplementary
180 FGEFGH
125180 y
55 y ∴=
Also, using the property, “an exterior angle of a triangle is equal to the sum of the two opposite interior angles”, we get, In EFG∆ with FGH∠ as its exterior angle
ext. FGHFE ∠∠∠ =+ 12555 x =+
12555 x
70 x
Thus, 70 x =
and 55 y =
31� (a) We represent the weights on the horizontal axis. We choose the scale on the horizontal axis
as 1 cm5 kg = . Also, since the first class interval is starting from 35 and not zero, we show it on the graph by marking a kink or a break on the axis.
(b) We represent the number of people (frequency) on the vertical axis. Since the maximum frequency is 28, we choose the scale as 1 cm5 = people.
(c) We now draw rectangles (or rectangular bars) of width equal to the class size and lengths according to the frequencies of the corresponding class intervals.
(a) Length of leaves are represented in a discontinuous class intervals having a difference of 1 mm in between them. So we have to add 1 0.5 mm 2 = to each upper class limit and also have to subtract 0.5 mm from the lower class limits so as to make our class intervals continuous.
3
5
4 171.5180.5 2
Now taking length of leaves on the x -axis and number of leaves on the y -axis, we can draw the histogram of this information as below:
Here 1 unit on the y-axis represents 2 leaves.
(b) Other suitable graphical representation of this data could be frequency polygon.
(c) No, as maximum numbers of leaves (i.e. 12) have their length in between of 144.5 mm and 153.5 mm. It is not necessary that all have their lengths as 153 mm.
32� Given that ∠EPA =∠DPB
⇒∠EPA +∠DPE =∠DPB +∠DPE
⇒∠DPA =∠EPB (1)
Now, in ∆DAP and ∆EBP
∠DAP =∠EBP (given)
AP = BP (P is the mid-point of AB)
∠DPA =∠EPB [From (1)]
∴∆DAP ≅∆EBP (ASA congruence rule)
∴ AD = BE (Corresponding parts of congruent triangle)
OR
In ∆APB and ∆AQB,
∠APB =∠AQB (each 90°)
∠PAB =∠QAB (l is angle bisector of ∠A)
AB = AB (common)
∴∆APB ≅∆AQB (by AAS congruence rule)
∴ BP = BQ (Corresponding parts of congruent triangle)
Or we can say that B is equidistant from the arms of ∠A.
33� We know that if ()px is divided by xa + , then the remainder ()pa=− .
Now, () 4322337pxxxxaxa =−+−+− is divided by 1 x + , then the remainder () 1 p =−
Now, ()() 432 1(1)2(1)3(1)137 paa −=−−−+−−−+−
()() 1213137 aa =−−+++−
12347 a =+++−
14a =−+
Also, remainder = 19 1449 a ∴−+= 420;2045 aa ⇒==÷=
Again, when ()px is divided by 2 x + , then Remainder ()() 432 2(2)2(2)3(2)237 paa =−=−−−+−−−+− 161612237 aa =++++−
375a =+
() 3755372562 =+=+=
34� (a) Height (h) of conical tent 10 m =
Radius (r) of conical tent 24 m =
Let the slant height of conical tent 1 = 222222 (10m)(24m)676m rhr=+=+=
26 m l ∴=
Thus, the slant height of the conical tent is 26 m. (b) CSA of a tent
22 22 13728 2426m m 77 rlπ ==××=
Cost of 2 1 m canvas = `70
Then, cost of 2 13728 m 7 canvas = ` 13728 70 7 ×= `137280.
Thus, the cost of canvas required to make the tent is `137280.
(Angles on the same side of a transversal line are supplementary)
Also, PFQCD ‖
PFGFGDr ∠=∠= (Alternate Angles) and EFPEFGPFGqr ∠=∠−∠=− putting the value of EFP∠ in equation (1) we get,
(a) We know that line joining mid points of two sides of triangle is half and parallel to third side.
Hence RQ is parallel to BC and half of BC.
28 14 cm 2 RQ ==
Length of RQ = 14 cm
(b) By mid-point theorem we know that line joining mid points of two sides of triangle is half and parallel to third side.
25 12.5 cm
Length of garland 12.5141339.5 cm PQQRRP =++=++=
Length of garland 39.5 cm = .
(c) (i) As R and P are mid-points of sides AB and BC of the triangle ABC, by mid point theorem, RPAC ‖ Similarly, RQBC ‖ and PQAB ‖ . Therefore , ARPQBRQP and RQCP are all parallelograms. Now RQ is a diagonal of the parallelogram ARPQ, therefore, ARQPQR∆≅∆ Similarly. CPQRQP∆≅∆ and BPRQRP∆≅∆ so, all the four triangles are congruent.
