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Exam questions · Maths

Geometry and Measures

  • 32 exam questions
  • 104 marks
  • 45 quick checks

Angles in Parallel Lines and Polygons

Just this lesson
  1. 1 Calculate [2 marks]

    Two straight lines cross. One of the angles is \(38^\circ\). (a) Write down the size of the angle directly opposite it. [1 mark] (b) Calculate the size of an angle next to it. [1 mark]

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    Model answer

    (a) Vertically opposite angles are equal, so \(38^\circ\). (b) Angles on a straight line add up to \(180^\circ\), so \(180 - 38 = 142^\circ\).

    Mark scheme

    • (a) \(38^\circ\) — B1
    • (b) \(142^\circ\) — B1
  2. 2 Calculate [4 marks]

    \(ABCDE\) is a regular pentagon. (a) Calculate the size of one interior angle of the pentagon. [2 marks] (b) Calculate the size of angle \(x\). [2 marks]

    A regular pentagon ABCDE with the diagonal AC drawn and the angle ACD marked x.
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    Model answer

    (a) The exterior angle is \(360 \div 5 = 72^\circ\), so the interior angle is \(180 - 72 = 108^\circ\). (b) Triangle \(ABC\) is isosceles, so angle \(BCA = (180 - 108) \div 2 = 36^\circ\). Then \(x = 108 - 36 = 72^\circ\).

    Mark scheme

    • (a) \(360 \div 5\) or \(540 \div 5\) — M1
    • (a) \(108^\circ\) — A1
    • (b) \((180 - 108) \div 2 = 36\) — M1
    • (b) \(72^\circ\) — A1
  3. 3 Calculate [3 marks]

    A regular polygon has 15 sides. (a) Calculate the size of one exterior angle. [1 mark] (b) Calculate the size of one interior angle. [2 marks]

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    Model answer

    (a) \(360 \div 15 = 24^\circ\). (b) \(180 - 24 = 156^\circ\).

    Mark scheme

    • (a) \(24^\circ\) — B1
    • (b) \(180 - 24\) — M1
    • (b) \(156^\circ\) — A1
  4. 4 Calculate [3 marks]

    \(ABCD\) is a parallelogram. Angle \(DAB = 70^\circ\). Calculate the size of angle \(ABC\). Give a reason for your answer.

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    Model answer

    Angles \(DAB\) and \(ABC\) are co-interior angles between the parallel sides \(AD\) and \(BC\), so they add up to \(180^\circ\). Angle \(ABC = 180 - 70 = 110^\circ\).

    Mark scheme

    • \(180 - 70\) — M1
    • \(110^\circ\) — A1
    • Reason: co-interior angles add up to 180 degrees — B1
  5. 5 Calculate [3 marks]

    The interior angles of a polygon add up to \(1260^\circ\). Calculate the number of sides of the polygon.

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    Model answer

    \((n - 2) \times 180 = 1260\), so \(n - 2 = 7\) and \(n = 9\).

    Mark scheme

    • \(1260 \div 180 = 7\) — M1
    • \(7 + 2\) — M1
    • 9 — A1
  6. 6 Calculate [4 marks]

    The angles of a pentagon are \(x^\circ\), \((x + 10)^\circ\), \(2x^\circ\), \((2x + 20)^\circ\) and \((3x - 30)^\circ\). Calculate the size of the largest angle of the pentagon.

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    Model answer

    The interior angles of a pentagon add up to \(540^\circ\), so \(x + x + 10 + 2x + 2x + 20 + 3x - 30 = 540\). That gives \(9x = 540\), so \(x = 60\). The angles are \(60^\circ\), \(70^\circ\), \(120^\circ\), \(140^\circ\) and \(150^\circ\), so the largest is \(150^\circ\).

    Mark scheme

    • \(540\) used as the total — M1
    • \(9x = 540\) — M1
    • \(x = 60\) — A1
    • \(150^\circ\) — A1

Quick check

  1. 1

    Three angles on a straight line are \(47^\circ\), \(68^\circ\) and \(x\). What is \(x\)?

    1. A\(115^\circ\)
    2. B\(65^\circ\)
    3. C\(75^\circ\)
    4. D\(245^\circ\)
    Show answerHide answer

    B: \(65^\circ\)

    Angles on a straight line add up to \(180^\circ\). \(180 - 47 - 68 = 65\).

  2. 2

    Three angles of a quadrilateral are \(80^\circ\), \(95^\circ\) and \(110^\circ\). What is the fourth angle?

