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Exam questions · Maths · Geometry and Measures

Pythagoras' Theorem

  • 7 exam questions
  • 24 marks
  • 9 quick checks
  1. 1 Work out [3 marks]

    A right-angled triangle has shorter sides of length 20 cm and 21 cm. Work out the length of the hypotenuse of the triangle.

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    Model answer

    \(20^2 + 21^2 = 400 + 441 = 841\), and \(\sqrt{841} = 29\) cm.

    Mark scheme

    • \(20^2 + 21^2\) — M1
    • \(841\) — M1
    • 29 cm — A1
  2. 2 Work out [3 marks]

    A right-angled triangle has a hypotenuse of length 25 cm. One of the shorter sides has length 7 cm. Work out the length of the other shorter side.

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    Model answer

    \(25^2 - 7^2 = 625 - 49 = 576\), and \(\sqrt{576} = 24\) cm.

    Mark scheme

    • \(25^2 - 7^2\) — M1
    • \(576\) — M1
    • 24 cm — A1
  3. 3 Work out [4 marks]

    A ladder is 6.5 m long. It leans against a vertical wall. The foot of the ladder is on horizontal ground, 2.5 m from the wall. (a) Work out the height, \(h\), that the ladder reaches up the wall. (3 marks) (b) A safety rule says that the distance from the foot of the ladder to the wall should be one quarter of the height reached. Does the ladder follow this rule? (1 mark)

    A 6.5 m ladder leaning against a vertical wall with its foot 2.5 m from the wall and the height up the wall marked h.
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    Model answer

    (a) \(h^2 = 6.5^2 - 2.5^2 = 42.25 - 6.25 = 36\), so \(h = 6\) m. (b) One quarter of 6 is 1.5 m, but the foot is 2.5 m from the wall, so the ladder does not follow the rule.

    Mark scheme

    • (a) \(6.5^2 - 2.5^2\) — M1
    • (a) \(36\) — M1
    • (a) 6 m — A1
    • (b) No, with \(6 \div 4 = 1.5\) compared with 2.5 — C1
  4. 4 Show that [3 marks]

    A triangle has sides of length 8 cm, 11 cm and 13 cm. Is the triangle right-angled? You must show your working.

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    Model answer

    The longest side is 13, so \(13^2 = 169\). The other two give \(8^2 + 11^2 = 64 + 121 = 185\). Since \(185 \neq 169\), the triangle is not right-angled.

    Mark scheme

    • \(8^2 + 11^2 = 185\) — M1
    • \(13^2 = 169\) — M1
    • Not right-angled, with the two values compared — C1
  5. 5 Work out [4 marks]

    A rectangle has a diagonal of length 25 cm and a width of 7 cm. Work out the area of the rectangle.

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    Model answer

    The length is \(\sqrt{25^2 - 7^2} = \sqrt{576} = 24\) cm, so the area is \(24 \times 7 = 168\) cm\(^2\).

    Mark scheme

    • \(25^2 - 7^2\) — M1
    • \(\sqrt{576} = 24\) — A1
    • \(24 \times 7\) — M1
    • 168 cm\(^2\) — A1
  6. 6 Work out [4 marks]

    An isosceles triangle has a base of 16 cm and two equal sides of 17 cm. Work out the area of the triangle.

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    Model answer

    The height splits the base into two lots of 8 cm, so \(h = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15\) cm. The area is \(\dfrac{1}{2} \times 16 \times 15 = 120\) cm\(^2\).

    Mark scheme

    • Half of the base \(= 8\) used — M1
    • \(17^2 - 8^2 = 225\) — M1
    • \(h = 15\) — A1
    • 120 cm\(^2\) — A1
  7. 7 Work out [3 marks]

    Point \(A\) has coordinates \((-3, 2)\) and point \(B\) has coordinates \((2, 14)\). Work out the length of the line \(AB\).

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    Model answer

    The horizontal difference is \(2 - (-3) = 5\) and the vertical difference is \(14 - 2 = 12\). So \(AB = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\).

    Mark scheme

    • Differences 5 and 12 found — M1
    • \(5^2 + 12^2 = 169\) — M1
    • 13 — A1

Quick check

  1. 1

    A right-angled triangle has shorter sides of 9 cm and 12 cm. What is the hypotenuse?

    1. A15 cm
    2. B21 cm
    3. C225 cm
    4. D10 cm
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    A: 15 cm

    \(9^2 + 12^2 = 81 + 144 = 225\), and \(\sqrt{225} = 15\).

  2. 2

    A right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. What is the other shorter side?

    1. A8 cm
    2. B18 cm
    3. C144 cm
    4. D12 cm
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    D: 12 cm

    \(13^2 - 5^2 = 169 - 25 = 144\), and \(\sqrt{144} = 12\).

  3. 3

    Which set of lengths makes a right-angled triangle?

    1. A5 cm, 7 cm, 9 cm
    2. B6 cm, 7 cm, 10 cm
    3. C6 cm, 8 cm, 10 cm
    4. D4 cm, 5 cm, 7 cm
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    C: 6 cm, 8 cm, 10 cm

    \(6^2 + 8^2 = 36 + 64 = 100 = 10^2\). The other sets do not satisfy \(a^2 + b^2 = c^2\).

  4. 4

    Which side of a right-angled triangle is the hypotenuse?

    1. AThe shortest side
    2. BThe longest side, opposite the right angle
    3. CThe side along the bottom
    4. DThe side next to the right angle
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    B: The longest side, opposite the right angle

    The hypotenuse is always the longest side, and it is opposite the right angle.

  5. 5

    A rectangle is 15 cm long and 8 cm wide. How long is its diagonal?

    1. A17 cm
    2. B23 cm
    3. C7 cm
    4. D289 cm
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    A: 17 cm

    \(15^2 + 8^2 = 225 + 64 = 289\), and \(\sqrt{289} = 17\).

  6. 6

    A ladder 6.5 m long leans against a wall. Its foot is 2.5 m from the wall. How high up the wall does it reach?

    1. A9 m
    2. B4 m
    3. C36 m
    4. D6 m
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    D: 6 m

    \(6.5^2 - 2.5^2 = 42.25 - 6.25 = 36\), and \(\sqrt{36} = 6\).

  7. 7

    An isosceles triangle has base 10 cm and equal sides of 13 cm. What is its height?

    1. A8 cm
    2. B9 cm
    3. C12 cm
    4. D7 cm
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    C: 12 cm

    The height splits the base into two lots of 5 cm. \(13^2 - 5^2 = 144\), so the height is 12 cm.

  8. 8

    A right-angled triangle has shorter sides of 2 cm and 4 cm. What is the hypotenuse?

    1. A\(2\sqrt{3}\) cm
    2. B\(2\sqrt{5}\) cm
    3. C6 cm
    4. D\(\sqrt{6}\) cm
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    B: \(2\sqrt{5}\) cm

    \(2^2 + 4^2 = 20\), and \(\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}\).

  9. 9

    What is the distance between the points \((1, 2)\) and \((7, 10)\)?

    1. A10
    2. B14
    3. C100
    4. D8
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    A: 10

    The horizontal difference is 6 and the vertical difference is 8, so the distance is \(\sqrt{6^2 + 8^2} = \sqrt{100} = 10\).