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Exam questions · Maths · Further Trigonometry

Area of a Triangle and Segments

  • 6 exam questions
  • 20 marks
  • 9 quick checks
  1. 1 Work out [2 marks]

    Diagram NOT accurately drawn. Work out the area of triangle \(ABC\). (2 marks)

    A triangle diagram showing a triangle with two sides and the angle between them.
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    Model answer

    Area \(= \dfrac{1}{2} \times 8 \times 5 \times \sin 30^\circ = 20 \times \dfrac{1}{2} = 10\) cm\(^2\).

    Mark scheme

    • \(\dfrac{1}{2} \times 8 \times 5 \times \sin 30^\circ\) — M1
    • \(10\) — A1
  2. 2 Work out [3 marks]

    A triangle has two sides of length 12 cm and 5 cm. Its area is 15 cm\(^2\). The angle between the two sides is acute. Work out the size of this angle. (3 marks)

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    Model answer

    \(15 = \dfrac{1}{2} \times 12 \times 5 \times \sin C = 30\sin C\), so \(\sin C = \dfrac{1}{2}\) and \(C = 30^\circ\).

    Mark scheme

    • \(15 = \dfrac{1}{2} \times 12 \times 5 \times \sin C\) — M1
    • \(\sin C = \dfrac{1}{2}\) — M1
    • \(30\) — A1
  3. 3 Work out [3 marks]

    In triangle \(ABC\), \(b = 9\) cm and angle \(C = 30^\circ\). The area of the triangle is 18 cm\(^2\). Work out the length of \(a\). (3 marks)

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    Model answer

    \(18 = \dfrac{1}{2} \times a \times 9 \times \sin 30^\circ = \dfrac{9a}{4}\), so \(a = 18 \times \dfrac{4}{9} = 8\) cm.

    Mark scheme

    • \(18 = \dfrac{1}{2} \times a \times 9 \times \sin 30^\circ\) — M1
    • \(18 = \dfrac{9a}{4}\) or equivalent — M1
    • \(8\) — A1
  4. 4 Work out [3 marks]

    Diagram NOT accurately drawn. The diagram shows a triangular plot of land \(PQR\). Work out the exact area of the plot. (3 marks)

    A triangle diagram showing a triangular plot of land PQR.
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    Model answer

    Area \(= \dfrac{1}{2} \times 10 \times 12 \times \sin 120^\circ = 60 \times \dfrac{\sqrt{3}}{2} = 30\sqrt{3}\) m\(^2\).

    Mark scheme

    • \(\dfrac{1}{2} \times 10 \times 12 \times \sin 120^\circ\) — M1
    • \(\sin 120^\circ = \dfrac{\sqrt{3}}{2}\) — M1
    • \(30\sqrt{3}\) — A1
  5. 5 Work out [5 marks]

    Diagram NOT accurately drawn. The diagram shows a circle, centre \(O\), with radius 4 cm. \(A\) and \(B\) are points on the circle and angle \(AOB = 120^\circ\). Work out the exact area of the segment bounded by the chord \(AB\) and the minor arc \(AB\). (5 marks)

    A circle with a chord AB and two radii enclosing an angle at the centre O, with the radius given.
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    Model answer

    Sector \(= \dfrac{120}{360} \times \pi \times 4^2 = \dfrac{16\pi}{3}\). Triangle \(= \dfrac{1}{2} \times 4 \times 4 \times \sin 120^\circ = 8 \times \dfrac{\sqrt{3}}{2} = 4\sqrt{3}\). Segment \(= \dfrac{16\pi}{3} - 4\sqrt{3}\) cm\(^2\).

    Mark scheme

    • \(\dfrac{120}{360} \times \pi \times 4^2\) — M1
    • \(\dfrac{16\pi}{3}\) — A1
    • \(\dfrac{1}{2} \times 4 \times 4 \times \sin 120^\circ\) — M1
    • \(4\sqrt{3}\) — A1
    • \(\dfrac{16\pi}{3} - 4\sqrt{3}\) — A1
  6. 6 Work out [4 marks]

    A regular hexagon has sides of length 4 cm. Work out the exact area of the hexagon. (4 marks)

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    Model answer

    The hexagon is made of 6 triangles with two sides of 4 cm and an angle of \(60^\circ\) between them. Each has area \(\dfrac{1}{2} \times 4 \times 4 \times \sin 60^\circ = 4\sqrt{3}\). So the total is \(6 \times 4\sqrt{3} = 24\sqrt{3}\) cm\(^2\).