    1. A\(75^\circ\)
    2. B\(85^\circ\)
    3. C\(105^\circ\)
    4. D\(285^\circ\)
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    A: \(75^\circ\)

    The angles of a quadrilateral add up to \(360^\circ\). \(80 + 95 + 110 = 285\) and \(360 - 285 = 75\).

  3. 3

    Which type of angles are equal and form a Z shape between parallel lines?

    1. ACorresponding angles
    2. BCo-interior angles
    3. CVertically opposite angles
    4. DAlternate angles
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    D: Alternate angles

    Alternate angles are on opposite sides of the transversal, and make a Z shape.

  4. 4

    Two co-interior angles lie between parallel lines. One is \(72^\circ\). What is the other?

    1. A\(72^\circ\)
    2. B\(18^\circ\)
    3. C\(108^\circ\)
    4. D\(118^\circ\)
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    C: \(108^\circ\)

    Co-interior angles add up to \(180^\circ\), so \(180 - 72 = 108\).

  5. 5

    What is the sum of the interior angles of a hexagon?

    1. A\(540^\circ\)
    2. B\(720^\circ\)
    3. C\(900^\circ\)
    4. D\(1080^\circ\)
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    B: \(720^\circ\)

    A hexagon has 6 sides, so the sum is \((6 - 2) \times 180 = 720\).

  6. 6

    What is each exterior angle of a regular octagon?

    1. A\(45^\circ\)
    2. B\(135^\circ\)
    3. C\(60^\circ\)
    4. D\(40^\circ\)
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    A: \(45^\circ\)

    \(360 \div 8 = 45\). The interior angle is \(135^\circ\).

  7. 7

    Each exterior angle of a regular polygon is \(24^\circ\). How many sides does it have?

    1. A14
    2. B16
    3. C156
    4. D15
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    D: 15

    \(360 \div 24 = 15\). The number 156 is the interior angle.

  8. 8

    An exterior angle of a triangle is \(118^\circ\). One of the interior opposite angles is \(54^\circ\). What is the other interior opposite angle?

    1. A\(62^\circ\)
    2. B\(72^\circ\)
    3. C\(64^\circ\)
    4. D\(172^\circ\)
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    C: \(64^\circ\)

    The exterior angle equals the sum of the two interior opposite angles, so \(118 - 54 = 64\).

  9. 9

    The angles of a triangle are \(x\), \(2x + 10\) and \(3x - 10\) degrees. What is the size of the largest angle?

    1. A\(30^\circ\)
    2. B\(80^\circ\)
    3. C\(70^\circ\)
    4. D\(90^\circ\)
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    B: \(80^\circ\)

    \(6x = 180\), so \(x = 30\). The angles are \(30^\circ\), \(70^\circ\) and \(80^\circ\).

Area, Perimeter and Circles

Just this lesson
  1. 1 Calculate [5 marks]

    The diagram shows a sector of a circle with radius 6 cm and angle \(120^\circ\). Give your answers in terms of \(\pi\). (a) Calculate the area of the sector. [2 marks] (b) Calculate the length of the arc. [2 marks] (c) Write down the perimeter of the sector. [1 mark]

    A sector of a circle with radius 6 cm and an angle of 120 degrees at the centre.
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    Model answer

    \(\dfrac{120}{360} = \dfrac{1}{3}\). (a) The area of the circle is \(\pi \times 36 = 36\pi\), so the sector is \(12\pi\) cm\(^2\). (b) The circumference is \(2\pi \times 6 = 12\pi\), so the arc is \(4\pi\) cm. (c) The perimeter is the arc plus two radii, \(4\pi + 6 + 6 = 12 + 4\pi\) cm.

    Mark scheme

    • (a) \(\dfrac{1}{3} \times \pi \times 6^2\) — M1
    • (a) \(12\pi\) — A1
    • (b) \(\dfrac{1}{3} \times 2 \times \pi \times 6\) — M1
    • (b) \(4\pi\) — A1
    • (c) \(12 + 4\pi\), with their arc if correct — B1
  2. 2 Calculate [2 marks]

    A triangle has a base of 15 cm and a perpendicular height of 8 cm. Calculate the area of the triangle.

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    Model answer

    \(\dfrac{1}{2} \times 15 \times 8 = 60\) cm\(^2\).

    Mark scheme

    • \(\dfrac{1}{2} \times 15 \times 8\) or \(15 \times 8 \div 2\) — M1
    • 60 cm\(^2\) — A1
  3. 3 Calculate [3 marks]

    A trapezium has an area of 72 cm\(^2\). Its parallel sides are 7 cm and 11 cm long. Calculate the distance between the parallel sides.

