Year 10 Geometry and Proofs - Questions and worked solutions - Level 1

Year 10

Geometry and Proofs Test Questions Level 1

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This document, including but not limited to all tests, questions, answers, and any accompanying materials, is protected by copyright law and is provided exclusively for the use of teaching staff.

The content herein is intended to assess specific curriculum outcomes as outlined in the Mathematics K –10 Syllabus (2023) and is aligned with the subtopics covered within the designated Mathspace textbook.

Permitted Use: - This material is to be used solely by the teaching staff of a Mathspace customer school for the purpose of assessing students who are officially enrolled and have procured the necessary textbook.

- Teachers are authorised to edit, copy, and adjust the test questions for the sole purpose of tailoring the assessment to their students' needs. All modifications must retain the original purpose of assessing curriculum-specific outcomes.

- Sharing of this material is restricted to teaching staff within a Mathspace customer school. It may not, under any circumstances, be disclosed or distributed to students or any third parties outside the teaching staff.

Prohibited Use: - Reproduction of this material, in part or whole, for any purposes other than those explicitly permitted herein is strictly prohibited.

- Distribution of this material to students or individuals not employed by a Mathspace customer school as teaching staff is forbidden.

- Use of this material for any commercial purposes, or in any setting outside of a Mathspace customer school, is not allowed.

Special Notes: - Each test contains a table that identifies the curriculum outcomes it addresses and references the relevant subtopic from the textbook. This ensures alignment with the educational goals and facilitates ease of use for teachers.

- Fully worked solutions and marking rubrics are included to aid in the accurate assessment of student performance.

- A table is provided at the end of each set of worked solutions, indicating the overlapping questions among tests of varying difficulty levels. This is intended to assist in scaling and equitably assessing students undertaking different levels of difficulty.

By using this material, you agree to adhere to the terms outlined above. Violation of these terms may result in disciplinary action and could lead to legal proceedings.

For any inquiries or further clarification on permitted use, please contact Mathspace at info@mathspace.com.au

Mathspace Year 10 2024 NSW Curriculum Chapter 14: Geometric Proofs Test 1

1 Core

Name:

Remember to show all your working!

Marks

For each of the biconditional statements, explain whether they are true or false. If false, justify your answer with a counterexample. a. A number is odd if and only if it is not a multiple of two.

7

b. A quadrilateral is a rectangle if and only if it has two pairs of parallel sides.

c. Angles are complementary if and only if they are adjacent.

d. A number is a perfect square if and only if its square root is a whole number.

e. A triangle is a right triangle if and only if the sum of the squares of the two shorter sides is equal to the square of the longest side.

2 Core

Complete the proof for the theorem: If ∠𝐴 and ∠𝐵 are equal and complementary, then ∠𝐴 = ∠𝐵 = 45° ∠𝐴 = ∠𝐵 (___________) ( ____________________ ) ∠𝐴 + ∠𝐵 = 90° ∠𝐴 + _____ = 90° (Substitute ∠𝐵 = ∠𝐴) _____ = 90° (Collect like terms) ( ____________________ ) ∠𝐴 = 45° _____ = 45° (Substitute ∠𝐴 = ∠𝐵) ∴∠𝐴 = ∠𝐵 = 45°

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3

3 Core

In the diagram below, 𝐴𝐶||𝐷𝐺, ∠𝐵𝐸𝐷 = 127° and ∠𝐶𝐵𝐹 = 62°. Prove that:

6

a. ∠𝐴𝐵𝐸 = 53°

b. ∠𝐸𝐵𝐹 = 65°

4 Core

Using the diagram:

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a. Determine if there is enough information to justify the conclusion that ∠2 = ∠12 if 𝑏||𝑐 and ∠1 = ∠5.

1

b. Prove that 𝑎 is parallel to 𝑏 if 𝑏||𝑐 and ∠3 = ∠11.

