IGCSE · Practice questions · Similar shapes

Find the scale factor.

Six original Cambridge-style questions on similar shapes: scale factors, missing lengths, nested and crossed triangles, and giving geometric reasons (Extended).

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Match, divide, multiply.

Match: pair up corresponding sides (opposite equal angles). Redraw triangles separately if they overlap.
Divide: \(k = \text{new} \div \text{old}\) from one known matching pair.
Multiply: new length = old length \(\times k\); old length = new length \(\div k\).
Nested triangles: use whole sides from the shared vertex.
Reasons (Extended): two pairs of equal angles, each with its geometric reason.

Recall

[2 marks]

State the two conditions for two polygons to be similar.

Corresponding angles are equal. ✓
Corresponding sides are in the same ratio (all multiplied by the same scale factor). ✓

Calculation

[3 marks]

A triangle has sides 4 cm, 6 cm and 7 cm. A similar triangle has longest side 17.5 cm. Find its other two sides.

The longest sides match: \(k = 17.5 \div 7 = 2.5\). ✓
\(4 \times 2.5 = 10\) cm and \(6 \times 2.5 = 15\) cm. ✓

Analysis

[3 marks]

Rectangle A is 8 cm by 12 cm, rectangle B is 10 cm by 15 cm and rectangle C is 12 cm by 16 cm. Which rectangles are similar to A? Show your working.

A to B: \(10 \div 8 = 1.25\) and \(15 \div 12 = 1.25\). Equal, so B is similar to A. ✓
A to C: \(12 \div 8 = 1.5\) but \(16 \div 12 = 1.33\). Not equal, so C is not similar to A. ✓

Calculation

[4 marks]

In triangle ABC, D lies on AB and E lies on AC, with DE parallel to BC. AD = 6 cm, DB = 4 cm, AE = 7.5 cm and DE = 9 cm.

Find (a) BC (b) EC.

AB = 6 + 4 = 10 cm, so \(k = 10 \div 6 = \dfrac{5}{3}\). ✓
(a) BC \(= 9 \times \dfrac{5}{3} = 15\) cm. ✓
(b) AC \(= 7.5 \times \dfrac{5}{3} = 12.5\) cm, so EC \(= 12.5 - 7.5 = 5\) cm. ✓

Proof

[4 marks]

(Extended) PQ is parallel to RS. Lines PS and QR cross at T. PT = 3 cm, TS = 7.5 cm and PQ = 4 cm.

(a) Show that triangles PQT and SRT are similar, giving reasons. (b) Find RS.

(a) Angle PTQ = angle STR (vertically opposite angles). ✓
Angle QPT = angle RST (alternate angles, PQ parallel to RS). ✓
Two pairs of equal angles, so the triangles are similar (P matches S, Q matches R). ✓
(b) \(k = TS \div TP = 7.5 \div 3 = 2.5\), so RS \(= 4 \times 2.5 = 10\) cm. ✓

Analysis

[4 marks]

A photo is 10 cm wide and 15 cm high. Mia wants a similar enlargement that fits inside a frame 30 cm wide and 40 cm high, as large as possible.

(a) Show that the photo cannot fill the frame exactly. (b) Find the width and height of the largest similar enlargement that fits.

(a) Width factor \(30 \div 10 = 3\); height factor \(40 \div 15 = 2.67\). Different, so no similar enlargement fills both. ✓
(b) The smaller factor, 2.666..., is the largest that fits both ways. ✓
Width \(= 10 \times \dfrac{40}{15} = 26.7\) cm (3 s.f.); height \(= 15 \times \dfrac{40}{15} = 40\) cm. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.