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Cambridge IGCSE Mathematics 0580 · Geometry

Similarity

What makes two shapes similar, how to find a scale factor and use it to calculate missing lengths, including similar triangles inside each other and across crossing lines. Extended: showing that triangles are similar, and how areas, surface areas and volumes of similar shapes and solids are related.

C4.4E4.4Core and ExtendedFree lesson

What the syllabus says

C4.4 (Core)

Calculate lengths of similar shapes.

E4.4 (Extended)

1 Calculate lengths of similar shapes.
2 Use the relationships between lengths and areas of similar shapes and lengths, surface areas and volumes of similar solids.
3 Solve problems and give simple explanations involving similarity.

Notes and examples

Includes use of scale factor, e.g. $\frac{Volume of A}{Volume of B} = \frac{(Length of A)^{3}}{(Length of B)^{3}}$ .
Includes showing that two triangles are similar using geometric reasons.

Quoted from the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027.

What similar means

Two shapes are similar when one is an enlargement of the other. That means:

all the corresponding angles are equal, and
all the corresponding lengths are in the same ratio.

That common ratio is the scale factor, $k$ :
$k = \frac{length on the new shape}{corresponding length on the original shape}$

If $k > 1$ the new shape is bigger; if $k < 1$ it is smaller; if $k = 1$ the shapes are congruent. The scale factor works on every length: sides, heights, diagonals, perimeters and radii all multiply by $k$ .

The key word is corresponding: lengths in the same position on the two shapes. In rectangle A below, the longer side 3 corresponds to the longer side 4.5 of B, and the shorter side 2 to the shorter side 3. Both ratios are $1.5$ , so B is similar to A with scale factor 1.5.

Similar is about multiplying, never adding. A 3 by 2 rectangle with 1.5 added to each side becomes 4.5 by 3.5, which is a different shape.

Finding the scale factor and missing lengths

The method for any pair of similar shapes:

Match up the corresponding sides. If the shapes are turned or flipped, use the equal angles to decide which side matches which.
Find a pair of corresponding sides where both lengths are known, and divide: $k = \frac{new}{original}$ .
To find a length on the new shape, multiply the matching original length by $k$ . To find a length on the original, divide by $k$ .

Always check the direction with common sense: the length on the bigger shape must come out bigger.

Scale factors are often fractions or decimals. Keeping them as fractions avoids rounding: if $k = \frac{7}{4}$ , a side of 6 cm becomes $6 \times \frac{7}{4} = 10.5$ cm exactly. On a calculator paper, store the scale factor in the memory rather than rounding it.

Worked example 1

Two similar picture frames are rectangles. The small frame is 15 cm by 12 cm. The longer side of the large frame is 40 cm. Find the shorter side of the large frame.

Corresponding sides: longer to longer. $k = \frac{40}{15} = \frac{8}{3}$ .
Shorter side $= 12 \times \frac{8}{3} = 32$ cm.
Check: $\frac{40}{32} = 1.25$ and $\frac{15}{12} = 1.25$ , so the large frame has the same shape.

Answer: 32 cm

Check yourself

Rectangles $A$ and $B$ are similar. $A$ is 8 cm by 5 cm. The shorter side of $B$ is 12 cm. Find the longer side of $B$ .

Answer it in the portal, where your answer is checked as you type →

Similar triangles inside each other

A very common exam diagram is a triangle with a line drawn across it parallel to one side. The small triangle at the top and the whole triangle are similar:

they share the angle at the top vertex, and
the parallel line makes corresponding angles equal at the other two corners.

The trap is the sides. The small triangle has sides $A P$ and $A Q$ ; the large triangle has sides $A B$ and $A C$ , which are the whole lengths, not the bits $P B$ and $Q C$ . If you are told $A P = 3$ and $P B = 6$ , then $A B = 3 + 6 = 9$ .

It helps to redraw the two triangles separately, side by side, and label every length you know on each.

Worked example 2

In the diagram above, $P Q$ is parallel to $B C$ . $A P = 3$ cm, $P B = 6$ cm and $B C = 12$ cm. (a) Find $P Q$ . (b) $A Q = 3.5$ cm. Find $Q C$ .

