Area of a triangle: ½ab sin C - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 37 exam-style questions on Area of a triangle: \\\\u00bdab sin C for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

37 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

The diagram shows triangle ABC.
AB = 7.2 cm, AC = 9.5 cm and angle BAC = 48°.

Work out the area of triangle ABC.
Give your answer correct to 3 significant figures.

Geometry diagram, marked "Diagram NOT accurately drawn". Triangle ABC. Side AB is labelled 7.2 cm. Side AC is labelled 9.5 cm. Angle BAC is marked 48°.

Model answer

Area \(= \frac{1}{2} \times 7.2 \times 9.5 \times \sin 48^\circ\)
\(= 34.2 \times 0.74314...\)
\(= 25.415...\)
Area = 25.4 cm² (3 s.f.)

Explanation

Use Area = \(\frac{1}{2}ab\sin C\), where \(a\) and \(b\) are two sides and \(C\) is the angle between them. Here the 48° angle is between AB and AC, so it is the included angle. Make sure your calculator is in degree mode, and only round at the end.

Mark scheme

1 mark
M1 for \(\frac{1}{2} \times 7.2 \times 9.5 \times \sin 48^\circ\)
oe
1 mark
A1 for 25.4 cm²
awrt 25.4 (25.415...); correct answer with no working scores full marks

Question 2

3 marks

The diagram shows parallelogram ABCD.
AB = 12.4 cm, AD = 7.8 cm and angle DAB = 64°.

Work out the area of the parallelogram.
Give your answer correct to 3 significant figures.

Geometry diagram, marked "Diagram NOT accurately drawn". Parallelogram ABCD. Side AB is labelled 12.4 cm. Side AD is labelled 7.8 cm. Angle DAB is marked 64°. AB and DC are each marked with an arrowhead (parallel lines). AD and BC are each marked with two arrowheads (parallel lines).

Model answer

Diagonal BD splits the parallelogram into two congruent triangles.
Area of triangle ABD \(= \frac{1}{2} \times 12.4 \times 7.8 \times \sin 64^\circ = 43.465...\)
Area of ABCD \(= 2 \times 43.465... = 86.931...\)
Area = 86.9 cm² (3 s.f.)

Explanation

A diagonal splits a parallelogram into two congruent triangles, so its area is \(2 \times \frac{1}{2}ab\sin C = ab\sin C\). The angle you use must be between the two sides. A common mistake is to stop after finding the area of one triangle.

Mark scheme

1 mark
M1 for the area of triangle ABD: \(\frac{1}{2} \times 12.4 \times 7.8 \times \sin 64^\circ\) (= 43.46...)
oe
1 mark
M1 for doubling, or for \(12.4 \times 7.8 \times \sin 64^\circ\)
oe
1 mark
A1 for 86.9 cm²
awrt 86.9 (86.931...); correct answer with no working scores full marks

Question 3

4 marks

The diagram shows a sector OAB of a circle with centre O.
The radius of the circle is 9 cm and angle AOB = 70°.
The shaded region is the segment between the chord AB and the arc AB.

Work out the area of the shaded region.
Give your answer correct to 3 significant figures.

Geometry diagram, marked "Diagram NOT accurately drawn". A minor arc from A to B with centre O (the region between the arc and AB is shaded). Line segments OA, OB and AB are drawn. Angle AOB is marked 70°. OA is labelled 9 cm.

Model answer

Area of sector \(= \frac{70}{360} \times \pi \times 9^2 = 49.480...\)
Area of triangle OAB \(= \frac{1}{2} \times 9 \times 9 \times \sin 70^\circ = 38.057...\)
Area of segment \(= 49.480... - 38.057... = 11.422...\)
Area = 11.4 cm² (3 s.f.)

Explanation

A segment is a sector with a triangle taken away. Find the sector area as a fraction of the circle, and the triangle area with \(\frac{1}{2}ab\sin C\), using the two radii and the angle between them. Keep full calculator values: rounding both areas early can change the final answer.

Mark scheme

1 mark
M1 for the area of the sector: \(\frac{70}{360} \times \pi \times 9^2\) (= 49.48...)
oe
1 mark
M1 for the area of triangle OAB: \(\frac{1}{2} \times 9 \times 9 \times \sin 70^\circ\) (= 38.05...)
oe
1 mark
M1 for area of sector − area of triangle
dependent on at least one previous M1
1 mark
A1 for 11.4 cm²
awrt 11.4 (11.42...); correct answer with no working scores full marks

Question 4

5 marks

A farmer has a field in the shape of a triangle.
Two sides of the field are 85 m and 120 m long, and the angle between these sides is 74°.

The farmer wants to spread fertiliser over the whole field.
One bag of fertiliser covers 500 m² and costs £18.50.
The farmer can only buy whole bags.

Work out the cost of the fertiliser the farmer needs to buy.

Model answer

Area \(= \frac{1}{2} \times 85 \times 120 \times \sin 74^\circ = 5100 \times 0.96126... = 4902.43...\) m²
Number of bags \(= 4902.43... \div 500 = 9.80...\)
9 bags are not enough, so the farmer needs 10 bags.
Cost \(= 10 \times £18.50 = £185\)

Explanation

First find the area with \(\frac{1}{2}ab\sin C\), using the angle between the two given sides. Then divide by the area one bag covers. Because only whole bags can be bought and 9.80 bags are needed, round up to 10: rounding to the nearest whole number is a common mistake in problems like this.

Mark scheme

1 mark
P1 for \(\frac{1}{2} \times 85 \times 120 \times \sin 74^\circ\)
oe; correct answer with no working scores full marks
1 mark
A1 for an area of 4902.4... m²
accept 4900 to 4903
1 mark
P1 for their area ÷ 500 (= 9.80...)
oe, e.g. 500 × 10 = 5000 compared with their area
1 mark
P1 for rounding up to 10 bags
ft their area, as long as it is not already a whole number of bags
1 mark
A1 for £185
cao

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