Therefore, Area of ARQ ∆= Area of CPQ ∆= Area of BPR ∆= Area of PQR∆
Area ABC ∆= Area of ARQ ∆+ Area of CPQ ∆+ Area of BPR ∆+ Area of PQR∆
Area of 4 ABC ∆= Area of PQR∆ () 1 4 PQRarABC∆= OR
(ii) As R and Q are mid-points of sides AB and AC of the triangle ABC . Similarly, P and Q are mid points of sides BC and AC by mid-point theorem, RQBC ‖ and PQAB ‖ . Therefore BRQP is parallelogram
37� (a) The radius of the ball 3.5 mm = Volume of the ball 3 4 3 r=π
(b) Radius of one ball 3.5 cm = The surface area of one ball
2 4 r=π 22 43.53.5 7 =××× 2 154 mm =
(c) (i) Radius of one ball 7 cm = Thus volume of 10 balls of radius 7 mm 3 4 10 3 r=×π 3 422 107 37 =××× 3 14373.3 mm = OR
(ii) Volume of 10 balls of 7mm = 14373.3 mm3
Volume of 1 ball of 3 3.5 mm179.66 mm =
Volume of 20 balls of 3 3.5 mm179.66203593.33 mm =×=
Total steel required to be melted 3 14373.33593.3317966 mm =+= (Approx)
Thus, steel left over 3 20,000179662034 mm =−=
38� (a) The distance from Raj’s house to the garden is 4 km.
(b) The distance from garden to Raj’s school is 4 km.
(c) (i) By applying Pythagoras theorem, we get the shortest distance from Raj’s house to his school 22 443216242 km =+==×= OR
(ii) Distance from Raj’s house to the garden 4 km = Auto rickshaw fare = `10/km
Hence, travelling cost from his house to garden = ` () 410×= `40
This section consists of 20 questions of 1 mark each�
1� How many linear equations in ‘x ’ and ‘y ’ can be satisfied by 1,2xy== ? (a) Infinitely many (b) Two (c) Only one (d) Three
2� After rationalising the denominator of 7 3322 , we get the denominator as (a) 5 (b) 35 (c) 19 (d) 13
3� The point which lies on x -axis at a distance of 3 units in the positive direction of x -axis is (a) (0, 3) (b) (0, 3) (c) (3, 0) (d) ( 3, 0)
4� ABCD is a parallelogram. If is produced to E such that ED bisects BC at O. Then which of the following is correct? (a) OCBE = (b) OEOC = (c) ABOE = (d) ABBE =
5� In the given figure, O is the centre of a circle and chords AC and BD intersect at E. If 110 AEB ∠= and CBE∠ 30 = , then ADB ∠=?
6� In the given figure PQ and RS are two equal chords of a circle with centre OOA and OB are perpendiculars on chords PQ and RS, respectively. If 140 AOB ∠= , then PAB∠ is equal to
60
(b)
7� The value of x in +=+ 1 1 3 2 32(64)(27) x is
(a) 14 (b) 8 (c) 5 (d) 3
8� Each side of an equilateral triangle measures 8 cm. The area of the triangle is (a) 2 323 cm (b) 2 48 cm (c) 2 163 cm (d) 2 83 cm
9� 3 is a polynomial of degree.
(a) 0 (b) 2 (c) 1 2 (d) 1
10� A point is at a distance of 3 units from the x -axis and 7 units from the y -axis. Which of the following may be the co-ordinates of the point?
(a) (7, 3) (b) (3,7) (c) (4, 5) (d) (0, 0)
11� In figure ABCD is a parallelogram. If 60 DAB ∠= and 80 DBC ∠= then CDB∠ is
(a) 60 (b) 40
12� 1 3 (125) = ? (a) 1 5 (b) –5 (c) 1 5 (d) 5
13� x co-ordinate is known as (a) Origin (b) Points (c) Abscissa (d) Ordinate
14� If a linear equation has solutions ()() 1,2,1,16 and () 0,7 , then it is of the form (a) 97yx=− (b) 970 xy−+= (c) 97xy−= (d) 97xy=−
15� In figure, what is the value of x ?
y °
y ° 7y ° x ° (a) 60 (b) 35 (c) 45 (d) 50
16� The value of 2 2 111 xxx xxx
17� In the given figure, O is the centre of a circle. If 100 AOB ∠= and 90 AOC ∠= , then BAC ∠= ?
(a) 75 (b) 95 (c) 85 (d) 80
18� If () 2,5 is a solution of 211 xmy+= , then the value of ‘m’ is (a) –2 (b) 2 (c) 3 (d) –3
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
19� Assertion (A): 26 + is an irrational number.
Reason (R): Sum of a rational number and an irrational number is always an irrational number.