Triangles $A P Q$ and $A B C$ are similar, because $P Q ∥ B C$ .
$A B = A P + P B = 3 + 6 = 9$ cm.
Scale factor from the small triangle to the large one: $k = \frac{A B}{A P} = \frac{9}{3} = 3$ .
(a) $P Q = B C \div 3 = 12 \div 3 = 4$ cm.
(b) $A C = 3.5 \times 3 = 10.5$ cm, so $Q C = A C - A Q = 10.5 - 3.5 = 7$ cm.

Answer: (a) 4 cm (b) 7 cm

Similar triangles across crossing lines

The other common picture is two straight lines crossing at a point $X$ , with the ends joined by two parallel lines, making an "hourglass" shape. The two triangles are similar:

the angles at $X$ are vertically opposite, so they are equal;
the parallel lines make alternate angles equal.

One triangle is upside down compared with the other, so match corresponding sides carefully: a side of one triangle corresponds to the side opposite the equal angle in the other, which here means the side on the same straight line through $X$ .

Worked example 3

In the diagram above, lines $L S$ and $M T$ cross at $X$ , and $L M$ is parallel to $T S$ . $L X = 5$ cm, $X S = 7.5$ cm and $L M = 7$ cm. Find $T S$ .

Triangles $L M X$ and $S T X$ are similar: angle $L X M =$ angle $S X T$ (vertically opposite), and the parallel lines give alternate angles equal.
$L X$ and $X S$ lie on the same straight line, so they correspond. $k = \frac{7.5}{5} = 1.5$ .
$L M$ corresponds to $S T$ : $T S = 7 \times 1.5 = 10.5$ cm.

Answer: 10.5 cm

Check yourself

Triangles $A B C$ and $P Q R$ are similar, with $A$ corresponding to $P$ , $B$ to $Q$ and $C$ to $R$ . $A B = 6$ cm, $B C = 9$ cm, $P Q = 10$ cm. Find $Q R$ .

Answer it in the portal, where your answer is checked as you type →

Showing that triangles are similar Extended

Extended candidates may be asked to show that two triangles are similar. For triangles, equal angles are enough: if two pairs of angles are equal, the third pair must be too, because the angles of each triangle add to $180^{°}$ . So you need to prove two pairs of equal angles, with a geometric reason for each.

Reasons you will use:

common angle (both triangles share the same angle);
vertically opposite angles are equal;
alternate angles are equal (parallel lines);
corresponding angles are equal (parallel lines);
a given equal angle.

Write it in a short list: "Angle $A$ is common. Angle $A B D =$ angle $A C B$ (given). So the third angles are equal and the triangles are similar." Name each angle with three letters and give each reason. A statement such as "they look the same shape" earns nothing.

Alternatively, all three pairs of corresponding sides in the same ratio also shows similarity.

Areas of similar shapes Extended

If lengths are multiplied by $k$ , areas are multiplied by $k^{2}$ . Area is made from two lengths multiplied together, and each of them is multiplied by $k$ : a 1 cm by 1 cm square becomes $k$ by $k$ , with area $k^{2}$ .

$\frac{area of B}{area of A} = (\frac{length of B}{length of A})^{2}$

For example, doubling every length of a shape multiplies its area by 4, not 2. If two similar triangles have bases 4 cm and 12 cm, then $k = 3$ and the area of the larger is $3^{2} = 9$ times the area of the smaller.

Volumes and surface areas of similar solids Extended

For similar solids with length scale factor $k$ :

every length is multiplied by $k$ ;
every area, including surface area, is multiplied by $k^{2}$ ;
the volume is multiplied by $k^{3}$ , because volume is three lengths multiplied together.

As the syllabus writes it:
$\frac{Volume of A}{Volume of B} = \frac{(Length of A)^{3}}{(Length of B)^{3}}$

Quantities that depend on volume scale the same way: the capacity of a container, and the mass of solids made of the same material. Doubling the height of a similar jug multiplies what it holds by 8.