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    Model answer

    \(\dfrac{1}{2}(7 + 11) \times h = 72\), so \(9h = 72\) and \(h = 8\) cm.

    Mark scheme

    • \(\dfrac{1}{2}(7 + 11) \times h = 72\) or \(9h = 72\) — M1
    • \(72 \div 9\) — M1
    • 8 cm — A1
  4. 4 Calculate [3 marks]

    A circle has diameter 16 cm. Give your answers in terms of \(\pi\). (a) Calculate the circumference of the circle. [1 mark] (b) Calculate the area of the circle. [2 marks]

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    Model answer

    (a) \(\pi \times 16 = 16\pi\) cm. (b) The radius is 8 cm, so the area is \(\pi \times 8^2 = 64\pi\) cm\(^2\).

    Mark scheme

    • (a) \(16\pi\) — B1
    • (b) \(\pi \times 8^2\) — M1
    • (b) \(64\pi\) — A1
  5. 5 Calculate [4 marks]

    A semicircle has radius 5 cm. Give your answers in terms of \(\pi\). (a) Calculate the area of the semicircle. [2 marks] (b) Calculate the perimeter of the semicircle. [2 marks]

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    Model answer

    (a) \(\dfrac{1}{2} \times \pi \times 5^2 = 12.5\pi\) cm\(^2\). (b) The curved edge is \(\dfrac{1}{2} \times \pi \times 10 = 5\pi\) and the straight edge is 10 cm, so the perimeter is \(10 + 5\pi\) cm.

    Mark scheme

    • (a) \(\dfrac{1}{2} \times \pi \times 5^2\) — M1
    • (a) \(12.5\pi\) or \(\dfrac{25\pi}{2}\) — A1
    • (b) \(5\pi\) or the diameter 10 found — M1
    • (b) \(10 + 5\pi\) — A1
  6. 6 Calculate [3 marks]

    A circular lawn has a radius of 8 m. A path 2 m wide goes all the way round the lawn. Calculate the area of the path. Give your answer in terms of \(\pi\).

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    Model answer

    The lawn and path together have radius 10 m, so their area is \(100\pi\). The lawn has area \(64\pi\). The path is \(100\pi - 64\pi = 36\pi\) m\(^2\).

    Mark scheme

    • \(\pi \times 10^2\) or \(\pi \times 8^2\) — M1
    • \(100\pi - 64\pi\) — M1
    • \(36\pi\) m\(^2\) — A1

Quick check

  1. 1

    A triangle has base 10 cm and perpendicular height 6 cm. What is its area?

    1. A60 cm\(^2\)
    2. B16 cm\(^2\)
    3. C30 cm\(^2\)
    4. D32 cm\(^2\)
    Show answerHide answer

    C: 30 cm\(^2\)

    \(\dfrac{1}{2} \times 10 \times 6 = 30\) cm\(^2\).

  2. 2

    A trapezium has parallel sides of 7 cm and 13 cm, and a height of 6 cm. What is its area?

    1. A120 cm\(^2\)
    2. B60 cm\(^2\)
    3. C26 cm\(^2\)
    4. D78 cm\(^2\)
    Show answerHide answer

    B: 60 cm\(^2\)

    \(\dfrac{1}{2}(7 + 13) \times 6 = 60\). Forgetting the half gives 120.

  3. 3

    A parallelogram has base 9 cm, slanted side 5 cm and perpendicular height 4 cm. What is its area?

    1. A36 cm\(^2\)
    2. B45 cm\(^2\)
    3. C18 cm\(^2\)
    4. D20 cm\(^2\)
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    A: 36 cm\(^2\)

    \(\text{base} \times \text{perpendicular height} = 9 \times 4 = 36\). The slanted side is not used.

  4. 4

    A circle has radius 5 cm. What is its area, in terms of \(\pi\)?

    1. A\(10\pi\) cm\(^2\)
    2. B\(5\pi\) cm\(^2\)
    3. C\(50\pi\) cm\(^2\)
    4. D\(25\pi\) cm\(^2\)
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    D: \(25\pi\) cm\(^2\)

    \(\pi r^2 = \pi \times 25 = 25\pi\).

  5. 5

    A circle has diameter 14 cm. What is its circumference, in terms of \(\pi\)?

    1. A\(7\pi\) cm
    2. B\(49\pi\) cm
    3. C\(14\pi\) cm
    4. D\(28\pi\) cm
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    C: \(14\pi\) cm

    \(C = \pi d = 14\pi\).