20� Assertion (A): In ∆ , ABCE and F are the midpoints of AC and AB respectively. The altitude AP at BC intersects FE at Q. Then, AQQP =
Reason (R): Q is the midpoint of AP.
SECTION B
This section has 5 very short answer type questions of 2 marks each�
21� In the given figure, we have ∠=∠∠=∠,43 ABCACB . Show that 12∠=∠ . A D
22� For what value of c , the linear equation 28 xcy+= has equal values of x and y for its solution?
23� If O is the centre of the circle, find the value of x in the given figure:
In figure, OA and OB are respectively perpendiculars to chords CD and EF of a circle whose centre is O If OA = OB, prove that ECDF ≅ .
24� A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
OR
The surface areas of two spheres are in the ratio of 4:25 . Find the ratio of their volumes.
25� In figure, if = ACBD , then prove that = ABCD .
SECTION C
This section has 6 short answer type questions of 3 marks each�
26� In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles.
OR
BN and CM are perpendiculars to a line passing through the vertex A of a triangle ABC. If L is the mid point of BC, Prove that LM = LN
BLC
27� One side of an equilateral triangle is 8 cm. Find its area by using Heron’s Formula. Find its altitude also.
OR
The sides of a triangle are in the ratio of 13:14:15 and its perimeter is 84 cm. Find the area of the triangle.
28� Factorize the polynomial: 81263322 ababab −−+
29� Find the remainder when 32331xxx+++ is divided by 52. x +
30� If 25 25 a + = and 25 25 b = + , then find the value of 22ab .
31� A random survey of the number of children of various age groups playing in a park was found as follows:
Draw a histogram to represent the data above.
SECTION D
This section has 4 long answer questions of 5 marks each�
32� The length of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.
OR
Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find the area of the field.
33� Verify that:
34� In a line segment AB point C is called a mid-point of line segment AB, prove that every line segment has one and only one mid-point.
35� What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m?
Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use = 3.14 π )
SECTION E
This section has 3 case-study based questions of 4 marks each�
36� Read the text carefully and answer the questions:
Ajay lives in Delhi, The city of Ajay’s father in laws residence is at Jaipur is 600 km from Delhi. Ajay used to travel this 600 km partly by train and partly by car.
He used to buy cheap items from Delhi and sell at Jaipur and also buying cheap items from Jaipur and sell at Delhi.
Once From Delhi to Jaipur in forward journey he covered 2x km by train and the rest y km by taxi.
But, while returning he did not get a reservation from Jaipur in the train. So first 2y km he had to travel by taxi and the rest x km by Train. From Delhi to Jaipur he took 8 hrs but in returning it took 10 hrs .
(a) Write the above information in terms of equation.
(b) Find the value of x and y ?
(c) Find the speed of Taxi?
Find the speed of Train?
OR
JAIPUR
37� Read the text carefully and answer the questions:
Child labour refers to any work or activity that deprives children of their childhood. It is a violation of children’s rights. This can them mentally or physically. It also exposes them to hazardous situations or stops them from going to school. Naman got data on the number of child labors (in million) in different country that is given below.
(a) What is the difference between highest no child labor and the minimum no of child labor?
(b) What is the percentage of no. of child labor in Peru over the no. of child labor in India?
(c) What is the total no. of child labor in the countries having child labor more than 2 million?
How many countries are having child labor more than Mexico?
38� Read the text carefully and answer the questions:
Once four friends Rahul, Arun, Ajay and Vijay went for a picnic at a hill station. Due to peak season, they did not get a proper hotel in the city. The weather was fine so they decided to make a conical tent at a park. They were carrying 2 300 m cloth with them. As shown in the figure they made the tent with height 10 m and diameter 14 m. The remaining cloth was used for the floor.
(a) How much Cloth was used for the floor?
(b) What was the volume of the tent?
(c) What was the area of the floor?
What was the total surface area of the tent?
Also, ABCD = (ABCD is parallelogram)
Sample Question Paper 3
SOLUTIONS
1� (a) Infinitely many
There are many linear equations in ‘x’ and ‘y’ can be satisfied by 1,2xy==
For example; 3 xy+= 1 xy−=−
24 xy+= and so on there are infinite number of examples
2� (d) 60
Explanation: Produce OP to intersect RQ at point N.
Now, OP‖RS and transversal RN intersects them at N and R respectively.
RNPSRN∴∠=∠ (Alternate interior angles)
130 RNP ⇒∠= 18013050 PNQ ∴∠=−= (Linear pair)
OPQPNQPQN ∠=∠+∠ (Exterior angle property)
11050 PQN⇒=+∠
1105060 PQNPQR⇒∠=−==∠
3� (c) (3, 0)
Since it lies on x-axis so ordinate will be zero because the value of the y-coordinate in the x-axis is equal to zero.
