Mathematics questions, chapter 13

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The Pythagoras' Theorem

40 marks. Use Practice to check each answer as you go, or Exam to answer everything first. You may use a calculator for square roots. Show your working for Section B.

Section A: objective questions

12 marks

Choose the correct answer.

Question 1

Diagram 1 shows a right-angled triangle PQR. Which side is the hypotenuse?

Triangle PQR with a right angle at Q — Diagram 1

PQ
QR
PR
The triangle has no hypotenuse

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The hypotenuse is the side opposite the right angle. The right angle is at Q, and the side that does not touch Q is PR. PQ and QR both meet at the right angle.

For the teacher: Remembering, low difficulty, standard 13.1.1

Textbook p. 294

Question 2

Which statement about the hypotenuse is correct?

It is the shortest side of a right-angled triangle.
It is the longest side of a right-angled triangle and is opposite the right angle.
It is one of the two sides that form the right angle.
Every triangle has a hypotenuse.

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The hypotenuse is the longest side and lies opposite the right angle. The two sides that form the right angle are the shorter sides, and only right-angled triangles have a hypotenuse.

For the teacher: Remembering, low difficulty, standard 13.1.1

Textbook p. 294

Question 3

Diagram 2 shows a right-angled triangle KLM. Which relationship is correct?

Triangle KLM with a right angle at L — Diagram 2

KM² = KL² + LM²
KL² = KM² + LM²
LM² = KL² + KM²
KM = KL + LM

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The right angle is at L, so KM is the hypotenuse and must be alone: KM² = KL² + LM². B and C put a shorter side alone, and D forgets to square the sides.

For the teacher: Understanding, low difficulty, standard 13.1.2

Textbook p. 296

Question 4

Diagram 3 shows squares drawn on the sides of a right-angled triangle. The areas of the two smaller squares are 36 cm² and 64 cm². What is the area of the largest square?

Right-angled triangle with a square on each side. The square on the shortest side is 36 cm², the square on the middle side is 64 cm², and the square on the hypotenuse is marked ? — Diagram 3

28 cm²
14 cm²
10 cm²
100 cm²

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The square on the hypotenuse equals the sum of the other two squares: 36 + 64 = 100 cm². 28 subtracts the areas, 14 adds the side lengths (6 + 8), and 10 is the length of the hypotenuse, not the area of its square.

For the teacher: Understanding, low difficulty, standard 13.1.2

Textbook p. 296

Question 5

Diagram 4 shows a right-angled triangle. Find the value of x.

Right-angled triangle with sides 9 cm and 12 cm meeting at the right angle and hypotenuse x cm — Diagram 4

21
15
225
7.94

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x is the hypotenuse: x² = 9² + 12² = 81 + 144 = 225, so x = √225 = 15. 21 adds the sides without squaring, 225 forgets the square root, and 7.94 subtracts instead of adding.

For the teacher: Applying, low difficulty, standard 13.1.3

Textbook p. 297

Question 6

Diagram 5 shows a right-angled triangle. Find the value of x.

Right-angled triangle with hypotenuse 17 cm, a side of 8 cm and a side of x cm — Diagram 5

18.79
9
15
225

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17 cm is the hypotenuse, so subtract: x² = 17² − 8² = 289 − 64 = 225, x = 15. 18.79 adds the squares (but x cannot be longer than the hypotenuse), 9 is 17 − 8, and 225 forgets the square root.

For the teacher: Applying, medium difficulty, standard 13.1.3

Textbook p. 297

Question 7

Diagram 6 shows a right-angled triangle. Find the value of x, correct to two decimal places.

Right-angled triangle with sides 5 cm and 7 cm meeting at the right angle and hypotenuse x cm — Diagram 6

8.60
4.90
12.00
74.00

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x² = 5² + 7² = 25 + 49 = 74, x = √74 = 8.60. 4.90 is √(49 − 25) (subtracting), 12 adds the sides, and 74 forgets the square root.

For the teacher: Applying, medium difficulty, standard 13.1.3

Textbook p. 297

Question 8

The size of a television screen is the length of its diagonal. Diagram 7 shows a screen 48 cm long and 36 cm high. What is the size of the screen?

Rectangular television screen 48 cm long and 36 cm high with a dashed diagonal — Diagram 7

84 cm
31.75 cm
3 600 cm
60 cm

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The corner of the screen is a right angle, so the diagonal is the hypotenuse: d² = 48² + 36² = 2 304 + 1 296 = 3 600, d = 60 cm. 84 adds the sides, 31.75 subtracts the squares, and 3 600 forgets the square root.

For the teacher: Applying, medium difficulty, standard 13.1.4

Textbook p. 298

Question 9

Which set of side lengths forms a right-angled triangle?