  6. 6

    A circle has diameter 10 cm. What is its area, in terms of \(\pi\)?

    1. A\(100\pi\) cm\(^2\)
    2. B\(25\pi\) cm\(^2\)
    3. C\(10\pi\) cm\(^2\)
    4. D\(50\pi\) cm\(^2\)
    Show answerHide answer

    B: \(25\pi\) cm\(^2\)

    The radius is \(10 \div 2 = 5\), so \(A = \pi \times 5^2 = 25\pi\). Using 10 as the radius gives \(100\pi\).

  7. 7

    A rectangle measures 12 cm by 7 cm. A rectangle 5 cm by 3 cm is cut from one corner. What is the area of the remaining shape?

    1. A69 cm\(^2\)
    2. B99 cm\(^2\)
    3. C15 cm\(^2\)
    4. D84 cm\(^2\)
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    A: 69 cm\(^2\)

    \(12 \times 7 = 84\) and \(5 \times 3 = 15\), so \(84 - 15 = 69\).

  8. 8

    A sector has radius 9 cm and angle \(80^\circ\). What is its area, in terms of \(\pi\)?

    1. A\(4\pi\) cm\(^2\)
    2. B\(36\pi\) cm\(^2\)
    3. C\(81\pi\) cm\(^2\)
    4. D\(18\pi\) cm\(^2\)
    Show answerHide answer

    D: \(18\pi\) cm\(^2\)

    \(\dfrac{80}{360} \times \pi \times 81 = \dfrac{2}{9} \times 81\pi = 18\pi\).

  9. 9

    A sector has radius 6 cm and angle \(60^\circ\). What is the length of its arc, in terms of \(\pi\)?

    1. A\(6\pi\) cm
    2. B\(\pi\) cm
    3. C\(2\pi\) cm
    4. D\(12\pi\) cm
    Show answerHide answer

    C: \(2\pi\) cm

    \(\dfrac{60}{360} \times 2 \times \pi \times 6 = \dfrac{1}{6} \times 12\pi = 2\pi\).

Volume and Surface Area

Just this lesson
  1. 1 Calculate [4 marks]

    The diagram shows a cone. Give your answers in terms of \(\pi\). (a) Calculate the volume of the cone. [2 marks] (b) Calculate the curved surface area of the cone. [2 marks]

    A cone with base radius 6 cm, vertical height 8 cm and slant height 10 cm.
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    Model answer

    (a) \(\dfrac{1}{3} \times \pi \times 6^2 \times 8 = \dfrac{1}{3} \times 288\pi = 96\pi\) cm\(^3\). (b) The curved surface area is \(\pi r l = \pi \times 6 \times 10 = 60\pi\) cm\(^2\).

    Mark scheme

    • (a) \(\dfrac{1}{3} \times \pi \times 6^2 \times 8\) — M1
    • (a) \(96\pi\) — A1
    • (b) \(\pi \times 6 \times 10\) — M1
    • (b) \(60\pi\) — A1
  2. 2 Calculate [3 marks]

    A cuboid measures 9 cm by 4 cm by 5 cm. Calculate the total surface area of the cuboid.

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    Model answer

    The faces are \(9 \times 4 = 36\), \(9 \times 5 = 45\) and \(4 \times 5 = 20\) cm\(^2\). The total area is \(2 \times (36 + 45 + 20) = 2 \times 101 = 202\) cm\(^2\).

    Mark scheme

    • \(36\), \(45\) and \(20\) found — M1
    • \(2 \times (36 + 45 + 20)\) — M1
    • 202 cm\(^2\) — A1
  3. 3 Calculate [3 marks]

    A prism has a cross-section that is a trapezium. The parallel sides of the trapezium are 6 cm and 10 cm and the distance between them is 4 cm. The prism is 12 cm long. Calculate the volume of the prism.

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    Model answer

    The area of the trapezium is \(\dfrac{1}{2}(6 + 10) \times 4 = 32\) cm\(^2\). The volume is \(32 \times 12 = 384\) cm\(^3\).

    Mark scheme

    • \(\dfrac{1}{2}(6 + 10) \times 4\) — M1
    • \(32 \times 12\) — M1
    • 384 cm\(^3\) — A1
  4. 4 Calculate [4 marks]

    A cylinder has radius 3 cm and height 7 cm. Give your answers in terms of \(\pi\). (a) Calculate the volume of the cylinder. [2 marks] (b) Calculate the curved surface area of the cylinder. [2 marks]

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    Model answer

    (a) \(\pi \times 3^2 \times 7 = 63\pi\) cm\(^3\). (b) \(2 \times \pi \times 3 \times 7 = 42\pi\) cm\(^2\).

