What is G23: Area of a triangle?

Higher tier only. Use A=1/2absin C when two sides a and b and their included angle C are known.

How many marks are G23 questions worth?

In verified past-paper appearances, G23 questions on Edexcel GCSE Maths 1MA1 have been worth 2–6 marks.

Which past exam papers has G23 appeared in?

G23 has appeared in 9 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

G23

Know and apply Area = ½ ab sin C to calculate the area, sides or angles of any triangle [Higher only]

Higher only

Area of a triangle

Worked answers, methods and verified real exam appearances for G23 on Edexcel GCSE Maths 1MA1.

Explanation

Higher tier only. Use $A=\frac12ab\sin C$ when two sides $a$ and $b$ and their included angle $C$ are known.
The formula comes from $\frac12\times\text{base}\times\text{perpendicular height}$ , with height $b\sin C$ .
Rearrange it to find a missing side or use inverse sine for an angle.
Because $\sin C=\sin(180^\circ-C)$ , an angle calculation may have both an acute and an obtuse solution; test both against the stated triangle.
Examiners expect the angle between the two substituted sides, correct square units for area, and all valid angle solutions unless the diagram or context excludes one.

The included angle

C

produces perpendicular height

b\sin C

Worked example

Two sides of a triangle are $9$ cm and $12$ cm, with included angle $40^\circ$ . Find the area to $1$ decimal place.

1.Substitute the two sides and included angle: $A=\frac12(9)(12)\sin40^\circ$ .
2. $A=54\sin40^\circ=34.710\ldots$ .
3.Round the final area to $1$ decimal place and use square units.

Answer: $34.7\text{ cm}^2$ .

Common mistakes

Don't substitute an angle that is not between the two chosen sides.
Don't find one inverse-sine angle and discard the supplementary solution without checking it.

Exam tip

On the diagram, circle the included angle between the two substituted sides before using $\frac12ab\sin C$ .

Worked practice

Tier 1 · Easy

Two sides of a triangle are

10\text{ cm}

and

7\text{ cm}

, and their included angle is

30^\circ

. Work out the area.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$17.5\text{ cm}^2$	2	$A=\frac12ab\sin C=\frac12(10)(7)\sin30^\circ=35\times\frac12=17.5\text{ cm}^2$ .

Tier 2 · Standard

A triangle has area

48\text{ cm}^2

. Two sides have lengths

12\text{ cm}

and

x\text{ cm}

, and their included angle is

30^\circ

. Work out

x

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$x=16$	3	Use $48=\frac12(12)(x)\sin30^\circ$ . Since $\sin30^\circ=\frac12$ , this becomes $48=3x$ . Therefore $x=16$ .

Tier 3 · Hard

Two sides of a triangle are

10\text{ cm}

and

12\text{ cm}

. Its area is

30\sqrt3\text{ cm}^2

. Find both possible values of the included angle.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$60^\circ$ or $120^\circ$	4	$30\sqrt3=\frac12(10)(12)\sin C=60\sin C$ , so $\sin C=\frac{\sqrt3}{2}$ . Between $0^\circ$ and $180^\circ$ , this occurs at $C=60^\circ$ and $C=120^\circ$ .

Tier 1 · Easy

Triangle P has sides of

6\text{ cm}

and

9\text{ cm}

with an included angle of

30^\circ

. Triangle Q has sides of

5\text{ cm}

and

8\text{ cm}

with an included angle of

90^\circ

. Which triangle has the larger area? You must show your working.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	Triangle Q; its area is $20\text{ cm}^2$ compared with $13.5\text{ cm}^2$ for triangle P.	2	Area of P $=\frac12(6)(9)\sin30^\circ=27\times\frac12=13.5\text{ cm}^2$ . Area of Q $=\frac12(5)(8)\sin90^\circ=20\times1=20\text{ cm}^2$ . Since $20>13.5$ , triangle Q has the larger area.

Tier 2 · Standard

Two sides of a triangle are

7\text{ cm}

and

11\text{ cm}

. Work out the greatest possible area of the triangle and the included angle that gives this area.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	Greatest area $=38.5\text{ cm}^2$ Included angle $=90^\circ$	3	The area is $\frac12(7)(11)\sin C=38.5\sin C$ . The greatest possible value of $\sin C$ is $1$ , which occurs at $C=90^\circ$ . Therefore the greatest area is $38.5\text{ cm}^2$ .