6 cm, 7 cm, 8 cm
5 cm, 12 cm, 14 cm
7 cm, 24 cm, 25 cm
4 cm, 5 cm, 6 cm

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25² = 625 and 7² + 24² = 49 + 576 = 625, so they are equal. For 5, 12, 14: 14² = 196 but 25 + 144 = 169 (close to the triple 5, 12, 13, but not equal). For 6, 7, 8: 64 ≠ 85, and for 4, 5, 6: 36 ≠ 41.

For the teacher: Understanding, low difficulty, standard 13.2.1

Textbook p. 301

Question 10

A triangle has sides 8 cm, 11 cm and 14 cm. What type of angle is opposite the 14 cm side?

A right angle
An acute angle
A reflex angle
An obtuse angle

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14² = 196 and 8² + 11² = 64 + 121 = 185. Since 196 > 185 (c² > a² + b²), the angle opposite the longest side is obtuse. A triangle can never have a reflex angle.

For the teacher: Analysing, medium difficulty, standard 13.2.1

Textbook p. 300

Question 11

In Diagram 8, PQR is a straight line. SP = 10 cm, PQ = 6 cm, SQ = 8 cm and ∠QSR = 40°. Find the value of x.

Triangle SPR with Q on the straight line PR. SP = 10 cm, PQ = 6 cm, SQ = 8 cm, angle QSR = 40° and angle x at R — Diagram 8

50°
40°
90°
140°

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6² + 8² = 36 + 64 = 100 = 10², so ∠SQP = 90° by the converse. PQR is a straight line, so ∠SQR = 180° − 90° = 90°. Then x = 180° − 90° − 40° = 50°. 140° forgets the right angle (180° − 40°).

For the teacher: Applying, medium difficulty, standard 13.2.2

Textbook p. 301

Question 12

A 2.5 m ladder leans against a vertical wall with its base 0.7 m from the wall (Diagram 9). The top slides down h m until the base is 1.5 m from the wall (Diagram 10). Find h.

Ladder 2.5 m long with its base 0.7 m from the wall — Diagram 9

The same ladder with its base 1.5 m from the wall — Diagram 9

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Before: height = √(2.5² − 0.7²) = √(6.25 − 0.49) = √5.76 = 2.4 m. After: height = √(2.5² − 1.5²) = √(6.25 − 2.25) = √4 = 2.0 m. h = 2.4 − 2.0 = 0.4. 0.8 is how far the base moved (1.5 − 0.7), which is not the same as how far the top moved.

For the teacher: Analysing, high difficulty, standard 13.1.4

Textbook p. 305

Section B: structured questions

28 marks

Answer all questions. Show your working.

Question 13

Diagram 11 shows quadrilateral PQRS. The diagonal PR divides it into two right-angled triangles.

Quadrilateral PQRS with diagonal PR. Triangle PQR has a right angle at Q; triangle PRS has a right angle at R — Diagram 11

[1 mark] (a)Name the hypotenuse of triangle PQR.

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PR [1]1

Tick each point you got.

[1 mark] (b)Name the hypotenuse of triangle PRS.

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PS [1]1

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[1 mark] (c)Write the Pythagoras' theorem for triangle PRS.

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PS² = PR² + RS² [1]1

Tick each point you got.

For the teacher: Understanding, low difficulty, standard 13.1.2

Textbook p. 295

Question 14

In Diagram 12, PQ = 9 cm, QR = 12 cm and RS = 8 cm. ∠PQR = ∠PRS = 90°.

Triangle PQR with PQ = 9 cm, QR = 12 cm and a right angle at Q, joined along PR to triangle PRS with RS = 8 cm and a right angle at R — Diagram 12

[2 marks] (a)Calculate the length of PR.

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PR² = 9² + 12² = 81 + 144 = 225 [1]1
PR = √225 = 15 cm [1]1

Tick each point you got.

[2 marks] (b)Calculate the length of PS.

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PS² = 15² + 8² = 225 + 64 = 289 [1]1
PS = √289 = 17 cm [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 13.1.3

Textbook p. 297

Question 15

Diagram 13 shows a rectangle PQRS, 20 cm by 15 cm. The corner triangle TQU is cut off. PT = 8 cm and UR = 6 cm.

Rectangle PQRS with PQ along the top, SR = 20 cm, SP = 15 cm. T is on PQ with PT = 8 cm and U is on QR with UR = 6 cm. The corner TQU is cut off along TU, leaving the shaded pentagon PTURS — Diagram 13

[2 marks] (a)Calculate the length of TU.

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TQ = 20 − 8 = 12 cm and QU = 15 − 6 = 9 cm [1]1
TU² = 12² + 9² = 144 + 81 = 225, TU = 15 cm [1]1

Tick each point you got.