3

5 Core

In the diagram below, 𝑀𝑃 is a straight line. What additional information is needed to prove ∆𝑀𝑁𝑄 and ∆𝑃𝑁𝑄 are congruent by:

2

a. Right angle-Hypotenuse-Side congruence?

b. Angle-Side-Angle congruence?

6 Core

Prove that ∆𝑇𝑈𝑉≡∆𝑊𝑋𝑉 if 𝑇𝑊 and 𝑈𝑋 are straight lines.

7 Core

Using the diagram: a. Explain why we can’t be certain that ∆𝐴𝐵𝐶~∆𝐷𝐸𝐹.

b. Give three examples of extra information which could be used to prove the triangles are similar, and the relevant similarity test in each case.

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4

2

3

8 Core

Show that the triangles are similar.

9 Core

Quadrilateral 𝑀𝑁𝑂𝑃 has 𝑀𝑁||𝑃𝑂 and 𝑀𝑃||𝑁𝑂. C is the point at which the diagonals 𝑀𝑂 and 𝑁𝑃 intersect.

i.

∠𝑀𝑁𝑂

ii.

∠𝑁𝑂𝑃

4

a. What type of quadrilateral is 𝑀𝑁𝑂𝑃?

1

b. If 𝑀𝑂 = 10 cm, find the length of 𝐶𝑂.

1

c. If ∠𝑀𝑃𝑂 = 112°, find:

2

d. Give three examples of extra information which would show that 𝑀𝑁𝑂𝑃 is a rhombus.

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3

10 Core

The diagram below shows a right triangle on a Cartesian plane. Point 𝑀 is the midpoint of 𝑂𝑄 and point 𝑁 is the midpoint of 𝑃𝑄. a. Find the co-ordinates of 𝑀 and 𝑁.

b. Show that ∆𝑁𝑂𝑃 is isosceles.

c. Show that 𝑀𝑁 is parallel to 𝑂𝑃

11 Opt

Show that 𝑥 = 5. 5

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2

4

3

7

11a

In the diagram, O is the centre of the circle. Solve for x. Show all reasoning.

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3

12 Opt

The diagram below shows quadrilateral 𝑂𝐽𝐾𝐿 on the Cartesian plane. Prove that 𝑂𝐽𝐾𝐿 is a parallelogram.

6

13

AG and GH are tangents to the circle with centre F. Solve for FH.

3

Total Marks = /70

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Question Subtopic Outcome 1

14.01

MA5-GEO-P-01.5

2

14.01

MA5-GEO-P-01.5

3

14.02

MA5-GEO-P-02.1

4

14.02

MA5-GEO-P-02.1

5

14.03

MA5-GEO-P-02.1

6

14.03

MA5-GEO-P-02.1

7

14.04

MA5-GEO-P-02.2

8

14.04

MA5-GEO-P-02.2

9

14.05

MA5-CIR-P-01.1

10

14.05

MA5-CIR-P-01.1

11

14.04

MA5-GEO-P-02.2

12

14.06

MA5-CIR-P-01.1

13

14.05

MA5-CIR-P-01.1

14

14.07

MA5-CIR-P-01.2

MA5-GEO-P-01.5 Apply logical reasoning to numerical problems involving plane shapes Contributes to: MA5-GEO-P-01 establishes conditions for congruent triangles and similar triangles and solves problems relating to properties of similar figures and plane shapes (Path: Ext) MA5-GEO-P-02.1

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Construct formal proofs involving congruent and similar triangles Contributes to: MA5-GEO-P-02 constructs proofs involving congruent triangles and similar triangles and proves properties of plane shapes (Path: Ext)

MA5-GEO-P-02.2 Apply logical reasoning to proofs involving plane shapes Contributes to: MA5-GEO-P-02 constructs proofs involving congruent triangles and similar triangles and proves properties of plane shapes (Path: Ext)

MA5-CIR-P-01.1 Prove and apply angle and chord properties of circles Contributes to: MA5-CIR-P-01 applies deductive reasoning to prove circle theorems and solve related problems (Path: Ext) MA5-CIR-P-01.1 Prove and apply angle and chord properties of circles Contributes to: MA5-CIR-P-01 applies deductive reasoning to prove circle theorems and solve related problems (Path: Ext) MA5-CIR-P-01.2 Prove and apply tangent and secant properties of circles Contributes to: MA5-CIR-P-01 applies deductive reasoning to prove circle theorems and solve related problems (Path: Ext)

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Year 10

Geometry and Proofs Worked Solutions Level 1

Copyright © 2024 Mathspace Pty Ltd All rights reserved.

