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Maths · Further Trigonometry

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Area of a Triangle and Segments

Using half ab sin C to find areas, sides and angles, and finding the area of a segment of a circle.

  • Higher
  • 9 key terms
  • All boards

Learning Objectives

  1. 1Use the formula \(\text{area} = \dfrac{1}{2}ab\sin C\) to find the area of a triangle.
  2. 2Find a missing side or angle from the area of a triangle.
  3. 3Find the area of a segment of a circle.
  4. 4Use exact values to give answers with surds, such as \(12\sqrt{3}\).

The area of any triangle

The usual formula, half the base times the perpendicular height, needs the height, which is often not given. If you know two sides and the angle between them, the height can be written using sine, and this gives a new formula for the area that does not need the height. The formula is also a quick way to find the area of a segment of a circle, which is the sector minus a triangle.

Using the formula

Choose two sides and the angle between them.

  • The formula

    \(\text{Area} = \dfrac{1}{2}ab\sin C\).

  • Included angle

    The angle \(C\) is between the sides \(a\) and \(b\).

  • Exact values

    Use \(\sin 30^\circ = \dfrac{1}{2}\), \(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\) and \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\).

  • Units

    The area is in square units.

The area of a triangle

In triangle \(ABC\), \(a = 6\) cm, \(b = 8\) cm and angle \(C = 60^\circ\). Work out the exact area of the triangle.

Show the solutionHide the solution
  1. 1 Choose the formula Two sides and the angle between them are known.
  2. 2 Substitute \(\text{Area} = \dfrac{1}{2} \times 6 \times 8 \times \sin 60^\circ\).
  3. 3 Exact value \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\), so the area is \(24 \times \dfrac{\sqrt{3}}{2}\).
  4. 4 Simplify \(12\sqrt{3}\) cm\(^2\).

Answer\(12\sqrt{3}\) cm\(^2\)

Finding an angle or a side from the area

If the area is given, substitute it and solve.

  • Write the equation

    \(\text{Area} = \dfrac{1}{2}ab\sin C\), with the area on the left.

  • Rearrange

    \(\sin C = \dfrac{2 \times \text{Area}}{ab}\).

  • Inverse sine

    Use an exact value to find \(C\). If \(C\) could be obtuse, there are two possible angles, \(C\) and \(180^\circ - C\).

  • Missing side

    Rearrange for \(a\) or \(b\) in the same way.

Finding an angle from the area

A triangle has sides of 10 cm and 8 cm, and an area of \(20\) cm\(^2\). The angle between the sides is acute. Work out the size of this angle.

Show the solutionHide the solution
  1. 1 Write the equation \(20 = \dfrac{1}{2} \times 10 \times 8 \times \sin C\).
  2. 2 Simplify \(20 = 40\sin C\), so \(\sin C = \dfrac{1}{2}\).
  3. 3 Exact value \(\sin 30^\circ = \dfrac{1}{2}\), so \(C = 30^\circ\).
  4. 4 Check The other solution of \(\sin C = \dfrac{1}{2}\) is \(150^\circ\), which is obtuse, so it is not used.

Answer\(30^\circ\)

The area of a segment

A segment is the part of a circle between a chord and an arc.

  • Sector

    Area of the sector \(= \dfrac{\theta}{360} \times \pi r^2\).

  • Triangle

    The two radii and the chord make a triangle, with area \(\dfrac{1}{2}r^2\sin\theta\).

  • Segment

    Area of the segment \(=\) area of the sector \(-\) area of the triangle.

  • Exact answers

    Leave \(\pi\) and surds in the answer.

The area of a segment

\(O\) is the centre of a circle of radius 6 cm. \(A\) and \(B\) are points on the circle and angle \(AOB = 60^\circ\). Work out the exact area of the segment cut off by the chord \(AB\).