    Mark scheme

    • (a) \(\pi \times 3^2 \times 7\) — M1
    • (a) \(63\pi\) — A1
    • (b) \(2 \times \pi \times 3 \times 7\) — M1
    • (b) \(42\pi\) — A1
  5. 5 Calculate [4 marks]

    A swimming pool is a cuboid, 25 m long, 10 m wide and 2 m deep. It is filled using a pump that delivers 25 000 litres of water each hour. 1 m\(^3\) \(= 1000\) litres. Calculate the time taken to fill the pool completely.

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    Model answer

    The volume is \(25 \times 10 \times 2 = 500\) m\(^3\), which is \(500 \times 1000 = 500\,000\) litres. The time is \(500\,000 \div 25\,000 = 20\) hours.

    Mark scheme

    • \(25 \times 10 \times 2 = 500\) — M1
    • \(500\,000\) litres — M1
    • \(500\,000 \div 25\,000\) — M1
    • 20 hours — A1
  6. 6 Calculate [3 marks]

    A sphere has a volume of \(288\pi\) cm\(^3\). Calculate the radius of the sphere.

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    Model answer

    \(\dfrac{4}{3}\pi r^3 = 288\pi\), so \(r^3 = 288 \times \dfrac{3}{4} = 216\) and \(r = 6\) cm.

    Mark scheme

    • \(\dfrac{4}{3}\pi r^3 = 288\pi\) — M1
    • \(r^3 = 216\) — M1
    • 6 cm — A1

Quick check

  1. 1

    A cuboid measures 8 cm by 5 cm by 3 cm. What is its volume?

    1. A16 cm\(^3\)
    2. B158 cm\(^3\)
    3. C240 cm\(^3\)
    4. D120 cm\(^3\)
    Show answerHide answer

    D: 120 cm\(^3\)

    \(8 \times 5 \times 3 = 120\) cm\(^3\).

  2. 2

    The cross-section of a prism has area 12 cm\(^2\). The prism is 15 cm long. What is its volume?

    1. A27 cm\(^3\)
    2. B90 cm\(^3\)
    3. C180 cm\(^3\)
    4. D12 cm\(^3\)
    Show answerHide answer

    C: 180 cm\(^3\)

    Volume \(=\) area of cross-section \(\times\) length \(= 12 \times 15 = 180\).

  3. 3

    A cylinder has radius 3 cm and height 10 cm. What is its volume, in terms of \(\pi\)?

    1. A\(30\pi\) cm\(^3\)
    2. B\(90\pi\) cm\(^3\)
    3. C\(60\pi\) cm\(^3\)
    4. D\(900\pi\) cm\(^3\)
    Show answerHide answer

    B: \(90\pi\) cm\(^3\)

    \(\pi r^2 h = \pi \times 9 \times 10 = 90\pi\).

  4. 4

    A cylinder has radius 3 cm and height 10 cm. What is its curved surface area, in terms of \(\pi\)?

    1. A\(60\pi\) cm\(^2\)
    2. B\(30\pi\) cm\(^2\)
    3. C\(90\pi\) cm\(^2\)
    4. D\(180\pi\) cm\(^2\)
    Show answerHide answer

    A: \(60\pi\) cm\(^2\)

    \(2\pi r h = 2 \times \pi \times 3 \times 10 = 60\pi\).

  5. 5

    What is the total surface area of a cube with side 4 cm?

    1. A64 cm\(^2\)
    2. B16 cm\(^2\)
    3. C24 cm\(^2\)
    4. D96 cm\(^2\)
    Show answerHide answer

    D: 96 cm\(^2\)

    A cube has 6 faces, each of area \(4 \times 4 = 16\). So \(6 \times 16 = 96\).

  6. 6

    A tank holds 2500 cm\(^3\) of water. How many litres is this?

    1. A25 litres
    2. B0.25 litres
    3. C2.5 litres
    4. D250 litres
    Show answerHide answer

    C: 2.5 litres

    \(1000\text{ cm}^3 = 1\) litre, so \(2500 \div 1000 = 2.5\).

  7. 7

    A cylinder has radius 2 cm and height 7 cm. What is its volume, in terms of \(\pi\)?

    1. A\(14\pi\) cm\(^3\)
    2. B\(28\pi\) cm\(^3\)
    3. C\(56\pi\) cm\(^3\)
    4. D\(98\pi\) cm\(^3\)
    Show answerHide answer

    B: \(28\pi\) cm\(^3\)

    \(\pi \times 2^2 \times 7 = 28\pi\). Using the diameter instead of the radius would give \(98\pi\).