Thus point will be (3, 0).
4� (d) ABBE = A DC 2 4 1 3 O BE
In the figure, ∆BCD is a parallelogram, where AB is produced to E such that OCOB =
Using identity () 333 ()3 xyxyxyxy −=−−− with respect to the expression () 33 (2)()322 ababab −−××− , we get (2)3 ab
Therefore, after factorizing the expression 81263322 ababab −−+ , we get (2)3 ab
29� We need to find the zero of the polynomial 52x + 520 x += 5 2 x ⇒=−
While applying the remainder theorem, we need to put the zero of the polynomial 52x + in the polynomial 32331xxx+++ , to get () 32331pxxxx=+++ 32 5555 331 2222 p
We need to find the zero of the polynomial 52x + 520 x += 5 2 x ⇒=−
While applying the remainder theorem, we need to put the zero of the polynomial 52x + in the polynomial 32331xxx+++ , to get () 32331pxxxx=+++ 32 5555 331 2222 p −=−+−+−+
1252515 31 842 =−+−+
1257515 1 842 =−+−+ 125150608 8 −+−+ = 27 8 =−
31� Age (in years)
Number of Children (frequency) Width of the Class Length of the Rectangle
32� Given, perimeter 144 cm = and ratio of the sides
3:4:5 =
Sum of ratio terms 34512=++=
∴ 1st side, =×= 3 14436 cm 12 a
IInd side, =×= 4 14448 cm 12 b
IIIrd side, c 5 14460 cm 12 =×=
Now, semi-perimeter of the triangle, ++++ ==== 364860144 72 cm 222 abc s
Area of triangle ()()() ssasbsc =−−− [by Heron’s formula]
()()() 72723672487260 =×−−− 22 72362412(36)(24)=×××=× 2 3624864 cm =×=
Hence, the area of the given triangle is 2 864 cm
Let height of a triangle be h cm.
Then, area of triangle 1 2 =× Base × Height
1
86460 2 h ⇒=×× [Since the longest side of a triangle is 60 cm, so we consider it as base of the triangle]
86430 h ⇒=
⇒= 28.8 cm h
Hence, the height corresponding to the longest side is 28.8 cm.
OR
Let: = 85 m a and = 154 m b
Given that perimeter 324 m =
Perimeter 2324 m s == 324 m 2 s ⇒=
324 abc++= r,
⇒=−−=3248515485 m c
By Herons’s formula, we have:
Area of triangle ()()() ssasbsc =−−− ()()() 1621628516215416285 =−−− 16277877=××× 12967777=×× 36777736=×××
3677=× 2 2772 m =
33� L.H.S. is 333 3 xyzxyz ++− and R.H.S. is () 222 1 ()()() 2 xyzxyyzzx++−+−+−
We know that ()++−=++333 3 xyzxyzxyz () ++−−−222 xyzxyyzzx
Therefore, we can conclude that the desired result is verified.
34� We need to prove that every line segment has one and only one mid-point. Let us consider the given below line segment AB and assume that C and D are the mid-points of the line segment AB. ACDB
If C is the mid-point of line segment AB, then = ACCB .
An axiom of the Euclid says that “If equals are added to equals, the wholes are equal.”
+=+ ACACCBAC (i)
From the figure, we can conclude that + CBAC will coincide with AB.
An axiom of the Euclid says that “Things which coincide with one another are equal to one another.” += ACACAB (ii)
An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.”
Let us compare equations (i) and (ii), to get += ACACAB, or = 2 ACAB . (iii)
If D is the mid-point of line segment AB, then = ADDB .
An axiom of the Euclid says that “If equals are added to equals, the wholes are equal.”
+=+ ADADDBAD (iv)
From the figure, we can conclude that + DBAD will coincide with AB
An axiom of the Euclid says that “Things which coincide with one another are equal to one another.”
+= ADADAB (v)
An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.”
Let us compare equations (iv) and (v), to get
+= ADADAB , or = 2 ADAB (vi)
An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.”
Let us compare equations (iii) and (vi), to get = 22ACAD .
An axiom of the Euclid says that “Things which are halves of the same things are equal to one another.” AC = AD.
Therefore, we can conclude that the assumption that we made previously is false and a line segment has one and only one midpoint.
35� Height of the conical tent () 8 m h = and Radius of the conical tent () 6 m r =
Slant height of the tent () 22lrh =+ (6)(8)22=+
3664=+
100 = 10 m =
Area of tarpaulin = Curved surface area of tent = =××= 2 3.14610188.4 m rlπ
Width of tarpaulin 3 m =
Let Length of tarpaulin L =
∴ Area of tarpaulin = Length × Breadth
33 LL=×=
Now According to question, 3L 188.4 = 188.4 62.8 m 3 L ⇒==
The extra length of the material required for stitching margins and cutting is 20 cm0.2 m = .