Show the solutionHide the solution
  1. 1 Sector \(\dfrac{60}{360} \times \pi \times 6^2 = \dfrac{1}{6} \times 36\pi = 6\pi\).
  2. 2 Triangle \(\dfrac{1}{2} \times 6 \times 6 \times \sin 60^\circ = 18 \times \dfrac{\sqrt{3}}{2} = 9\sqrt{3}\).
  3. 3 Subtract Segment \(= 6\pi - 9\sqrt{3}\).
  4. 4 Units The answer is in \(\text{cm}^2\).

Answer\((6\pi - 9\sqrt{3})\) cm\(^2\)

Test yourself

  1. 1

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

    Show answerHide answer

    \(\dfrac{1}{2}ab\sin C\).

  2. 2

    Which angle is used?

    Show answerHide answer

    The angle between the two sides.

  3. 3

    What is \(\sin 60^\circ\)?

    Show answerHide answer

    \(\dfrac{\sqrt{3}}{2}\).

  4. 4

    What is the area of a segment?

    Show answerHide answer

    Area of the sector minus area of the triangle.

  5. 5

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

    Show answerHide answer

    1.

Exam technique: area questions

Check the angle, then write each step.

  • Angle between

    Make sure the angle is between the two sides that you are using.

  • Exact answers

    If the question says exact, leave surds and \(\pi\).

  • Two possible angles

    If an angle is found from \(\sin\), check whether it could be obtuse.

  • Show every step

    The formula with the numbers in earns the first mark.

Summary and exam focus

  • \(\text{Area} = \dfrac{1}{2}ab\sin C\) with \(C\) between \(a\) and \(b\).
  • Rearrange the formula to find an angle or a side.
  • A segment is a sector minus a triangle.
  • Use exact values to give answers in surd form.

Exam focus

Triangle \(ABC\) has \(AB = 10\) cm, \(AC = 12\) cm and angle \(BAC = 30^\circ\). Work out the area of the triangle. (2 marks) (2 marks)

\(\text{Area} = \dfrac{1}{2} \times 10 \times 12 \times \sin 30^\circ = 60 \times \dfrac{1}{2} = 30\) cm\(^2\). Show the formula with the numbers substituted.

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Area
The amount of surface inside a shape.
Included angle
The angle between two given sides.
Segment
The part of a circle cut off by a chord.
Sector
The part of a circle between two radii.
Exact value
A value written with fractions, surds and \(\pi\).
Chord
A straight line joining two points on a circle.
Perpendicular height
The shortest distance from a vertex to the opposite side.
Surd
A root that cannot be written as a whole number or fraction.
Radius
The distance from the centre of a circle to its edge.

Questions and answers

15 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Calculate 2 marks Core

The diagram is not drawn to scale. Calculate the exact area of triangle \(ABC\). [2 marks]

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

Mark scheme — 2 marks available

  • \(\dfrac{1}{2} \times 10 \times 12 \times \sin 45^\circ\) — M1
  • \(30\sqrt{2}\) — A1

Model answer

Area \(= \dfrac{1}{2} \times 10 \times 12 \times \sin 45^\circ = 60 \times \dfrac{\sqrt{2}}{2} = 30\sqrt{2}\) cm\(^2\).

2. Exam question Calculate 3 marks Core

A triangle has two sides of length 10 cm and 4 cm. Its area is \(10\sqrt{2}\) cm\(^2\). The angle between the two sides is acute. Calculate the size of this angle. [3 marks]

Mark scheme — 3 marks available

  • \(10\sqrt{2} = \dfrac{1}{2} \times 10 \times 4 \times \sin C\) — M1
  • \(\sin C = \dfrac{\sqrt{2}}{2}\) — M1
  • \(45\) — A1

Model answer

\(10\sqrt{2} = \dfrac{1}{2} \times 10 \times 4 \times \sin C = 20\sin C\), so \(\sin C = \dfrac{\sqrt{2}}{2}\) and \(C = 45^\circ\).