This document, including but not limited to all tests, questions, answers, and any accompanying materials, is protected by copyright law and is provided exclusively for the use of teaching staff. The content herein is intended to assess specific curriculum outcomes as outlined in the Australian Curriculum v9 and is aligned with the subtopics covered within the designated Mathspace textbook.

Permitted Use: - This material is to be used solely by the teaching staff of a Mathspace customer school for the purpose of assessing students who are officially enrolled and have procured the necessary textbook.

- Teachers are authorized to edit, copy, and adjust the test questions for the sole purpose of tailoring the assessment to their students' needs. All modifications must retain the original purpose of assessing curriculum-specific outcomes.

- Sharing of this material is restricted to teaching staff within a Mathspace customer school. It may not, under any circumstances, be disclosed or distributed to students or any third parties outside the teaching staff.

Prohibited Use: - Reproduction of this material, in part or whole, for any purposes other than those explicitly permitted herein is strictly prohibited.

- Distribution of this material to students or individuals not employed by a Mathspace customer school as teaching staff is forbidden.

- Use of this material for any commercial purposes, or in any setting outside of a Mathspace customer school, is not allowed.

Special Notes: - Each test contains a table that identifies the curriculum outcomes it addresses and references the relevant subtopic from the textbook. This ensures alignment with the educational goals and facilitates ease of use for teachers.

- Fully worked solutions and marking rubrics are included to aid in the accurate assessment of student performance.

- A table is provided at the end of each set of worked solutions, indicating the overlapping questions among tests of varying difficulty levels. This is intended to assist in scaling and equitably assessing students undertaking different levels of difficulty.

By using this material, you agree to adhere to the terms outlined above. Violation of these terms may result in disciplinary action and could lead to legal proceedings.

For any inquiries or further clarification on permitted use, please contact Mathspace at info@mathspace.com.au

Mathspace Year 10 2024 NSW Curriculum Chapter 14: Geometric Proofs Test 1 Worked Solutions

1

a. True

(1 mark)

State that it is true.

b. False (1 mark) A parallelogram has two pairs of parallel sides but is not a rectangle.

(1 mark)

State that it is false.

Give a valid counterexample.

c. False (1 mark) In the diagrams below (not to scale): ● ∠𝐴𝐵𝐶 is adjacent to ∠𝐶𝐵𝐷 but they are not complementary ● ∠𝐴𝐵𝐶 is complementary to ∠𝐸𝐹𝐺 but they are not adjacent

State that it is false.

Give a valid counterexample.

(1 mark for a single counterexample) State that it is true.

2

d. True

(1 mark)

e. True

(1 mark)

∠𝐴 = ∠𝐵 (Given) ∠𝐴 + ∠𝐵 = 90° (Complementary angles) ∠𝐴 + ∠𝐴 = 90° (Substitute ∠𝐵 = ∠𝐴) 2∠𝐴 = 90° (Collect like terms) ∠𝐴 = 45° (Divide both sides by 2) ∠𝐵 = 45° (Substitute ∠𝐴 = ∠𝐵) ∴∠𝐴 = ∠𝐵 = 45° (0.5 mark for each completed blank) Copyright © 2024 Mathspace Pty Ltd

State that it is true.

