Areas of 2D Shapes Revision Notes | GCSE AQA Higher Maths

1. Why Area Formulae Matter

Every time an architect calculates how much flooring a room needs, or a farmer works out how much seed to buy for a field, they are using area. At GCSE, you need three specific formulae — triangles, parallelograms, and trapezia — and examiners test them regularly on both calculator-allowed questions and questions without a calculator.

The good news is that all three formulae are connected by one idea: base times perpendicular height. Once you see that thread, the formulae stop feeling like separate things to memorise and start feeling like variations on a single theme.

2. The Three Formulae

Before working through examples, pin down each formula precisely. The variable $h$ always means the perpendicular height — the distance measured at a right angle between the relevant parallel sides. This is the single most tested distinction in area questions.

Notice the pattern: the parallelogram formula is $b \times h$; the triangle is half of that ($\frac{1}{2} b h$) because a triangle is exactly half a parallelogram. The trapezium averages the two parallel sides first — $(a + b)$ — then treats that average as the effective base, halving the result.

3. Worked Examples

Theory clicks into place through practice. Work through each example below by covering the steps and attempting each one yourself before revealing the solution. These examples mirror the style and difficulty of GCSE Higher exam questions.

Worked example: Calculating the Area of a Triangle

A triangle has a base of 9 cm and a perpendicular height of 6 cm. Calculate its area. [2 marks]

1. $A = \frac{1}{2} \times b \times h$ — Always start by stating the formula — this earns the method mark [M1] even if arithmetic goes wrong later.
2. $A = \frac{1}{2} \times 9 \times 6$ — $b = 9$ cm and $h = 6$ cm. Substitution earns [M1].
3. $9 \times 6 = 54$, then $\frac{1}{2} \times 54 = 27$ — Multiply base by height first, then halve. Expanding the arithmetic avoids errors.
4. $A = 27 \text{ cm}^2$ — Square units are essential — missing units costs a mark [A1] on the answer.

Answer: $A = 27 \text{ cm}^2$

Worked example: Calculating the Area of a Parallelogram

A parallelogram has a base of 11 cm and a perpendicular height of 5 cm. Its slant side is 7 cm. Calculate its area. [2 marks]

1. $A = b \times h$ — The formula for a parallelogram uses base and perpendicular height only [M1].
2. $b = 11$ cm, $h = 5$ cm (perpendicular height). The slant side of 7 cm is not used. — Examiners include the slant side as a deliberate trap. The perpendicular height is always the one measured at 90° to the base.
3. $A = 11 \times 5 = 55$ — $11 \times 5 = 55$. Straightforward multiplication.
4. $A = 55 \text{ cm}^2$ — Include cm² — area is always in square units [A1].

Answer: $A = 55 \text{ cm}^2$

Worked example: Calculating the Area of a Trapezium

A trapezium has parallel sides of 8 cm and 14 cm, and a perpendicular height of 7 cm. Calculate its area. [2 marks]

1. $A = \frac{1}{2}(a + b) \times h$ — State the formula first to secure the method mark [M1].
2. $A = \frac{1}{2}(8 + 14) \times 7$ — $a = 8$ cm, $b = 14$ cm (the two parallel sides), $h = 7$ cm.
3. $8 + 14 = 22$ — Work inside the bracket first — this is the sum of the two parallel sides.
4. $22 \times 7 = 154$, then $\frac{1}{2} \times 154 = 77$ — $22 \times 7 = 154$. Halving: $154 \div 2 = 77$. You can halve either factor first to keep numbers manageable.
5. $A = 77 \text{ cm}^2$ — Square units confirm this is an area answer [A1].

Answer: $A = 77 \text{ cm}^2$

4. Finding a Missing Dimension

GCSE Higher exam questions do not always ask you to find the area. Sometimes the area is given and you need to work backwards to find a missing length. This tests whether you can rearrange the formula — a skill worth practising deliberately.

Worked example: Finding the Height of a Triangle Given Its Area

A triangle has an area of 40 cm² and a base of 10 cm. Find its perpendicular height. [3 marks]

1. $A = \frac{1}{2} \times b \times h$ — Start with the standard formula [M1].
2. $40 = \frac{1}{2} \times 10 \times h$ — $A = 40$ cm² and $b = 10$ cm are both known. $h$ is the unknown.
3. $40 = 5h$ — $\frac{1}{2} \times 10 = 5$. The equation is now straightforward to solve [M1].
4. $h = 40 \div 5 = 8$ — Divide both sides by 5. $40 \div 5 = 8$.
5. $h = 8 \text{ cm}$ — Height is a length, so the unit is cm (not cm²) [A1].

Answer: $h = 8 \text{ cm}$

5. Deliberate Practice

These questions cover all three shapes and include a reverse problem. Attempt each one before checking the solution — working under that slight pressure is what builds exam confidence.

6. Exam Technique and Common Mistakes

Knowing the formulae is only part of the challenge. The marks students drop most often come from a small set of predictable errors — and every one of them is avoidable once you know what to watch for.

These three formulae — triangle, parallelogram, and trapezium — are the core of the higher-tier mensuration specification, and every one of them rewards the same disciplined approach: write the formula, substitute carefully, use the perpendicular height, and include square units in your answer.

Area questions are most reliable when you slow down and choose the correct structure before calculating. For composite shapes, split the shape into familiar pieces such as rectangles and triangles, or subtract a missing part from a larger enclosing shape. Keep every length in the same unit before squaring, and check whether the question expects an exact answer or a rounded decimal. In exam conditions, a labelled sketch with the chosen decomposition written on it prevents accidental double-counting and helps you justify why your area calculation uses addition in one problem and subtraction in another.