So the total length of tarpaulin bought is () 62.80.2m63 m +=
36� (a) Delhi to Jaipur: += 2600 xy
Jaipur to Delhi: += 2600 yx
Let 1S and 2 S be the speeds of Train and Taxi respectively, then
Dehli to Jaipur: += 12 2 8 xy SS (i)
Jaipur to Delhi: 12 2 10 xy SS += (b) += 2600 xy
+=2600xy (i)
Solving (1) and (2) 2 ×
+−−=− 2246001200 xyxy
⇒−=−3600 y
⇒= 200 y
Put = 200 y in (1)
2200600 x +=
⇒== 400 200 2 x
(c) We know that speed = Distance Time ⇒ Time Distance Speed =
Let 1S and 2 S are speeds of train and taxi respectively.
Delhi to Jaipur: += 12 2 8 xy SS (i)
Jaipur to Delhi: += 12 2 10 xy SS (ii)
Solving (i) and (ii) 2 × 1212 4 22 82012 yyxx SSSS ⇒+−−=−=−
2 3 12 y S ⇒=−
We know that y200 km =
2 3200 S 50 km/hr 12 × ⇒==
Hence speed of Taxi 50 km/hr = OR
We know that = 200 km x
Put = 2 S50 km/hr (i)
1 400200 8 50 S += 1 400 844 S ⇒=−=
1 400 S100 km/hr 4 ⇒==
Hence speed of Train 100 km/hr =
37� (a) The highest no child labor are in India and the lowest no child labor are in United states
No of child labor in India = 20,000,000
No of child labor in United states 8,00,000 =
The difference 20,000,0008,00,000 = 19,200,000 =
(b) No. of child labor in Peru 4,000,000 = No. of child labor in India 20,00,000 =
The percentage 4000000 10020% 20000000 =×=
(c) The countries having child labor more than 2 million are
Egypt 3 = million
Brazil 3.5 = million
Peru = 4 million
Bangladesh = 4.4 million
India = 20 million
Total no of these labor child 33.544.42034.9 =++++= million. OR
The countries having child labor more than Mexico are:
Philippines = 1.8 million
Egypt = 3 million
Brazil = 3.5 million
Peru = 4 million
Bangladesh = 4.4 million
India = 20 million
Thus 6 countries are having child labor more than Mexico.
38� (a) Height of the tent = 10 m h
Radius = 7 cm r
Thus Latent height =+=+==222271014912.20 m lrh
Curved surface of tent ==××= 2 22 712.2268.4 m 7 rlπ
Thus the length of the cloth used in the tent 2 268.4 m = The remaining cloth 2 300268.431.6 m =−= Hence the cloth used for the floor 2 31.6 m =
(b) Height of the tent = 10 m h
Radius = 7 cm r
Thus the volume of the tent 2 1 3 rh=π 122 7710 37 =×××× 3 513.3 m =
(c) Radius of the floor 7 m =
Area of the floor 2 22 77 7 rπ ==×× 2 154 m = OR
Radius of the floor = 7 m r
Latent height of the tent l12.2 m =
Thus total surface area of the tent ()rrlπ =+ () 22 7712.2 7 =×+ 2219.2=× 2 422.4 m =
This section consists of 20 questions of 1 mark each�
1� The histogram shows information about the weights, in grams, of some plums.
The estimate for the proportion of these plums with a weight of less than 100 grams is
2� The value of 1.9999 in the form p q , where ‘p’ and ‘q’ are integers and q ≠ 0, is
3� If a diagonal AC and BD of a quadrilateral ABCD bisect each other, then ABCD is a (a) Parallelogram (b) Rhombus (c) Rectangle (d) Triangle
4� In the given figure, AB || CD, CD || EF and y : z = 3 : 7, then x = ?
(a) 63° (b) 126° (c) 108° (d) 162°
5� In the figure, ∠DAB = 60°, ∠ABD = 50°, then ∠ACB is equal to:
(a) 80° (b) 60° (c) 50° (d) 70°
6� In a histogram, which of the following is proportional to the frequency of the corresponding class? (a) Width of the rectangle (b) Length of the rectangle (c) Perimeter of the rectangle (d) Area of the rectangle
7� Which of the following point does not lie on the line y = 2x + 3?