3

a. 𝐴𝐶||𝐷𝐺 (Given) ∠𝐵𝐸𝐷 = 127° (Given) (0.5 mark) ∠𝐴𝐵𝐸 + ∠𝐵𝐸𝐷 = 180° (Co-interior angles) (1 mark) ∠𝐴𝐵𝐸 + 127° = 180° (Substitute ∠𝐵𝐸𝐷 = 127°) (1 mark) ∠𝐴𝐵𝐸 = 53° (Subtract 127° from both sides) (0.5 mark) b. ∠𝐶𝐵𝐹 = 62° (Given) ∠𝐴𝐵𝐸 + ∠𝐸𝐵𝐹 + ∠𝐶𝐵𝐹 = 180° (Supplementary angles) (1 mark) 53° + ∠𝐸𝐵𝐹 + 62° = 180° (Substitute ∠𝐴𝐵𝐸 = 53° and ∠𝐶𝐵𝐹 = 62°) (1 mark) ∠𝐸𝐵𝐹 + 115° = 180° (Evaluate the sum) (0.5 mark) ∠𝐸𝐵𝐹 = 65° (Subtract 115° from both sides) (0.5 mark)

4

5

6

a. Yes

(1 mark)

b. 𝑏||𝑐 (Given) ∠3 = ∠11 (Given) ∠7 = ∠11 (Corresponding angles) ∠3 = ∠7 (Substitute ∠7 = ∠11) ∴𝑎||𝑏 (Corresponding angles)

(0.5 mark) (1 mark) (1 mark) (0.5 mark)

a. 𝑀𝑁 = 𝑁𝑃

(1 mark)

b. ∠𝑀𝑁𝑄 = ∠𝑃𝑁𝑄

(1 mark)

𝑈𝑉 = 𝑋𝑉 𝑇𝑈||𝑋𝑊 ∠𝑇𝑉𝑈 = ∠𝑊𝑉𝑋 ∠𝑈𝑇𝑉 = ∠𝑋𝑊𝑉

(Given) (Given) (1 mark) (Vertically opposite angles) (1 mark) (Alternate angles) (1 mark)

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State the relevant given information.

Use parallel lines relationships to make links between known and unknown angles.

Use known angle(s) to write an equation which can be solved for ∠𝐴𝐵𝐸.

Make the conclusion.

State the relevant given information.

Use angle relationships to make links between known and unknown angles.

Use known angle(s) to find ∠𝐸𝐵𝐹, showing working.

Make the conclusion.

By opposite angles ∠2 = ∠4. ∠1 = ∠5 means that 𝑎||𝑏 (corresponding angles).

Given 𝑏||𝑐 then 𝑎||𝑐 and ∠4 = ∠12 (corresponding angles). By substitution ∠2 = ∠12.

State the relevant given information.

Use parallel lines relationships and given information to make links between angles.

Make the conclusion.

From the diagram ∠𝑀𝑄𝑁 = ∠𝑃𝑄𝑁 = 90° (R) and 𝑁𝑄 is common (S), so to satisfy the RHS test the hypotenuses must be of equal length.

From the diagram ∠𝑀𝑄𝑁 = ∠𝑃𝑄𝑁 = 90° (A) and 𝑁𝑄 is common (S), so to satisfy the ASA test the angles on the other side of the common side must be equal.

State the relevant given information.

Use angle relationships and parallel lines relationships to make links between angles.

Alternatively, ∠𝑇𝑈𝑉 = ∠𝑊𝑋𝑉 ∴∆𝑇𝑈𝑉≡∆𝑊𝑋𝑉 (AAS)

(Alternate angles) (1 mark)

Conclude that the triangles are congruent, stating the relevant test.

a. There is information given for only one pair of corresponding sides and one pair of angles.

Note that information is given for 2 pairs (1 mark) of sides/angles only.

Each similarity test requires information about three Note that each test requires information pairs of sides or angles.

(1 mark) for 3 pairs of sides/angles.

7

b.

