Learn | Grasp Maths

Lesson overview

Geometry — mixed problems and methods

GCSE geometry exam questions often combine multiple skills: area formulas, circle properties, angle relationships, trigonometry (sin, cos, tan rules), vector methods, and similarity or congruence. Success requires: quickly identifying which properties apply, choosing the most efficient method, and maintaining accuracy through multi-step calculations. Common structures: finding an unknown side/angle using one method, then using that result in another (e.g., find area using trigonometry, then use area formula). Sketching accurate diagrams and labelling known and unknown values is essential for clarity and avoiding errors.

Combined problem: area and trigonometry

Triangle ABC has AB = 10 cm, angle A = 40°, angle B = 60°. Find the area.

Multi-step: find missing sides/angles first using sine/cosine rules, then use area formula.

Combined problem: circles and coordinates

A circle has centre (3, 4) and radius 5. Point P = (6, 8). Is P on, inside, or outside the circle?

Compare the distance from centre to the point with the radius.

Combined problem: area and compound shapes

A shape consists of a rectangle 8 cm × 5 cm with a semicircle of diameter 5 cm attached to one side. Find the total area.

Break the compound shape into simpler parts, find the area of each, then add.

Combined problem: similarity and scale

Two similar triangles have corresponding sides in ratio 2:3. If the smaller triangle has area 40 ${cm}^{2}$ , what is the area of the larger triangle?

Smaller (area 40 cm²)

Larger

For similar shapes, if side ratio is a:b, area ratio is ${a}^{2}$ : ${b}^{2}$ .

Worked example — multi-step geometry problem

A trapezium ABCD has parallel sides AB and DC. AB = 12 cm, DC = 8 cm, and the perpendicular height is 6 cm. Find the area. If angle A = 50°, find the length AD.

Identify the trapezium formula: Area = ( $\frac{1}{2}$ )(a + b)h where a and b are parallel sides and h is height.
Substitute: Area = ( $\frac{1}{2}$ )(12 + 8)(6) = ( $\frac{1}{2}$ )(20)(6) = 60 ${cm}^{2}$ .
For the slant side AD, use the right-angled triangle formed by the height.
The height is the opposite side to angle A = 50°.
Use sin(50°) = opposite / hypotenuse = 6 / AD.
Rearrange: AD = 6 / sin(50°) ≈ 7.83 cm.

Try it

Use the examples carefully, then choose the answer.

Question 1

A rectangle has length 10 cm and width 8 cm. A circle is inscribed (fits inside touching all sides). What is the area of the circle?

💡 The inscribed circle's diameter equals the smaller dimension: 8 cm. Radius = 4 cm.

Question 2

Two triangles are similar with side ratio 3:5. If the smaller triangle has area 45 ${cm}^{2}$ , what is the larger triangle's area?

Smaller (area 45 cm²)

Larger

💡 Area ratio = ( $\frac{3}{5}$ ${)}^{2}$ = $\frac{9}{25}$ . So 45 = ( $\frac{9}{25}$ ) × large → large = 125 ${cm}^{2}$ .

Question 3

A trapezium has parallel sides 6 cm and 10 cm, with height 4 cm. What is its area?

💡 Area = ( $\frac{1}{2}$ )(6 + 10)(4) = ( $\frac{1}{2}$ )(16)(4) = 32 ${cm}^{2}$ .

Question 4

A compound shape is a rectangle 10 cm × 6 cm with a triangle (base 10 cm, height 4 cm) on top. Total area?

💡 Rectangle area = 60 ${cm}^{2}$ . Triangle area = ( $\frac{1}{2}$ ) × 10 × 4 = 20 ${cm}^{2}$ . Total = 80 ${cm}^{2}$ .

Question 5

A sector of a circle has radius 8 cm and angle 45°. Find the area (to 1 d.p.).

💡 Area = (angle/360) × π ${r}^{2}$ = ( $\frac{45}{360}$ ) × π × 64.

Common mistakes

Watch for these when working through the lesson.

Forgetting to add areas of separate parts in a compound shape.
Using the wrong area formula (e.g., trapezium formula instead of triangle).
Confusing the side ratio with the area ratio in similar shapes (area ratio is the square of the side ratio).
Not identifying which method to use first; carefully read the given information.

Combined geometry exam questions

Lesson overview

Worked example — multi-step geometry problem

Try it

Common mistakes

Related topics

Independent practice will plug in here