(a) (-5, -7) (b) (-1, 1) (c) (3, 9) (d) (3, 7)
8� x = 2, y = 5 is a solution of the linear equation (a) 5x + y = 7 (b) x + y = 7 (c) 5x + 2y = 7 (d) x + 2y = 7
9� The congruence rule, by which the two triangles in the given figure are congruent is. 8cm 8cm
Q P (a) ASA (b) SAS (c) SSS (d) RHS
10� The curved surface area of a cylinder and a cone is equal. If their base radius is same, then the ratio of the slant height of the cone to the height of the cylinder is (a) 1 : 1 (b) 2 : 3 (c) 1 : 2 (d) 2 : 1
11� AOB is the diameter of the circle. If ∠AOE = 150°, then the measure of ∠CBE is
(a) 115° (b) 125° (c) 120° (d) 105°
12� The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, then ∠DPC = (a) 70° (b) 80° (c) 90° (d) 100°
13� In the figure AB and CD are two straight lines intersecting at O, OP is a ray. What is the measure of ∠AOD.
14� The zeros of the polynomial p(x) = 3x 2 - 1 are
(a) 1 and 3 3 (b) () 1 1 and 33 (c) () 1 and 3 3 (d) 1 and 3 3
15� The graph of the linear equation 2x + 3y = 6 is a line which meets the x-axis at the point (a) (0, 3) (b) (3, 0) (c) (2, 0) (d) (0, 2)
16� To draw a histogram to represent the following frequency distribution: Class interval 5-10 10-15 15-25 25-45 45-75 Frequency 6 12 10 8 15 The adjusted frequency for the class 25–45 is (a) 6 (b) 5 (c) 2 (d) 3
17� The linear equation 3x - 5y = 15 has
(a) no solution (b) infinitely many solutions (c) a unique solution (d) two solutions
18� The cost of a notebook is twice the cost of a pen. The equation to represent this statement is (a) 2x = 3y (b) x = 3y (c) x + 2y = 0 (d) x - 2y = 0
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
19� Assertion (A): If 3 , 2 k is a solution of the linear equation 2x + 3y = 0, then k = 1
Reason (R): The linear equationa x + b = 0 can be expressed as a linear equation in two variables as ax + y + b = 0.
20� Assertion (A): Two diameters of a circle intersect each other at right angles. Then, the quadrilateral formed by joining their end-point is a rectangle.
Reason (R): Angle in a semi-circle is a right angle.
SECTION B
This section has 5 very short answer type questions of 2 marks each�
21� The curved surface area of a conical vessel is 10 times its slant height. Find the diameter of the vessel.
22� Find whether (2, 0) is the solution of the equation x - 2y = 4 or not? OR
Find whether the given equation have x = 2, y = 1 as a solution: 2x - 3y = 1
23� In the given figure, we have BX = 1 2 AB and BY = 1 2 BC and AB = BC. Show that BX = BY
24� Read the bar graph. Find the percentage of excess expenditure on wheat than pulses and ghee taken together.
25� Point C is called a mid point of line segment AB, prove that every line segment has one and only one mid-point.
SECTION C
This section has 6 short answer type questions of 3 marks each�
26� Represent the following frequency distribution by means of a histogram.
27� AB is a line segment and P is the mid-point. D and E are points on the same side of AB such that ∠BAD =∠ABE and ∠EPA =∠DPB. Show that:
ABC is an isosceles triangle in which AB = ACBE and CF are its two medians. Show that BE = CF
28� If x + y + z = 0, show that x 3 + y 3 + z 3 = 3xyz.
29� ABCD is a square and DEC is an equilateral triangle. Prove that AE = BE.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
30� Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
31� The total cost of making a spherical ball is `33,957 at the rate of `7 per cubic metre. What will be the radius of this ball? 22 Use 7 π
This section has 4 long answer questions of 5 marks each�
32� Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR (see the given figure). Show that:
(a) ∆ABM ≅∆PQN
(b) ∆ABC ≡∆PQR
33� Draw a histogram to represent the following grouped frequency.
Also draw frequency polygon.
34� In the adjoining figure, name:
(a) Six points
(b) Five line segments
(c) Four rays
(d) Four lines
(e) Four collinear points
35� Represent each of the numbers 5,6 and 7 on the real line. OR
Simplify: 732532 . 103651532 +++
SECTION E
This section has 3 case-study based questions of 4 marks each�
36� Decimal form of rational numbers can be classified into two types. Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form p q , where p and q are co-prime and the prime factorization of q is of the form 2n 5m, where n, m are non-negative and vice-versa.
Let x = p q be a rational number, such that the prime factorization of q is not of the form 2n 5m, where n and m are non-negative integers. Then x has a nonterminating repeating decimal expansion. Base on the above information, answer the following questions.
(a) Check whether 23 81 523 ×× has terminating decimal expansion or not?
(b) Check whether 222 441 257 ×× has terminating decimal expansion or not?
