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Trigonometry - Area of a Triangle (1/2 ab sin C)

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The area formula is used to find the area of any triangle when two sides and the included angle are known (SAS - Side Angle Side).

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It is specifically useful for non-right-angled triangles where the vertical height is not provided.

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The angle used in the formula MUST be the angle trapped between the two sides being used.

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In standard triangle notation, side 'a' is opposite angle 'A', side 'b' is opposite angle 'B', and side 'c' is opposite angle 'C'.

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Ensure your calculator is set to 'DEG' (Degrees) mode for IGCSE questions unless radians are specified.

📐Formulae

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

Area=12bcsin⁡A\text{Area} = \frac{1}{2}bc \sin A

Area=12acsin⁡B\text{Area} = \frac{1}{2}ac \sin B

💡Examples

Problem 1:

In triangle ABC, side AB=7 cmAB = 7\text{ cm}, side AC=10 cmAC = 10\text{ cm}, and angle BAC=42∘BAC = 42^\circ. Calculate the area of the triangle correct to 3 significant figures.

Solution:

Area=12×7×10×sin⁡(42∘)≈23.419...≈23.4 cm2\text{Area} = \frac{1}{2} \times 7 \times 10 \times \sin(42^\circ) \approx 23.419... \approx 23.4\text{ cm}^2

Explanation:

Identify the two sides (b=10,c=7b=10, c=7) and the included angle (A=42∘A=42^\circ). Substitute these values into the formula 12bcsin⁡A\frac{1}{2}bc \sin A and calculate.

Problem 2:

A triangle has an area of 30 cm230\text{ cm}^2. Two of its sides are 8 cm8\text{ cm} and 11 cm11\text{ cm}. Find the size of the acute angle between these two sides.

Solution:

30=12×8×11×sin⁡θ⇒30=44sin⁡θ⇒sin⁡θ=3044⇒θ=sin⁡−1(0.6818)≈43.0∘30 = \frac{1}{2} \times 8 \times 11 \times \sin \theta \Rightarrow 30 = 44 \sin \theta \Rightarrow \sin \theta = \frac{30}{44} \Rightarrow \theta = \sin^{-1}(0.6818) \approx 43.0^\circ

Explanation:

Rearrange the area formula to solve for the unknown angle θ\theta. Divide the area by the product of 0.50.5 and the two sides, then use the inverse sine (sin⁡−1\sin^{-1}) function.

Problem 3:

Calculate the area of an equilateral triangle with side lengths of 6 cm6\text{ cm}.

Solution:

Area=12×6×6×sin⁡(60∘)=18×32=93≈15.6 cm2\text{Area} = \frac{1}{2} \times 6 \times 6 \times \sin(60^\circ) = 18 \times \frac{\sqrt{3}}{2} = 9\sqrt{3} \approx 15.6\text{ cm}^2

Explanation:

In an equilateral triangle, all sides are equal (6 cm6\text{ cm}) and all internal angles are 60∘60^\circ. Using a=6,b=6,a=6, b=6, and C=60∘C=60^\circ allows the use of the sine area formula.