Tier 3 · Hard

Two triangles each have sides of lengths

8\text{ cm}

and

13\text{ cm}

. The included angle is

35^\circ

in the first triangle and

145^\circ

in the second triangle. Show that the triangles have equal areas.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	Each area is $52\sin35^\circ\text{ cm}^2$ , because $\sin145^\circ=\sin35^\circ$ .	3	The first area is $\frac12(8)(13)\sin35^\circ=52\sin35^\circ\text{ cm}^2$ . The second area is $\frac12(8)(13)\sin145^\circ=52\sin145^\circ\text{ cm}^2$ . Since $145^\circ=180^\circ-35^\circ$ and $\sin(180^\circ-35^\circ)=\sin35^\circ$ , the two areas are equal.

Tier 2 · Standard

Convex quadrilateral

ABCD

has diagonal

AC

, with

B

and

D

on opposite sides of

AC

. Given that

AB=7\text{ cm}

AC=12\text{ cm}

AD=9\text{ cm}

\angle BAC=48^\circ

and

\angle CAD=35^\circ

, work out the area of the quadrilateral, correct to

1

decimal place.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$62.2\text{ cm}^2$	4	The area of triangle $ABC$ is $\frac12(7)(12)\sin48^\circ=31.2120\ldots\text{ cm}^2$ . The area of triangle $ACD$ is $\frac12(9)(12)\sin35^\circ=30.9731\ldots\text{ cm}^2$ . Since the triangles are on opposite sides of $AC$ , their areas add to $62.1852\ldots\text{ cm}^2$ , which is $62.2\text{ cm}^2$ correct to $1$ decimal place.

Tier 3 · Hard

Two sides of a triangle are

7.5\text{ cm}

and

20\text{ cm}

. The area of the triangle is

60\text{ cm}^2

and the included angle is acute. Work out the included angle and the length of the third side, giving each value correct to

1

decimal place.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	Included angle $=53.1^\circ$ Third side $=16.6\text{ cm}$	5	$60=\frac12(7.5)(20)\sin C$ , so $\sin C=0.8$ . The angle is acute, so $C=\sin^{-1}(0.8)=53.1301\ldots^\circ$ . By the cosine rule, the third side $c$ satisfies $c^2=7.5^2+20^2-2(7.5)(20)\cos C$ , giving $c=16.6207\ldots\text{ cm}$ . Therefore the answers are $53.1^\circ$ and $16.6\text{ cm}$ correct to $1$ decimal place.

Tier 3 · Hard

Two sides of a hinged triangular frame have fixed lengths

8\text{ cm}

and

15\text{ cm}

. The included angle is increased from

30^\circ

45^\circ

. Work out the exact increase in the area of the triangle.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$30(\sqrt2-1)\text{ cm}^2$	4	With included angle $30^\circ$ , the area is $\frac12(8)(15)\sin30^\circ=30\text{ cm}^2$ . With included angle $45^\circ$ , the area is $\frac12(8)(15)\sin45^\circ=30\sqrt2\text{ cm}^2$ . The exact increase is $30\sqrt2-30=30(\sqrt2-1)\text{ cm}^2$ .

Q10

Tier 3 · Hard

A triangle has area

15\sqrt3\text{ cm}^2

. Two of its sides have lengths

10\text{ cm}

and

x\text{ cm}

, and the angle between them is

60^\circ

. Work out the value of

x

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$x=6$	4	The area is $\frac12(10)(x)\sin60^\circ=5x\times\dfrac{\sqrt3}{2}=\dfrac{5\sqrt3}{2}x$ . Setting $\dfrac{5\sqrt3}{2}x=15\sqrt3$ gives $x=15\sqrt3\times\dfrac{2}{5\sqrt3}=6$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-06	3H	Q18	5	Allowed	Higher	QP MS
2022-06	1H	Q21	5	Non-calculator	Higher	QP MS
2019-06	3H	Q14	4	Allowed	Higher	QP MS
2023-11	2H	Q16	4	Allowed	Higher	QP MS
2019-11	2H	Q23	6	Allowed	Higher	QP MS
2024-11	3H	Q17	4	Allowed	Higher	QP MS
2022-11	2H	Q18	2	Allowed	Higher	QP MS
2022-11	3H	Q26	5	Allowed	Higher	QP MS
2021-11	2H	Q15	5	Allowed	Higher	QP MS

Other points in G Geometry and measures

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