Area of a triangle Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Area of a triangle questions. See exactly how to solve problems on area of a triangle, half ab sin C, identifying the included angle, exact surd answers.

area of a trianglehalf ab sin Cidentifying the included angleexact surd answersequilateral triangleisosceles triangle
GCSE Higher70 questionsStep-by-step solutions
Question 1
2 markseasy
In triangle ABCABC, AB=8AB = 8 cm, AC=5AC = 5 cm and angle BAC=30∘BAC = 30^\circ. Work out the area of triangle ABCABC.

Worked solution

  1. State the formula for the area of a triangle.

    Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

    For any triangle the area is 12absin⁡C\frac{1}{2}ab\sin C, where aa and bb are two sides of the triangle and CC is the angle between them.

  2. Identify the angle between the two given sides.

    a=AB,b=AC,C=∠BACa = AB, \quad b = AC, \quad C = \angle BAC

    The sides ABAB and ACAC both start at AA, so the angle between them is the angle at AA, which is ∠BAC\angle BAC. This is the included angle, and it is the only angle that may be used with these two sides.

  3. State the area of triangle ABC.

    Area=12×8×5×sin⁡30∘=10 cm2\text{Area} = \frac{1}{2} \times 8 \times 5 \times \sin 30^\circ = 10\text{ cm}^2

    The area of triangle ABCABC is 1010 cm2^2.

Answer
Area=10 cm2\text{Area} = 10\text{ cm}^2
Question 2
2 markseasy
In triangle PQRPQR, QP=12QP = 12 cm, QR=7QR = 7 cm and angle PQR=30∘PQR = 30^\circ. Work out the area of triangle PQRPQR.

Worked solution

  1. State the formula for the area of a triangle.

    Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

    For any triangle the area is 12absin⁡C\frac{1}{2}ab\sin C, where aa and bb are two sides of the triangle and CC is the angle between them.

  2. Identify the angle between the two given sides.

    a=QP,b=QR,C=∠PQRa = QP, \quad b = QR, \quad C = \angle PQR

    The sides QPQP and QRQR both start at QQ, so the angle between them is the angle at QQ, which is ∠PQR\angle PQR. This is the included angle, and it is the only angle that may be used with these two sides.

  3. State the area of triangle PQR.

    Area=12×12×7×sin⁡30∘=21 cm2\text{Area} = \frac{1}{2} \times 12 \times 7 \times \sin 30^\circ = 21\text{ cm}^2

    The area of triangle PQRPQR is 2121 cm2^2.

Answer
Area=21 cm2\text{Area} = 21\text{ cm}^2
Question 3
2 markseasy
In triangle XYZXYZ, ZX=6ZX = 6 cm, ZY=9ZY = 9 cm and angle XZY=90∘XZY = 90^\circ. Work out the area of triangle XYZXYZ.

Worked solution

  1. State the formula for the area of a triangle.

    Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

    For any triangle the area is 12absin⁡C\frac{1}{2}ab\sin C, where aa and bb are two sides of the triangle and CC is the angle between them.

  2. Identify the angle between the two given sides.

    a=ZX,b=ZY,C=∠XZYa = ZX, \quad b = ZY, \quad C = \angle XZY

    The sides ZXZX and ZYZY both start at ZZ, so the angle between them is the angle at ZZ, which is ∠XZY\angle XZY. This is the included angle, and it is the only angle that may be used with these two sides.

  3. State the area of triangle XYZ.

    Area=12×6×9×sin⁡90∘=27 cm2\text{Area} = \frac{1}{2} \times 6 \times 9 \times \sin 90^\circ = 27\text{ cm}^2

    The area of triangle XYZXYZ is 2727 cm2^2.

Answer
Area=27 cm2\text{Area} = 27\text{ cm}^2
Question 4
2 markseasy
In triangle DEFDEF, DE=10DE = 10 cm, DF=6DF = 6 cm and angle EDF=150∘EDF = 150^\circ. Work out the area of triangle DEFDEF.

Worked solution

  1. State the formula for the area of a triangle.

    Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

    For any triangle the area is 12absin⁡C\frac{1}{2}ab\sin C, where aa and bb are two sides of the triangle and CC is the angle between them.

  2. Identify the angle between the two given sides.

    a=DE,b=DF,C=∠EDFa = DE, \quad b = DF, \quad C = \angle EDF

    The sides DEDE and DFDF both start at DD, so the angle between them is the angle at DD, which is ∠EDF\angle EDF. This is the included angle, and it is the only angle that may be used with these two sides.

  3. State the area of triangle DEF.

    Area=12×10×6×sin⁡150∘=15 cm2\text{Area} = \frac{1}{2} \times 10 \times 6 \times \sin 150^\circ = 15\text{ cm}^2

    The area of triangle DEFDEF is 1515 cm2^2.

Answer
Area=15 cm2\text{Area} = 15\text{ cm}^2
Question 5
2 markseasy
In triangle LMNLMN, ML=8ML = 8 cm, MN=5MN = 5 cm and angle LMN=60∘LMN = 60^\circ. Work out the area of triangle LMNLMN. Give your answer in its simplest surd form.

Worked solution

  1. State the formula for the area of a triangle.

    Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

    For any triangle the area is 12absin⁡C\frac{1}{2}ab\sin C, where aa and bb are two sides of the triangle and CC is the angle between them.

  2. Identify the angle between the two given sides.

    a=ML,b=MN,C=∠LMNa = ML, \quad b = MN, \quad C = \angle LMN

    The sides MLML and MNMN both start at MM, so the angle between them is the angle at MM, which is ∠LMN\angle LMN. This is the included angle, and it is the only angle that may be used with these two sides.

  3. State the area of triangle LMN.

    Area=12×8×5×sin⁡60∘=103 cm2\text{Area} = \frac{1}{2} \times 8 \times 5 \times \sin 60^\circ = 10\sqrt{3}\text{ cm}^2

    The area of triangle LMNLMN is 10310\sqrt{3} cm2^2.

Answer
Area=103 cm2\text{Area} = 10\sqrt{3}\text{ cm}^2

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