Pythagorean Theorem Practice Pythagorean Theorem Practice The Pythagorean theorem is related to the study of sides of a right angled triangle. It is also called as pythagoras theorem. The pythagorean theorem states that, In a right triangle, (length of the hypotenuse)2 = {(1st side)2 + (2nd side)2}. In a right angled triangle, there are three sides: hypotenuse, perpendicular and base. The base and the perpendicular make an angle of 90 degree with eachother. So, according to pythagorean theorem: (Hypotenuse)2 = (Perpendicular)2 + (Base)2 In the above figure1, c2 = a2 + b2 Therefore, Hypotenuse (c) = √ (a2 + b2) Pythagorus Theorem ProofBack to Top From the above figure 2, Δ ABC is a right angled triangle at angle C. From C put a perpendicular to AB at H. Now consider the two triangles Δ ABC and Δ ACH, these two triangles are similar to each other because of AA similarity. This is because both the triangle have a right angle and one common angle at A. Know More About Riemann Sum Calculator

Tutorvista.com

Page No. :- 1/5

So by these similarity, ac = ea and bc = db a2 = c*e and b2 = c*d Sum the a2 and b2, we get a2 + b2 = c*e + c*d a2 + b2 = c(e + d) a2 + b2 = c2 (since e + d = c) Hence Proved. Euclid Proof of Pythagorean Theorem According to Euclid, if the triangle had a right angle (90 degree), the area of the square formed with hypotenuse as the side will be equal to the sum of the area of the squares formed with the other two sides as the side of the squares. From the above figure 3, the sum of the area covered by the two small squares is equal to the area of the third square. Here, a2 is the area of the square ABDE, b2 is the area of the square BCFG and c2 is the area of the square ACHI. Therefore, a2 + b2 = c2 Hence Proved. Pythagorean Theorem Example Problems Below are example problems on Pythagorean theorem Example Problem 1: In a right triangle, the hypotenuse is 5 cm and the perpendicular is 4 cm. Find the length of the base of the triangle? Solution: By using Pythagoras theorem, h2 =p2 + b2 52 = 42 + b2, 25 = 16 + b2, 9 = b2, b = 3

Base is 3 cm Learn More Calculus Calculator

Tutorvista.com

Page No. :- 2/5

Definition of Circle Definition of Circle Circle is defined as the set of points that is at an equal distant from the centre of the circle.There are a number of terminologies involved in a Circle. Some of them are as follows: Centre: The predetermined point from which the surface of the circle is at an equidistant is called the centre of a circle. Radius: The constant distance from the centre to a point on the surface of the circle is called its radius . Circumference: The boundary of a circle is called its circumference. Chord: A line segment whose end points is present on the circumference of a circle is called a chord . Diameter: A chord crossing through the midpoint of a circle is called its diameter. Circle Formulas Diameter of a Circle: Diameter = 2 X Radius

Tutorvista.com

Page No. :- 3/5

Radius of a Circle: Radius(R) = Diameter / 2 Area of a Circle: Area = pi X R2 Circumference of a Circle: Circumference = 2 X pi X R Circle Theorem Theorem 1: A perpendicular from the centre of a circle to a chord bisects the chord. Given : AB is a chord in a circle with centre O. OC ⊥ AB. To prove: The point C bisects the chord AB. Construction: Join OA and OB Proof: In triangles OAC and OBC, m∠OCA = m∠OCB = 90 (Given) OA = OB (Radii) OC = OC (common side) ∠OAC = ∠OBC (RHS) CA = CB (corresponding sides) The point C bisects the chord AB. Hence the theorem is proved. Theorem 2: AB and CD are equal chords of a circle whose centre is O. OM ⊥ AB and ON ⊥ CD. Prove that m∠OMN = m∠ONM. Given : In a circle with centre O chords AB and CD are equal OM ⊥ AB, ON ⊥ CD (Fig.6.11). To prove : ∠OMN = ∠ONM Proof : AB = CD (given) OM ⊥ AB (given); ON ⊥ CD (given) Read More About Decimal Calculator Tutorvista.com

Page No. :- 4/5

ThankÂ You

TutorVista.com