    Mark scheme

    • Splits the hexagon into 6 triangles, each with an angle of \(60^\circ\) at the centre — M1
    • \(\dfrac{1}{2} \times 4 \times 4 \times \sin 60^\circ\) — M1
    • \(4\sqrt{3}\) for each triangle — A1
    • \(24\sqrt{3}\) — A1

Quick check

  1. 1

    What is the formula for the area of a triangle using a sine?

    1. A\(\dfrac{1}{2}ab\sin C\)
    2. B\(ab\sin C\)
    3. C\(\dfrac{1}{2}ab\cos C\)
    4. D\(\dfrac{1}{2}a\sin C\)
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    A: \(\dfrac{1}{2}ab\sin C\)

    The area is half the product of two sides and the sine of the angle between them.

  2. 2

    Which angle is used in \(\dfrac{1}{2}ab\sin C\)?

    1. AThe angle opposite \(a\)
    2. BThe largest angle
    3. CThe right angle
    4. DThe angle between sides \(a\) and \(b\)
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    D: The angle between sides \(a\) and \(b\)

    \(C\) is the included angle.

  3. 3

    What is \(\sin 90^\circ\)?

    1. A0
    2. B\(\dfrac{1}{2}\)
    3. C1
    4. D\(-1\)
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    C: 1

    This is on the sine graph at its maximum.

  4. 4

    A triangle has sides 8 cm and 5 cm with an angle of \(30^\circ\) between them. What is the area?

    1. A20 cm\(^2\)
    2. B10 cm\(^2\)
    3. C\(20\sqrt{3}\) cm\(^2\)
    4. D40 cm\(^2\)
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    B: 10 cm\(^2\)

    \(\dfrac{1}{2} \times 8 \times 5 \times \dfrac{1}{2} = 10\).

  5. 5

    A triangle has sides 6 cm and 8 cm with an angle of \(60^\circ\) between them. What is the exact area?

    1. A\(12\sqrt{3}\) cm\(^2\)
    2. B\(24\) cm\(^2\)
    3. C\(24\sqrt{3}\) cm\(^2\)
    4. D\(12\) cm\(^2\)
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    A: \(12\sqrt{3}\) cm\(^2\)

    \(\dfrac{1}{2} \times 6 \times 8 \times \dfrac{\sqrt{3}}{2} = 12\sqrt{3}\).

  6. 6

    A triangle has sides of 10 cm and 8 cm and an area of 20 cm\(^2\). The included angle is acute. What is it?

    1. A\(60^\circ\)
    2. B\(45^\circ\)
    3. C\(150^\circ\)
    4. D\(30^\circ\)
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    D: \(30^\circ\)

    \(20 = 40\sin C\), so \(\sin C = \dfrac{1}{2}\) and \(C = 30^\circ\).

  7. 7

    What is the area of a segment of a circle?

    1. AArea of the sector plus area of the triangle
    2. BArea of the circle minus the sector
    3. CArea of the sector minus area of the triangle
    4. DHalf the area of the sector
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    C: Area of the sector minus area of the triangle

    The segment is the sector with the triangle removed.

  8. 8

    A sector of a circle with radius 6 cm has an angle of \(60^\circ\). What is the area of the sector?

    1. A\(36\pi\) cm\(^2\)
    2. B\(6\pi\) cm\(^2\)
    3. C\(12\pi\) cm\(^2\)
    4. D\(\pi\) cm\(^2\)
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    B: \(6\pi\) cm\(^2\)

    \(\dfrac{60}{360} \times \pi \times 36 = 6\pi\).

  9. 9

    A triangle has two sides of 4 cm and 6 cm and an area of 6 cm\(^2\). What could the included angle be?

    1. A\(30^\circ\) or \(150^\circ\)
    2. B\(30^\circ\) only
    3. C\(60^\circ\) or \(120^\circ\)
    4. D\(45^\circ\) or \(135^\circ\)
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    A: \(30^\circ\) or \(150^\circ\)

    \(6 = 12\sin C\), so \(\sin C = \dfrac{1}{2}\), which gives \(30^\circ\) or \(150^\circ\).