37� Read the following text carefully and answer the questions that follow: Once upon a time in Ghaziabad was a corn cob seller. During the lockdown period in the year 2020, his business was almost lost. So, he started selling corn grains online through Amazon and Flipcart. Just to understand how many grains he will have from one corn cob, he started counting them. Being a student of mathematics let’s calculate it mathematically. Let’s assume that one corn cob (see figure), shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length as 20 cm.
(a) Find the curved surface area of the corn cub.
(b) What is the volume of the corn cub?
(c) (i) If each 1 cm2 of the surface of the cob carries an average of four grains, find how many grains you would find on the entire cob?
OR
(ii) How many such cubs can be stored in a cartoon of size 20 cm × 25 cm × 20 cm.
38� Read the following text carefully and answer the questions that follow: Ajay is writing a test which consists of ‘True’ or ‘False’ questions. One mark is awarded for every correct answer while 1 4 mark is deducted for every wrong answer. Ajay knew answers to some of the questions. Rest of the questions he attempted by guessing.
(a) If he answered 110 questions and got 80 marks and answer to all questions, he attempted by guessing were wrong, then how many questions did he answer correctly?
(b) If he answered 110 questions and got 80 marks and answer to all questions, he attempted by guessing were wrong, then how many questions did he guess?
(c) (i) If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks he got?
OR
(ii) If answer to all questions he attempted by guessing were wrong, then how many questions answered correctly to score 95 marks?
This section consists of 20 questions of 1 mark each�
1� The curved surface area of one cone is twice that of the other while the slant height of the later is twice that of the former. The ratio of their radii is (a) 2 : 1 (b) 4 : 1 (c) 8 : 1
2� In the figure, ABCD is a Parallelogram. The values of x and y are
3� If a linear equation has solutions (-2, 2), (0, 0) and (2, -2), then it is of the form:
4� The decimal form of a rational number 13 20 is (a) 0.655 (b) 0.560 (c) 0.620 (d) 0.650
5� The sides BC, CA and AB of ∆ABC have been produced to D, E and F respectively. ∠BAE +∠CBF +∠ACD = ?
6� The value of () 2 4 64 is (a) 1 2 (b) 1 8 (c) 1 16 (d) 1 4
7� The total surface area of a sphere with radius 7 cm is (a) 616 cm2 (b) 626 cm2 (c) 516 cm2 (d) 416 cm2
8� The simplest form of 0.57 is (a) 26 45 (b) 57
(c) 57
(d) 57
9� In the given, AB is side of regular five sided polygon and AC is a side of a regular six sided polygon inscribed in the circle with centre 0. AO and CB intersect at P, then ∠APB is equal to
(a) 90° (b) 72° (c) 86° (d) 96°
10� A semicircular thin sheet of a metal of diameter 28 cm is bent and an open conical cup is made. The capacity of the cup is S
11� Which of the following identity can be applied for evaluating 203 × 205? (a) (x + a)(x - a) (b) a 2 - b 2 (c) (x + a)(x + b) (d) (a - b)2
12� In ∆PQR, ∠R =∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is (a) 2.5 cm (b) 4 cm (c) 5 cm (d) 2 cm
13� In the figure AB, CD and EF are three Straight lines intersecting at O. The measure of ∠AOF is
(a) 82° (b) 152° (c) 54° (d) 98°
14� If the area of an isosceles right triangle is 8 cm2, what is the perimeter of the triangle? (a) 8 + 4 2 cm2 (b) 8 + 2 cm2 (c) 12 2 cm2 (d) 4 + 8 2 cm2
15� In a parallelogram, (a) diagonals bisect each other (b) diagonals are perpendicular to each other (c) diagonals are of equal length (d) diagonals are angle bisectors
16� For the equation 5x - 7y = 35, if y = 5, then the value of ‘x’ is (a) 12 (b) -12 (c) -14 (d) 14
17� Ordinate of all points on the y-axis is (a) 0 (b) -1 (c) any number (d) 1
18� Let ABCD be a parallelogram. The diagonals bisect at E. If DE = 5 cm and AE = 7 cm, then find AC
(a) 10 cm (b) 14 cm (c) 5 cm (d) 7 cm
Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
19� Assertion (A): If a = 3 and b = 4, then the value of ba is 64.
Reason (R): If a = 4 and b = 5, then the value of a 2 + b 2
20� Assertion (A): In given figure, if AB || CD, ∠EAB = 110° and ∠AEC = 30°, then ∠DCE = 140°.
Reason (R): If two parallel lines are intersected by a transversal, then each pair of alternate angles are equal.
SECTION B
This section has 5 very short answer type questions of 2 marks each�
21� Study the following graph and answer the question given below.
(a) In which years is the highest and lowest ever voters turn-out (in %)?
(b) For which two years, the numeric difference in voters turn-out (in %) was nearly equal to 10%?
22� Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find the radius of the base.