3. Exam question Calculate 3 marks Core

In triangle \(ABC\), \(b = 8\) cm and angle \(C = 45^\circ\). The area of the triangle is \(20\sqrt{2}\) cm\(^2\). Calculate the length of \(a\). [3 marks]

Mark scheme — 3 marks available

  • \(20\sqrt{2} = \dfrac{1}{2} \times a \times 8 \times \sin 45^\circ\) — M1
  • \(20\sqrt{2} = 2\sqrt{2}a\) or equivalent — M1
  • \(10\) — A1

Model answer

\(20\sqrt{2} = \dfrac{1}{2} \times a \times 8 \times \sin 45^\circ = 2\sqrt{2}a\), so \(a = 10\) cm.

4. Exam question Calculate 3 marks Core

The diagram is not drawn to scale. The diagram shows a triangular plot of land \(PQR\). Calculate the exact area of the plot. [3 marks]

A triangle diagram showing a triangular plot of land PQR.

Mark scheme — 3 marks available

  • \(\dfrac{1}{2} \times 6 \times 10 \times \sin 135^\circ\) — M1
  • \(\sin 135^\circ = \dfrac{\sqrt{2}}{2}\) — M1
  • \(15\sqrt{2}\) — A1

Model answer

Area \(= \dfrac{1}{2} \times 6 \times 10 \times \sin 135^\circ = 30 \times \dfrac{\sqrt{2}}{2} = 15\sqrt{2}\) m\(^2\).

5. Exam question Calculate 5 marks Stretch

The diagram is not drawn to scale. The diagram shows a circle, centre \(O\), with radius 8 cm. \(A\) and \(B\) are points on the circle and angle \(AOB = 60^\circ\). Calculate 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.

Mark scheme — 5 marks available

  • \(\dfrac{60}{360} \times \pi \times 8^2\) — M1
  • \(\dfrac{32\pi}{3}\) — A1
  • \(\dfrac{1}{2} \times 8 \times 8 \times \sin 60^\circ\) — M1
  • \(16\sqrt{3}\) — A1
  • \(\dfrac{32\pi}{3} - 16\sqrt{3}\) — A1

Model answer

Sector \(= \dfrac{60}{360} \times \pi \times 8^2 = \dfrac{32\pi}{3}\). Triangle \(= \dfrac{1}{2} \times 8 \times 8 \times \sin 60^\circ = 32 \times \dfrac{\sqrt{3}}{2} = 16\sqrt{3}\). Segment \(= \dfrac{32\pi}{3} - 16\sqrt{3}\) cm\(^2\).

6. Exam question Calculate 4 marks Stretch

A regular hexagon has sides of length 2 cm. Calculate the exact area of the hexagon. [4 marks]

Mark scheme — 4 marks available

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

Model answer

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

7. Multiple choice 1 mark Easier

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

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

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

8. Multiple choice 1 mark Easier

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

  1. A The angle opposite \(a\)
  2. B The largest angle
  3. C The right angle
  4. D The angle between sides \(a\) and \(b\) Correct

Why: \(C\) is the included angle.

9. Multiple choice 1 mark Easier

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

  1. A 0
  2. B \(\dfrac{1}{2}\)
  3. C 1 Correct
  4. D \(-1\)

Why: This is on the sine graph at its maximum.

10. Multiple choice 1 mark Core

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

  1. A 20 cm\(^2\)
  2. B 10 cm\(^2\) Correct
  3. C \(20\sqrt{3}\) cm\(^2\)
  4. D 40 cm\(^2\)

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

11. Multiple choice 1 mark Core

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

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

12. Multiple choice 1 mark Core

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 \(30^\circ\) Correct

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

13. Multiple choice 1 mark Core

What is the area of a segment of a circle?

  1. A Area of the sector plus area of the triangle
  2. B Area of the circle minus the sector
  3. C Area of the sector minus area of the triangle Correct
  4. D Half the area of the sector

Why: The segment is the sector with the triangle removed.

14. Multiple choice 1 mark Stretch

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\(^2\) Correct
  3. C \(12\pi\) cm\(^2\)
  4. D \(\pi\) cm\(^2\)

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

15. Multiple choice 1 mark Stretch

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

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