𝐴𝐶 𝐷𝐹 𝐴𝐵 𝐷𝐸

=2

(RHS)

=2

(SAS)

∠𝐵𝐴𝐶 = ∠𝐸𝐷𝐹 ∠𝐴𝐶𝐵 = ∠𝐷𝐹𝐸

(AAA) (AAA)

(1 mark each for any three of the above) 8

15 10 45 30 75 50

= 1. 5

(1 mark)

Find the ratio of the shortest sides

= 1. 5

(1 mark)

Find the ratio of the middle sides

= 1. 5

(1 mark) So the triangles are similar by SSS.

9

a. Parallelogram

Find the ratio of the longest sides (1 State the similarity test. mark)

(1 mark)

1

b. 𝐶𝑂 = 2 𝑀𝑂 1

= 2 (10) = 5 cm

(1 mark)

c. ∠𝑀𝑁𝑂 = 112°

i.

(1 mark)

∠𝑁𝑂𝑃 = 180° − 112° = 68° d. 𝑀𝑁 = 𝑀𝑃 𝑁𝑂 = 𝑂𝑃 𝑀𝑂 ⊥ 𝑁𝑃 𝑀𝑂 bisects ∠𝑁𝑀𝑃 or ∠𝑁𝑂𝑃 𝑁𝑃 bisects ∠𝑀𝑁𝑂 or ∠𝑀𝑃𝑂 ii.

(1 mark)

A quadrilateral with both pairs of opposite sides are parallel is a parallelogram.

Diagonals of a parallelogram bisect each other Substitute 𝑀𝑂 = 10 cm Evaluate Opposite angles in a parallelogram are congruent Co-interior angles in a parallelogram are supplementary State any two adjacent sides are equal State the diagonals are perpendicular State a diagonal bisects one of the angles it passes through

(1 mark for any three of the above) 10

(

a. 𝑀

𝑥1+𝑥2 2

,

𝑦1+𝑦2 2

)= (

0+0 0+2𝑏 , 2 2

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)

Use the midpoint formula for 𝑀 Simplify

= (0, 𝑏)

(

𝑁

𝑥1+𝑥2 2

= (𝑎, 𝑏)

b. 𝑂𝑁 =

(1 mark) ,

𝑦1+𝑦2 2

)= (

2𝑎+0 0+2𝑏 , 2 2

) (1 mark)

2

2

𝑎 +𝑏

=

𝑁𝑃 =

2

2

2

2

(1 mark) 2

(2𝑎 − 𝑎) + (0 − 𝑏)

𝑎 + (− 𝑏)

=

2

2

(1 mark) (1 mark)

𝑦 −𝑦

c. Slope of 𝑀𝑁 = 𝑥2−𝑥1 2

𝑏−𝑏 𝑎−0

(1 mark)

2

= 𝑎 +𝑏 𝑂𝑁 = 𝑁𝑃 so ∆𝑁𝑂𝑃 is isosceles =

Use the distance formula for 𝑂𝑁 Substitute the co-ordinates Simplify

(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2

(𝑎 − 0) + (𝑏 − 0)

=

Use the midpoint formula for 𝑁 Simplify

1

(0.5 mark)

=0 (0.5 mark) 𝑂𝑃 is horizontal so also has a slope of 0 (1 mark) The slopes of 𝑀𝑁 and 𝑂𝑃 are equal so 𝑀𝑁||𝑂𝑃

Substitute the co-ordinates for 𝑂𝑁

Simplify Make the conclusion, linking to previous working Use the gradient formula for 𝑀𝑁 Substitute the co-ordinates Simplify State the slope of 𝑂𝑃 Make the conclusion, linking to previous working

(1 mark)

11

𝐴𝐶 𝑋𝑍 𝐵𝐶 𝑌𝑍

=

5

= 4 = 5 4

∠𝐴𝐶𝐵 = ∠𝑋𝑍𝑌 = 92° ∆𝐴𝐵𝐶~∆𝑋𝑌𝑍 (SAS) 5 𝐴𝐵 So 𝑋𝑌 = 4 20 2𝑥+5 2𝑥+5 20

(1 mark)

Write the ratio of one pair of known sides

(1 mark)