[2 marks] (b)Calculate the perimeter of the shaded region PTURS.

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8 + 15 + 6 + 20 + 15 [1]1
= 64 cm [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 13.1.4

Textbook p. 299

Question 16

Mei Ling swims across a river 25 m wide, aiming from P straight to Q. The current carries her downstream, so she lands at R, 9 m from Q (Diagram 14).

River 25 m wide. P on the near bank, Q directly opposite on the far bank, R on the far bank 9 m downstream of Q. Angle PQR is a right angle and PR is drawn in orange — Diagram 14

[1 mark] (a)Which side of triangle PQR is the distance she actually swam? Give a reason.

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PR, because she swam from P to R; it is the hypotenuse, opposite the right angle at Q [1]1

Tick each point you got.

[2 marks] (b)Calculate the distance she actually swam, correct to two decimal places.

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PR² = 25² + 9² = 625 + 81 = 706 [1]1
PR = √706 = 26.57 m [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 13.1.4

Textbook p. 304

Question 17

Use the converse of Pythagoras' theorem for each triangle. Show your working.

[2 marks] (a)Determine whether a triangle with sides 10 cm, 24 cm and 26 cm is a right-angled triangle.

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26² = 676 and 10² + 24² = 100 + 576 = 676 [1]1
Equal, so it is a right-angled triangle [1]1

Tick each point you got.

[1 mark] (b)Determine whether a triangle with sides 9 cm, 12 cm and 16 cm is a right-angled triangle.

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16² = 256 but 9² + 12² = 81 + 144 = 225; not equal, so it is not a right-angled triangle [1]1

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[1 mark] (c)For the triangle in (b), what type of angle is opposite the 16 cm side? Give a reason.

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Obtuse, because 256 > 225 (c² > a² + b²) [1]1

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For the teacher: Analysing, medium difficulty, standard 13.2.1

Textbook p. 301

Question 18

In Diagram 15, PQ = 8 cm, QS = 15 cm, PS = 17 cm, ∠QRS = 70° and ∠QSR = 45°.

Quadrilateral PQRS with diagonal QS. PQ = 8 cm, QS = 15 cm, PS = 17 cm, angle QRS = 70° and angle QSR = 45° — Diagram 15

[2 marks] (a)Show that ∠PQS = 90°.

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8² + 15² = 64 + 225 = 289 and 17² = 289 [1]1
Equal, so by the converse of Pythagoras' theorem the angle opposite PS, ∠PQS, is 90° [1]1

Tick each point you got.

[2 marks] (b)Find ∠PQR.

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∠SQR = 180° − 70° − 45° = 65° [1]1
∠PQR = 90° + 65° = 155° [1]1

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 13.2.2

Textbook p. 302

Question 19

Farid was asked to find x in Diagram 16. Diagram 17 shows his working.

Right-angled triangle with hypotenuse 13 cm, a side of 5 cm and a side of x cm — Diagram 16

Farid's working. Line 1: x² = 13² + 5². Line 2: x² = 169 + 25. Line 3: x² = 194. Line 4: x = √194 = 13.93 cm — Diagram 16

[1 mark] (a)Farid's answer cannot be correct. Without calculating, explain how you know.

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13 cm is the hypotenuse, the longest side, so x must be less than 13 cm; 13.93 cm is longer [1]1

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[2 marks] (b)State his mistake and find the correct value of x.

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Line 1: x is not the hypotenuse, so he should subtract: x² = 13² − 5² = 169 − 25 = 144 [1]1
x = √144 = 12 [1]1

Tick each point you got.

For the teacher: Evaluating, high difficulty, standard 13.1.3

Textbook p. 297

Question 20

Aina has a 60 cm piece of wire. She wants to bend all of it into a right-angled triangle frame whose sides are whole numbers of centimetres.

[2 marks] (a)Suggest three side lengths she could use. Show that they add up to 60 cm and that they form a right-angled triangle.

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The only whole-number answers are 15, 20, 25 and 10, 24, 26.

A set with sum 60, e.g. 15 + 20 + 25 = 60 (or 10 + 24 + 26 = 60) [1]1
Converse check, e.g. 15² + 20² = 225 + 400 = 625 = 25² (or 10² + 24² = 100 + 576 = 676 = 26²) [1]1

Tick each point you got.

[1 mark] (b)Her friend suggests 20 cm, 20 cm and 20 cm. Explain why this does not meet Aina's plan.

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20² = 400 but 20² + 20² = 800; not equal, so there is no right angle (it is an equilateral triangle with 60° angles) [1]1

Tick each point you got.

For the teacher: Creating, high difficulty, standard 13.2.2

Textbook p. 305

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