  8. 8

    A sphere has radius 3 cm. What is its volume, in terms of \(\pi\)?

    1. A\(36\pi\) cm\(^3\)
    2. B\(12\pi\) cm\(^3\)
    3. C\(27\pi\) cm\(^3\)
    4. D\(108\pi\) cm\(^3\)
    Show answerHide answer

    A: \(36\pi\) cm\(^3\)

    \(\dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times \pi \times 27 = 36\pi\).

  9. 9

    A cone has radius 3 cm and vertical height 4 cm. What is its volume, in terms of \(\pi\)?

    1. A\(36\pi\) cm\(^3\)
    2. B\(15\pi\) cm\(^3\)
    3. C\(9\pi\) cm\(^3\)
    4. D\(12\pi\) cm\(^3\)
    Show answerHide answer

    D: \(12\pi\) cm\(^3\)

    \(\dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \pi \times 9 \times 4 = 12\pi\). Using the slant height 5 gives \(15\pi\), which is wrong.

Pythagoras' Theorem

Just this lesson
  1. 1 Calculate [4 marks]

    The diagram shows a right-angled triangle \(ABC\). (a) Calculate the length of \(AC\). [3 marks] (b) Calculate the perimeter of the triangle. [1 mark]

    A right-angled triangle ABC with the right angle at B, BC = 12 cm and AB = 9 cm.
    Show answerHide answer

    Model answer

    (a) \(AC^2 = 9^2 + 12^2 = 81 + 144 = 225\), so \(AC = 15\) cm. (b) The perimeter is \(9 + 12 + 15 = 36\) cm.

    Mark scheme

    • (a) \(9^2 + 12^2\) — M1
    • (a) 225 — M1
    • (a) 15 cm — A1
    • (b) 36 cm — B1
  2. 2 Calculate [3 marks]

    A right-angled triangle has shorter sides of length 8 cm and 15 cm. Calculate the length of the hypotenuse.

    Show answerHide answer

    Model answer

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

    Mark scheme

    • \(8^2 + 15^2\) — M1
    • 289 — M1
    • 17 cm — A1
  3. 3 Calculate [3 marks]

    A right-angled triangle has a hypotenuse of 41 cm and one shorter side of 9 cm. Calculate the length of the other shorter side.

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    Model answer

    \(41^2 - 9^2 = 1681 - 81 = 1600\), and \(\sqrt{1600} = 40\) cm.

    Mark scheme

    • \(41^2 - 9^2\) — M1
    • 1600 — M1
    • 40 cm — A1
  4. 4 Calculate [3 marks]

    A vertical flagpole stands on level ground. A wire 13 m long joins the top of the flagpole to a point on the ground 5 m from the base of the pole. Calculate the height of the flagpole.

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    Model answer

    \(h^2 = 13^2 - 5^2 = 169 - 25 = 144\), so \(h = 12\) m.

    Mark scheme

    • \(13^2 - 5^2\) — M1
    • 144 — M1
    • 12 m — A1
  5. 5 Calculate [3 marks]

    A rhombus has diagonals of length 16 cm and 12 cm. Calculate the length of one side of the rhombus.

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    Model answer

    The diagonals of a rhombus cross at right angles and bisect each other, so a right-angled triangle has shorter sides of 8 cm and 6 cm. The side is \(\sqrt{8^2 + 6^2} = \sqrt{100} = 10\) cm.

    Mark scheme

    • Half-diagonals 8 and 6 used — M1
    • \(8^2 + 6^2 = 100\) — M1
    • 10 cm — A1
  6. 6 Calculate [4 marks]

    An isosceles triangle has a base of 12 cm and two equal sides of 10 cm. Calculate the area of the triangle.

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    Model answer

    The height splits the base into two lots of 6 cm, so \(h = \sqrt{10^2 - 6^2} = \sqrt{64} = 8\) cm. The area is \(\dfrac{1}{2} \times 12 \times 8 = 48\) cm\(^2\).

    Mark scheme

    • Half of the base \(= 6\) used — M1
    • \(10^2 - 6^2 = 64\) — M1
    • \(h = 8\) — A1
    • 48 cm\(^2\) — A1
  7. 7 Calculate [3 marks]

    \(P\) is the point \((2, -1)\) and \(Q\) is the point \((9, 23)\). Calculate the length of \(PQ\).

