Area of any triangle Quiz (Grade 7)

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Topic:

Score: 0 / 10

Area of any triangle — Notes & Worked Examples Show / Hide

Explanation:

The area of any triangle can be found using Area = 12ab sin C, where a and b are two sides and C is the included angle sandwiched directly between them. If you are given two angles instead of the included angle, use angles in a triangle (180°) to find the included angle first.

When working backwards from a given area to find an unknown angle C, remember that sin(180° − θ) = sin θ: if the question states the angle is obtuse, subtract the calculator's acute angle from 180°.

Example 1: Find the area of a triangle with sides 8.5 cm and 11.2 cm and included angle 64° (3 s.f.).

Solution:

Substitute a = 8.5, b = 11.2, C = 64°:

Area = 12 × 8.5 × 11.2 × sin 64°

Area = 47.6 × sin 64° = 42.7826... cm²

Answer: 42.8 cm² (3 s.f.)

Example 2: A triangle has sides 9 cm and 12 cm and area 45 cm². Given angle C is obtuse, find C (1 d.p.).

Solution:

12 × 9 × 12 × sin C = 45 ⇒ 54 sin C = 45

sin C = 4554 = 56

Acute angle = sin⁻¹(56) = 56.442...°

Obtuse angle C = 180° − 56.442...° = 123.557...°

Answer: 123.6° (1 d.p.)

One Page Guide (PDF) Area of any triangle Worksheets

Question 1

[2 marks]

Calculate the area of triangle ABC where AC = 9 cm, AB = 10 cm, and angle BAC = 30°.

cm²

Worked Solution & Answer

Area = 12ab\sin C = 12 × 10 × 9 × \sin(30°) = 22.5 cm² B1

Answer: 22.5 cm² cm²

Question 2

[2 marks]

Calculate the area of triangle ABC where AC = 10 cm, AB = 8 cm, and angle BAC = 30°.

cm²

Worked Solution & Answer

Area = 12ab\sin C = 12 × 8 × 10 × \sin(30°) = 20 cm² B1

Answer: 20 cm² cm²

Question 3

[2 marks]

Here is triangle ABC.

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

cm²

Worked Solution & Answer

Area = 1/2 × b × c × sin A = 1/2 × 8.3 × 9.8 × sin(35°) = 40.67 × 0.573576... = 23.327... M1
⇒ 23.3 cm² (3 s.f.). A1

Answer: 23.3 cm²

Question Paper (PDF) • Mark Scheme (PDF)

Question 4

[4 marks]

The diagram shows a rectangle, ABDE, and two congruent triangles, AFE and BCD.

area of rectangle ABDE = area of triangle AFE + area of triangle BCD
AB : AE = 1 : 3
Work out the length of AE.

Worked Solution & Answer

Area of 2 triangles = 2 × (1/2 × 24 × AE × sin 30°) = 12 × AE M2
Area of rectangle = AB × AE = 12 × AE ⇒ AB = 12 cm M1
Since AB : AE = 1 : 3, AE = 3 × 12 = 36 cm A1

Answer: 36 cm

Question Paper (PDF) • Mark Scheme (PDF)

Question 5 AQA Nov 2023 3H Q24

[4 marks]

Work out the area of triangle ABC.

Answer cm²

Worked Solution & Answer

Angle A = 180° – (73° + 56°) = 51° M2
By the sine rule: BC / sin 51° = 24 / sin 73° ⇒ BC = (24 × sin 51°) / sin 73° = 19.5035... cm M1
Area = 1/2 × 24 × 19.5035... × sin 56° = 193.53... ≈ 193.5 cm². A1

Answer: Answer 193.5 cm²

Question Paper (PDF) • Mark Scheme (PDF)

Question 6 IGCSE Nov 2024 1H Q16

[4 marks]

OAPB is a sector of a circle, centre O

Angle AOB = 50°
Area of triangle OAB = 120 cm²
Work out the area of the sector OAPB
Give your answer correct to 3 significant figures.

cm²

Worked Solution & Answer

(The area of a sector is (angle ÷ 360°) × πr²)
½ r² sin 50° = 120 ⇒ r² = 240 / sin 50° = 313.2977.. M3
Sector area = (50 / 360) × π × r² = (50 / 360) × π × 313.2977 = 136.701... ≈ 137 cm² A1

Answer: 137 cm²

Question Paper (PDF) • Mark Scheme (PDF)

Question 7

[5 marks]

The area of triangle ABC is 6√(2) m².
Calculate the value of x.
Give your answer correct to 3 significant figures.