OR
If the radius and slant height of a cone are in the ratio 7 : 13 and its curved surface area is 286 cm2, find its radius.
23� In a given figure, two congruent circles with centres O and O' intersect at A and B. If ∠AOB = 50°, then find
24� Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D, P, Q respectively (see figure). Prove that ACP = QCD.
25� Find two rational and two irrrational numbers between 0.5 and 0.55. OR
Express the decimal 0.235 in the form p q , where p, q are integers and q ≠ 0.
SECTION C
This section has 6 short answer type questions of 3 marks each�
26� The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface at the rate of `210 per 100 m2 14 m 25m OR
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
27� Draw the graphs of y = x and y =-x in the same graph. Also find the co-ordinates of the point where the two lines intersect.
28� Find at least 3 solutions for the linear equation 2x - 3y + 7 = 0.
29� ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that,
(a) SR || AC and SR = 1 2 AC
(b) PQ = SR
(c) PQRS is a parallelogram. S R C D
30� If AE = AD and BD = CE. Prove that
31� The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of `5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
SECTION D
This section has 4 long answer questions of 5 marks each�
32� Draw a histogram to represent the following frequency distribution.
Marks
33� In each of the figures given below, AB || CD. Find the value of x° in each other case.
In figure, ∠ABC = 65°, ∠BCE = 30°, ∠DCE = 35° and ∠CFE = 145°. Prove that AB || EF.
34� The perimeter of a triangular field is 420 m and its sides are in the ratio 6 : 7 : 8. Find the area of the triangular field.
The perimeter of a right triangle is 24 cm . If its hypotenuse is 10 cm, find the other two sides. Find its area by using the formula area of a right triangle. Verify your result by using Heron’s formula.
35� A hemispherical dome, open at base is made of sheet of fiber. If the diameter of hemispherical dome is 80 cm and 13 170 of sheet actually used was wasted in making the dome, then find the cost of dome at the rate of `35 per 100 cm2
SECTION E
This section has 3 case-study based questions of 4 marks each
36� Two new roads, Road E and Road F were constructed between Society 4 and 1 and Society 1 and 2, respectively. Society 1
Based on the above information, answer the following questions:
(a) What would be the measure of the sum of angles formed by the straight roads at Society 1 and Society 3?
(b) Krish says, ‘The distance to go from Society 4 to Society 2 using Road D will be longer than the distance using Road E’. Is Krish correct? Justify your answer with examples.
(c) (i) If the length of Road B is equal to Road D, what can you say about the arcs they subtend? OR
(ii) Priya said, ‘Minor arc corresponding to Road B is congruent to minor arc corresponding to Road D. Do you agree with Priya? Give reason to support your answer.’
37� A vendor sells ice cream in a cone-shaped wafer with a hemispherical scoop of ice cream on top. The radius of the cone and scoop is 3.5 cm, and the height of the cone is 10 cm.
Answer these question based on above information:
(a) What is the slant height of the cone used in the above ice cream?
(b) (i) What is the total surface area of the ice cream cone?
OR
(ii) What is the total volume of ice cream, including the cone and the scoop?
(c) If the cost of 1 cm3 of ice cream is `0.50, what is the total cost of one filled ice cream (Round to nearest whole number)?
38� Read the following text carefully and answer the questions that follow:
Peter, Kevin James, Reeta and Veena were students of Class 9th B at Govt Sr. Sec School, Sector 5, Gurgaon.
Once the teacher told Peter to think a number x and to Kevin to think another number y so that the difference of the numbers is 10(x > y).
Now the teacher asked James to add double of Peter’s number and that three times of Kevin’s number, the total was found 120. Reeta just entered in the class, she did not know any number.
The teacher said Reeta to form the 1st equation with two variables x and y
Now Veena just entered the class so the teacher told her to form 2nd equation with two variables x and y.
Now teacher told Reeta to find the values of x and y. Peter and kelvin were told to verify the numbers x and y.
(a) What are the equation formed by Reeta and Veena?
(b) What was the equation formed by Veena?
(c) (i) Which number did Peter think?
(ii) Which number did Kelvin think?
OR
About the Book
LUMO is a comprehensive and student-friendly one-stop resource for Mathematics (CBSE subject code 041) for Grades IX–X. It offers a brief overview of the theory and a wide variety of practice questions, designed to promote conceptual clarity, core skill development and exam readiness. The book presents the entire NCERT curriculum in a clear, accessible, and exam-oriented format to support students at every stage of their learning journey.
Special Features
• Integrated AI tutor for evaluation, effective feedback and doubt clarification
• Includes latest solved CBSE papers for 2024–25
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• Concise chapter summaries and key concepts for quick revision
• Objective, short answer, long answer and competency-based questions
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• Suitable for classroom instruction, self-study and revision