Working back from areas and volumes Extended

Often you are given two areas or two volumes and need a length, or you are given areas and need a volume. Always go through the length scale factor:

from an area ratio: $k = \sqrt{area ratio}$ ;
from a volume (or mass, or capacity) ratio: $k = \sqrt[3]{volume ratio}$ ;
then use $k$ , $k^{2}$ or $k^{3}$ as the question needs.

For example, if two similar jugs hold 250 ml and 2 000 ml, the volume factor is 8, so $k = \sqrt[3]{8} = 2$ and the surface area factor is $2^{2} = 4$ . Going straight from one ratio to another without passing through $k$ is the commonest error in this topic.

Worked example 4 Extended

Two similar triangles have areas $20 {cm}^{2}$ and $45 {cm}^{2}$ . The smaller has a base of 6 cm. Find the base of the larger.

Area factor $= \frac{45}{20} = \frac{9}{4}$ .
Length factor $k = \sqrt{\frac{9}{4}} = \frac{3}{2}$ .
Base of the larger $= 6 \times \frac{3}{2} = 9$ cm.

Answer: 9 cm

Worked example 5 Extended

Two similar solid cones are made of the same metal. The smaller has height 8 cm and mass 640 g. The larger has height 12 cm. (a) Find the mass of the larger cone. (b) The larger cone has a curved surface area of $270 {cm}^{2}$ . Find the curved surface area of the smaller cone.

Length factor from small to large: $k = \frac{12}{8} = 1.5$ .
(a) Mass follows volume: $640 \times {1.5}^{3} = 640 \times 3.375 = 2 160$ g.
(b) Area factor $= {1.5}^{2} = 2.25$ . Going from large to small, divide: $270 \div 2.25 = 120 {cm}^{2}$ .

Answer: (a) 2 160 g (b) $120 {cm}^{2}$

Check yourself Extended

Two similar bottles have heights 10 cm and 15 cm. The smaller bottle holds 400 ml. How much does the larger bottle hold?

Answer it in the portal, where your answer is checked as you type →

Key points

Similar: equal angles, corresponding lengths in the same ratio.
Scale factor $k = \frac{new length}{original length}$ , from a pair of corresponding sides.
Multiply by $k$ to go to the new shape; divide to go back.
Inside triangles with a parallel side: use the whole side of the large triangle.
Hourglass triangles: vertically opposite and alternate angles; sides on the same line correspond.
Extended: areas (and surface areas) $\times k^{2}$ ; volumes, capacities and masses $\times k^{3}$ .
Extended: from areas $k = \sqrt{ratio}$ ; from volumes $k = \sqrt[3]{ratio}$ .
Extended: show similarity with two pairs of equal angles and a reason for each.

Where marks are lost

Adding the same amount to each side instead of multiplying by a scale factor.
Matching the wrong sides in turned or flipped triangles; use the equal angles.
Using the part $P B$ instead of the whole side $A B$ in nested triangles.
Dividing when you should multiply, giving a length on the bigger shape that is smaller.
Rounding the scale factor too early.
Extended: multiplying an area or volume by $k$ instead of $k^{2}$ or $k^{3}$ .
Extended: square-rooting a volume ratio instead of cube-rooting it.
Extended: "showing" similarity without a reason for each pair of equal angles.

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C4.5 · E4.5

All IGCSE Mathematics 0580 lessons →

An original lesson written to the Cambridge IGCSE Mathematics 0580 syllabus (2025 to 2027). The worked examples and questions are our own, not Cambridge questions. Back to every IGCSE Mathematics 0580 lesson.

Similarity

What similar means

Finding the scale factor and missing lengths

Worked example 1

Check yourself

Similar triangles inside each other

Worked example 2

Similar triangles across crossing lines

Worked example 3

Check yourself

Showing that triangles are similar Extended

Areas of similar shapes Extended

Volumes and surface areas of similar solids Extended

Working back from areas and volumes Extended

Worked example 4 Extended

Worked example 5 Extended

Check yourself Extended

Key points

Where marks are lost

Related IGCSE Mathematics 0580 lessons

Symmetry

Scale drawings

Angles

Geometrical constructions