Worked Solution & Answer

Area = 1/2 × (x + 3)(2x – 1) × sin 45° = 6√2. Since sin 45° = √2 / 2, (1/2)(√2 / 2)(2x² + 5x – 3) = 6√2 M2
⇒ 2x² + 5x – 3 = 24 M2
⇒ 2x² + 5x – 27 = 0. Using quadratic formula: x = (–5 + √(25 – 4(2)(–27))) / 4 = (–5 + √241) / 4 = 2.631... = 2.63. A1

Answer: 2.63

Question Paper (PDF) • Mark Scheme (PDF)

Question 8

[5 marks]

ABCD is a quadrilateral.

The area of triangle ABC is 54 cm²
Calculate the area of triangle ACD.
Give your answer correct to 3 significant figures.

cm²

Worked Solution & Answer

In ΔABC: ½ × 17 × AC × sin 35° = 54 ⇒ AC = 108 / (17 sin 35°) = 11.076... cm M2
In ΔACD: ∠CAD = 180° – (48° + 57°) = 75° M1
By Sine Rule: CD / sin 75° = AC / sin 57° ⇒ CD = 11.076... × sin 75° / sin 57° = 12.755... cm M1
Area of ΔACD = ½ × AC × CD × sin 48° = ½ × 11.076... × 12.755... × sin 48° = 52.496... = 52.5 cm² (3 sf). A1

Answer: 52.5 cm²

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Question 9

[5 marks]

Here is triangle ABC.

(a) Find the length of BC.
Give your answer correct to 3 significant figures. [3 marks]

(b) Find the area of triangle ABC.
Give your answer correct to 3 significant figures. [2 marks]

cm²

Worked Solution & Answer

(a)

BC² = 8² + 11² – 2(8)(11) cos 72° = 185 – 54.387 = 130.613 M2
⇒ BC = 11.428... = 11.4 cm A1

Answer (a): 11.4 cm

(b)

Area = ½ × 8 × 11 × sin 72° = 41.846... = 41.8 cm² M1 A1

Answer (b): 41.8 cm²

Question Paper (PDF) • Mark Scheme (PDF)

Question 10 IGCSE Jan 2020 1H Q14

[5 marks]

The diagram shows parallelogram EFGH.
EF = 9.3 cm
FG = 14.7 cm
Angle EFG = 106°

(a) Work out the area of the parallelogram.
Give your answer correct to 3 significant figures. [2 marks]

cm²

(b) Work out the length of the diagonal EG of the parallelogram.
Give your answer correct to 3 significant figures. [3 marks]

Worked Solution & Answer

(a)

(The area of a parallelogram is base × perpendicular height)
Area = 2 × ½ × 9.3 × 14.7 × sin 106° = 131.4... = 131 cm² M1 A1

Answer (a): 131 cm²

(b)

EG² = 9.3² + 14.7² − 2(9.3)(14.7) cos 106° = 377.95... M2
⇒ EG = 19.4 cm A1

Answer (b): 19.4 cm

Question Paper (PDF) • Mark Scheme (PDF)

Score: 0 / 10 (0 / 38 marks)

Area of any triangle Worksheets Next: The sine rule →

Grade 7 Quizzes

19 Topics

7A - Surds	7B - Bounds
7C - Proportion (harder)	7D - Algebraic fractions
7E - The quadratic formula	7F - Factorising harder quadratics
7G - Trigonometric & exp. Graphs	7H - Rearranging (harder)
7J - Functions (harder)	7K - Iteration
7L - Area of any triangle	7M - The sine rule
7N - The cosine rule	7P - Congruent triangles
7Q - 3D pythagoras / trig.	7R - Histograms
7S - Conditional probability	7T - Exact trig. values (harder)
7U - Area of a segment