OpenRevise

Exam questions · Maths · Further Trigonometry

Area of a Triangle and Segments

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

    Not drawn accurately. Work out the exact area of triangle \(ABC\). [2 marks]

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

    Model answer

    Area \(= \dfrac{1}{2} \times 6 \times 8 \times \sin 60^\circ = 24 \times \dfrac{\sqrt{3}}{2} = 12\sqrt{3}\) cm\(^2\).

    Mark scheme

    • \(\dfrac{1}{2} \times 6 \times 8 \times \sin 60^\circ\) — M1
    • \(12\sqrt{3}\) — A1
  2. 2 Work out [3 marks]

    A triangle has two sides of length 8 cm and 6 cm. Its area is \(12\sqrt{3}\) cm\(^2\). The angle between the two sides is acute. Work out the size of this angle. [3 marks]

    Show answerHide answer

    Model answer

    \(12\sqrt{3} = \dfrac{1}{2} \times 8 \times 6 \times \sin C = 24\sin C\), so \(\sin C = \dfrac{\sqrt{3}}{2}\) and \(C = 60^\circ\).

    Mark scheme

    • \(12\sqrt{3} = \dfrac{1}{2} \times 8 \times 6 \times \sin C\) — M1
    • \(\sin C = \dfrac{\sqrt{3}}{2}\) — M1
    • \(60\) — A1
  3. 3 Work out [3 marks]

    In triangle \(ABC\), \(b = 10\) cm and angle \(C = 60^\circ\). The area of the triangle is \(15\sqrt{3}\) cm\(^2\). Work out the length of \(a\). [3 marks]

    Show answerHide answer

    Model answer

    \(15\sqrt{3} = \dfrac{1}{2} \times a \times 10 \times \sin 60^\circ = \dfrac{5\sqrt{3}a}{2}\), so \(a = 15\sqrt{3} \times \dfrac{2}{5\sqrt{3}} = 6\) cm.

    Mark scheme

    • \(15\sqrt{3} = \dfrac{1}{2} \times a \times 10 \times \sin 60^\circ\) — M1
    • \(15\sqrt{3} = \dfrac{5\sqrt{3}a}{2}\) or equivalent — M1
    • \(6\) — A1
  4. 4 Work out [3 marks]

    Not drawn accurately. 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.
    Show answerHide answer

    Model answer

    Area \(= \dfrac{1}{2} \times 8 \times 9 \times \sin 120^\circ = 36 \times \dfrac{\sqrt{3}}{2} = 18\sqrt{3}\) m\(^2\).

    Mark scheme

    • \(\dfrac{1}{2} \times 8 \times 9 \times \sin 120^\circ\) — M1
    • \(\sin 120^\circ = \dfrac{\sqrt{3}}{2}\) — M1
    • \(18\sqrt{3}\) — A1
  5. 5 Work out [5 marks]

    Not drawn accurately. The diagram shows a circle, centre \(O\), with radius 6 cm. \(A\) and \(B\) are points on the circle and angle \(AOB = 90^\circ\). 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.
    Show answerHide answer

    Model answer

    Sector \(= \dfrac{90}{360} \times \pi \times 6^2 = 9\pi\). Triangle \(= \dfrac{1}{2} \times 6 \times 6 \times \sin 90^\circ = 18\). Segment \(= 9\pi - 18\) cm\(^2\).

    Mark scheme

    • \(\dfrac{90}{360} \times \pi \times 6^2\) — M1
    • \(9\pi\) — A1
    • \(\dfrac{1}{2} \times 6 \times 6 \times \sin 90^\circ\) — M1
    • \(18\) — A1
    • \(9\pi - 18\) — A1
  6. 6 Show that [3 marks]

    An equilateral triangle has sides of length 10 cm. Show that its area is \(25\sqrt{3}\) cm\(^2\). [3 marks]

    Show answerHide answer

    Model answer

    Area \(= \dfrac{1}{2} \times 10 \times 10 \times \sin 60^\circ = 50 \times \dfrac{\sqrt{3}}{2} = 25\sqrt{3}\) cm\(^2\).

    Mark scheme

    • \(\dfrac{1}{2} \times 10 \times 10 \times \sin 60^\circ\) — M1
    • \(50 \times \dfrac{\sqrt{3}}{2}\) — M1
    • \(25\sqrt{3}\), with a clear conclusion — Q1

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\(\dfrac{1}{2}a\sin C\)
    Show answerHide answer

    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 \(b\)
    Show answerHide answer

    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\(-1\)
    Show answerHide answer

    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\(^2\)
    Show answerHide answer

    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\(^2\)
    Show answerHide answer

    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\(30^\circ\)
    Show answerHide answer

    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
    Show answerHide answer

    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\(^2\)
    Show answerHide answer

    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 \(135^\circ\)
    Show answerHide answer

    A: \(30^\circ\) or \(150^\circ\)

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