Area of a Triangle Revision Notes | GCSE AQA Higher Further Maths

1. Why the Standard Area Formula Falls Short

You already know that the area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. That formula is reliable — but only when you actually know the perpendicular height. For a scalene triangle with no right angle, that height is often hidden.

Picture a triangle where you know two sides are 8 cm and 11 cm, and the angle between them is $35^\circ$. There is no obvious height to measure. This is exactly the situation the sine area formula was built for.

2. The Sine Area Formula

The key insight is that the perpendicular height of a triangle can always be expressed using a sine ratio. If you draw the height $h$ from one vertex down to the opposite side, it satisfies $h = b\sin C$. Substituting that into $\frac{1}{2} \times a \times h$ gives the formula below.

The letter $C$ is not fixed to a particular vertex — it labels whichever angle sits between the two sides you are using. You could equally write $\frac{1}{2}bc\sin A$ or $\frac{1}{2}ac\sin B$, as long as the angle is the one sandwiched between the two stated sides.

3. Worked Examples

Theory is clearest when you see it applied to a full question. The two examples below cover the standard calculation and then a reverse problem where the area is given and a side must be found.

Worked example: Finding the Area Given Two Sides and the Included Angle

Triangle $PQR$ has $PQ = 9\text{ cm}$, $PR = 6\text{ cm}$, and angle $P = 48^\circ$. Calculate the area of the triangle. Give your answer to 3 significant figures.

1. $a = PQ = 9\text{ cm}$, $b = PR = 6\text{ cm}$, $C = 48^\circ$ — Angle $P$ is between sides $PQ$ and $PR$, so it is the included angle. [M1 — correct identification]
2. $\text{Area} = \frac{1}{2}ab\sin C$ — State the formula before substituting — this earns the method mark even if arithmetic slips later.
3. $\text{Area} = \frac{1}{2} \times 9 \times 6 \times \sin 48^\circ$ — [M1 — correct substitution]
4. $\sin 48^\circ \approx 0.7431$ — Use your calculator in degree mode. Keep the full decimal for now — rounding here causes errors.
5. $\text{Area} = \frac{1}{2} \times 9 \times 6 \times 0.7431 = \frac{1}{2} \times 54 \times 0.7431 = \frac{1}{2} \times 40.1274 = 20.0637...$ — Work left to right: $9 \times 6 = 54$, then $54 \times 0.7431 = 40.1274$, then halve.
6. $\text{Area} = 20.1\text{ cm}^2$ (3 s.f.) — [A1 — correct answer with units]. Always include cm² for area.

Answer: $\text{Area} = 20.1\text{ cm}^2$

Now consider a reverse problem. The area formula can be rearranged to find a missing side when the area and the other measurements are known. The structure is the same — you are solving for a different unknown.

Worked example: Finding a Missing Side Given the Area

Triangle $ABC$ has $AB = 12\text{ cm}$, angle $A = 55^\circ$, and area $= 40\text{ cm}^2$. Find the length $AC$.

1. $\text{Area} = \frac{1}{2} \times AB \times AC \times \sin A$ — Angle $A$ is between sides $AB$ and $AC$, so it is the included angle. $AC$ is the unknown.
2. $40 = \frac{1}{2} \times 12 \times AC \times \sin 55^\circ$ — [M1 — correct substitution with area on the left]
3. $\sin 55^\circ \approx 0.8192$ — Keep the full decimal value to avoid rounding errors mid-calculation.
4. $40 = \frac{1}{2} \times 12 \times 0.8192 \times AC = 6 \times 0.8192 \times AC = 4.9152 \times AC$ — $\frac{1}{2} \times 12 = 6$, then $6 \times 0.8192 = 4.9152$.
5. $AC = \frac{40}{4.9152} = 8.138...$ — [M1 — correct rearrangement and division]
6. $AC = 8.14\text{ cm}$ (3 s.f.) — [A1 — correct answer with units]

Answer: $AC = 8.14\text{ cm}$

5. Exam Technique and Common Mistakes

The sine area formula is reliable, but there are a handful of errors that appear repeatedly in exam scripts. Knowing them in advance is the most efficient way to protect your marks.

The sine area formula is the foundation for more advanced triangle work at GCSE. Once you can apply $\frac{1}{2}ab\sin C$ fluently — forwards to find area and in reverse to find a missing side or angle — you have the core skill that underpins the sine rule and cosine rule questions that follow.