    Show answerHide answer

    Model answer

    The horizontal difference is 7 and the vertical difference is \(23 - (-1) = 24\). So \(PQ = \sqrt{7^2 + 24^2} = \sqrt{625} = 25\).

    Mark scheme

    • Differences 7 and 24 found — M1
    • \(7^2 + 24^2 = 625\) — M1
    • 25 — 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
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    A: 10

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

Trigonometry in Right-Angled Triangles

Just this lesson
  1. 1 Calculate [4 marks]

    The diagram shows a right-angled triangle with an angle of \(45^\circ\). (a) Calculate the value of \(x\). [2 marks] (b) Calculate the length of the hypotenuse. Give your answer in the form \(a\sqrt{2}\). [2 marks]

    A right-angled triangle with a 45 degree angle, a base of 6 cm and the opposite side labelled x.
    Show answerHide answer

    Model answer

    (a) \(\tan 45^\circ = \dfrac{x}{6}\) and \(\tan 45^\circ = 1\), so \(x = 6\). (b) \(\text{hypotenuse}^2 = 6^2 + 6^2 = 72\), so the hypotenuse is \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\) cm.

    Mark scheme

    • (a) \(\tan 45^\circ = \dfrac{x}{6}\) with \(\tan 45^\circ = 1\) — M1
    • (a) 6 — A1
    • (b) \(6^2 + 6^2 = 72\) — M1
    • (b) \(6\sqrt{2}\) — A1
  2. 2 Write down [2 marks]

    (a) Write down the exact value of \(\sin 60^\circ\). [1 mark] (b) Write down the exact value of \(\cos 30^\circ\). [1 mark]

    Show answerHide answer

    Model answer

    (a) \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\). (b) \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\).

    Mark scheme

    • (a) \(\dfrac{\sqrt{3}}{2}\) — B1
    • (b) \(\dfrac{\sqrt{3}}{2}\) — B1
  3. 3 Calculate [3 marks]

    A right-angled triangle has a hypotenuse of 18 cm and an angle of \(30^\circ\). Calculate the length of the side opposite the \(30^\circ\) angle.

    Show answerHide answer

    Model answer

    \(\sin 30^\circ = \dfrac{x}{18}\) and \(\sin 30^\circ = \dfrac{1}{2}\), so \(x = 9\) cm.

    Mark scheme

    • \(\sin 30^\circ = \dfrac{x}{18}\) — M1
    • \(\dfrac{1}{2} \times 18\) — M1
    • 9 cm — A1
  4. 4 Calculate [3 marks]

    In a right-angled triangle, the side opposite angle \(\theta\) is 4 cm and the hypotenuse is 8 cm. Calculate the size of angle \(\theta\).

    Show answerHide answer

    Model answer

    \(\sin\theta = \dfrac{4}{8} = \dfrac{1}{2}\), so \(\theta = 30^\circ\).

    Mark scheme

    • \(\sin\theta = \dfrac{4}{8}\) — M1
    • \(\dfrac{1}{2}\) — M1
    • \(30^\circ\) — A1
  5. 5 Calculate [3 marks]

    A vertical flagpole is 10 m tall. On level ground it casts a shadow 10 m long. Calculate the angle of elevation of the Sun.

    Show answerHide answer

    Model answer

    The flagpole and the shadow form a right-angled triangle, so \(\tan\theta = \dfrac{10}{10} = 1\) and \(\theta = 45^\circ\).

    Mark scheme

    • \(\tan\theta = \dfrac{10}{10}\) — M1
    • \(\tan\theta = 1\) — M1
    • \(45^\circ\) — A1
  6. 6 Calculate [3 marks]

    A right-angled triangle has an angle of \(60^\circ\) and a hypotenuse of 12 cm. Calculate the length of the side opposite the \(60^\circ\) angle. Give your answer in the form \(a\sqrt{3}\).

    Show answerHide answer

    Model answer

    \(\sin 60^\circ = \dfrac{x}{12}\), so \(x = 12 \times \dfrac{\sqrt{3}}{2} = 6\sqrt{3}\) cm.

    Mark scheme

    • \(\sin 60^\circ = \dfrac{x}{12}\) — M1
    • \(12 \times \dfrac{\sqrt{3}}{2}\) — M1
    • \(6\sqrt{3}\) cm — A1
  7. 7 Show that [3 marks]

    An equilateral triangle has sides of length 2 cm. By splitting the triangle in half, show that \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\).