Write the ratio of another pair of known sides State that the included angles are the same Write the congruency statement and test Write an equation for the ratio of the final pair of sides Substitute the values of the sides

15 12

5

= 4 =

(1 mark) (1 mark) (0.5 mark) (0.5 mark)

4 5 4

2𝑥 + 5 = 5 ×20 2𝑥 + 5 = 16 2𝑥 = 11 𝑥 = 5. 5 12

(1 mark) (1 mark)

Take the reciprocal of both sides Multiply both sides by 20 Evaluate the product Subtract 5 from both sides Divide both sides by 2

∠𝐴𝑂𝐶 = 2(46) = 92 (1 mark) (Angle at centre is twice angle at circumference)

Use centre angle theorem

2𝑥 + 92 = 180 (Angle sum of triangle)

Identifies isosceles triangle and use angle sum to equate to 180

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(1 mark)

2𝑥 = 98 𝑥 = 49 13

(1 mark)

Solves for x

Gradient of 𝑂𝐽: 𝑦 −𝑦

𝑚𝑂𝐽 = 𝑥2−𝑥1 2

𝑏−0

1

= 𝑎−0 𝑏

= 𝑎

Gradient of 𝐿𝐾: 𝑑−(𝑑−𝑏) 𝑚𝐿𝐾 = 𝑐−(𝑐−𝑎)

=

=

𝑑−𝑑+𝑏 𝑐−𝑐+𝑎 𝑏 𝑎

𝑚𝑂𝐽 = 𝑚𝐿𝐾∴𝑂𝐽 and 𝐿𝐾 are parallel

Gradient of 𝑂𝐿: (𝑑−𝑏)−0 𝑚𝑂𝐿 = (𝑐−𝑎)−0 𝑑−𝑏

= 𝑐−𝑎

Gradient of 𝐽𝐾: 𝑑−𝑏 𝑚𝐽𝐾 = 𝑐−𝑎

𝑚𝑂𝐿 = 𝑚𝐽𝐾∴𝑂𝐿 and 𝐽𝐾 are parallel

(1 mark) (1 mark)

(1 mark) (0.5 mark)

(0.5 mark) (1 mark) (0.5 mark)

Both pairs of opposite sides are parallel, so 𝑂𝐽𝐾𝐿 is a parallelogram.

(0.5 mark)

14

𝐹𝐴 = 𝐹𝐻 (𝑟𝑎𝑑𝑖𝑖) (0.5 marks) 𝐺𝐴 = 𝐺𝐻 (𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠 𝑓𝑟𝑜𝑚 𝑎 𝑐𝑜𝑚𝑚𝑜𝑛 𝑝𝑜𝑖𝑛𝑡) (0.5 marks) ∠𝐹𝐴𝐺 = ∠𝐹𝐻𝐺 = 90 (𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑟𝑎𝑑𝑖𝑖) (0.5 marks) ∆𝐺𝐴𝐹~∆𝐺𝐻𝐹 (𝑅𝐻𝑆) 𝐺𝐴 = 𝐺𝐻 2𝑥 − 3 = 𝑥 + 7 𝑥 = 10

𝐹𝐻 = 𝐹𝐴 𝐹𝐻 =− 2 + 𝑥 =− 2 + 10 = 8 Total Marks: 70

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Write the gradient of 𝑂𝐽 Use the gradient formula for 𝐿𝐾 Find the gradient of 𝑂𝐽 Conclude that 𝑂𝐽 and 𝐿𝐾 are parallel Find the gradient of 𝑂𝐿 Find the gradient of 𝐽𝐾

Conclude that 𝑂𝐿 and 𝐽𝐾 are parallel

Conclude that 𝑂𝐽𝐾𝐿 is a parallelogram

Use properties to prove congruence of the two triangles using RHS method

(0.5 marks)

(0.5 mark)

(0.5 mark)

Use congruence of triangles to equate corresponding side lengths and solve for x

Equate corresponding side lengths and substitute x to find FH

Common questions Level 1

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