GCSE Mathematics (9-1)

20.8 The area of any triangle — Learn

Prerequisites: triangle area, sine ratios, and matching sides to angles.

Objectives: use two sides and their included angle to find a triangle's area, or rearrange the formula to find a missing side or angle.

When perpendicular height is unknown, use Area=12absin⁡C\text{Area}=\frac12 ab\sin C. The angle CC must lie between the known sides aa and bb.

Example 1. Find the area when two sides are 88 cm and 55 cm, with included angle 30∘30^\circ.

Area=12(8)(5)sin⁡30∘\text{Area}=\frac12(8)(5)\sin30^\circ Area=20×12\text{Area}=20\times\frac12 Area=10 cm2\text{Area}=10\text{ cm}^2

Example 2. A triangle has sides 77 cm and 99 cm with included angle 40∘40^\circ. Find its area to 3 significant figures.

Area=12(7)(9)sin⁡40∘\text{Area}=\frac12(7)(9)\sin40^\circ Area≈20.25 cm2\text{Area}\approx20.25\text{ cm}^2 Area=20.3 cm2 (3 significant figures)\text{Area}=20.3\text{ cm}^2\text{ (3 significant figures)}

Example 3. A triangle has area 2020 cm2^2 and sides 88 cm and 1010 cm. Find its acute included angle CC.

20=12(8)(10)sin⁡C20=\frac12(8)(10)\sin C sin⁡C=2(20)(8)(10)\sin C=\frac{2(20)}{(8)(10)} sin⁡C=12\sin C=\frac12 C=30∘C=30^\circ

The other angle with sine 12\frac12 is 150∘150^\circ. The question asks for the acute angle, so 30∘30^\circ is the answer.

Example 4. A triangle has area 2020 cm2^2, one side is 1010 cm, and the included angle is 30∘30^\circ. Find the other side bb.

20=12(10)bsin⁡30∘20=\frac12(10)b\sin30^\circ 20=52b20=\frac52b b=8 cmb=8\text{ cm}

Watch out. The angle must be between the two given sides. If it is opposite one of them, mark the included angle before using the formula.

Progress check

For the garden rectangle, use base times perpendicular height for its area. For a slanted triangle, the sine factor supplies the perpendicular height.

You should now be able to

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