    Show answerHide answer

    Model answer

    Splitting the triangle gives a right-angled triangle with hypotenuse 2 cm and a base of 1 cm. The height is \(\sqrt{2^2 - 1^2} = \sqrt{3}\) cm. The angle at the top of the original triangle is \(60^\circ\), so the angle opposite the height is \(60^\circ\), and \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\), as required.

    Mark scheme

    • Right-angled triangle with hypotenuse 2 and base 1 — M1
    • Height \(= \sqrt{3}\) — M1
    • \(\sin 60^\circ = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{\sqrt{3}}{2}\) — A1

Quick check

  1. 1

    What is the formula for \(\sin\theta\) in a right-angled triangle?

    1. A\(\dfrac{\text{Adjacent}}{\text{Hypotenuse}}\)
    2. B\(\dfrac{\text{Opposite}}{\text{Hypotenuse}}\)
    3. C\(\dfrac{\text{Opposite}}{\text{Adjacent}}\)
    4. D\(\dfrac{\text{Hypotenuse}}{\text{Opposite}}\)
    Show answerHide answer

    B: \(\dfrac{\text{Opposite}}{\text{Hypotenuse}}\)

    SOH: sine is opposite over hypotenuse.

  2. 2

    Which ratio links the opposite side and the adjacent side?

    1. ATangent
    2. BSine
    3. CCosine
    4. DPythagoras
    Show answerHide answer

    A: Tangent

    TOA: tangent is opposite over adjacent.

  3. 3

    What is the exact value of \(\sin 30^\circ\)?

    1. A\(\dfrac{\sqrt{3}}{2}\)
    2. B1
    3. C\(\dfrac{\sqrt{2}}{2}\)
    4. D\(\dfrac{1}{2}\)
    Show answerHide answer

    D: \(\dfrac{1}{2}\)

    This is one of the exact values you must learn: \(\sin 30^\circ = \dfrac{1}{2}\).

  4. 4

    What is the exact value of \(\tan 45^\circ\)?

    1. A0
    2. B\(\dfrac{1}{2}\)
    3. C1
    4. D\(\sqrt{3}\)
    Show answerHide answer

    C: 1

    At \(45^\circ\) the opposite and adjacent sides are equal, so \(\tan 45^\circ = 1\).

  5. 5

    A right-angled triangle has a hypotenuse of 8 cm and an angle of \(30^\circ\). What is the length of the side opposite the \(30^\circ\) angle?

    1. A16 cm
    2. B4 cm
    3. C\(4\sqrt{3}\) cm
    4. D2 cm
    Show answerHide answer

    B: 4 cm

    \(\sin 30^\circ = \dfrac{x}{8}\), so \(x = \dfrac{1}{2} \times 8 = 4\).

  6. 6

    A right-angled triangle has opposite side 5 cm and adjacent side 5 cm. What is the angle?

    1. A\(45^\circ\)
    2. B\(30^\circ\)
    3. C\(60^\circ\)
    4. D\(90^\circ\)
    Show answerHide answer

    A: \(45^\circ\)

    \(\tan\theta = \dfrac{5}{5} = 1\), so \(\theta = 45^\circ\).

  7. 7

    In a right-angled triangle, \(\sin\theta = \dfrac{5}{13}\) and the hypotenuse is 39 cm. What is the opposite side?

    1. A3 cm
    2. B195 cm
    3. C5 cm
    4. D15 cm
    Show answerHide answer

    D: 15 cm

    \(x = \dfrac{5}{13} \times 39 = 5 \times 3 = 15\).

  8. 8

    A right-angled triangle has an angle of \(60^\circ\) and an adjacent side of 5 cm. What is the opposite side? (\(\tan 60^\circ = \sqrt{3}\))

    1. A\(\dfrac{5\sqrt{3}}{3}\) cm
    2. B\(\dfrac{5}{2}\) cm
    3. C\(5\sqrt{3}\) cm
    4. D\(5\sqrt{2}\) cm
    Show answerHide answer

    C: \(5\sqrt{3}\) cm

    \(\tan 60^\circ = \dfrac{x}{5}\), so \(x = 5\sqrt{3}\).

  9. 9

    A right-angled triangle has an angle of \(45^\circ\) and a hypotenuse of 10 cm. What is the opposite side?

    1. A\(10\sqrt{2}\) cm
    2. B\(5\sqrt{2}\) cm
    3. C\(5\sqrt{3}\) cm
    4. D5 cm
    Show answerHide answer

    B: \(5\sqrt{2}\) cm

    \(x = 10 \sin 45^\circ = 10 \times \dfrac{\sqrt{2}}{2} = 5\sqrt{2}\).