Similar Shapes: Length, Area and Volume GCSE Maths

Similar shapes have the same shape but may be different sizes. In similar shapes: length, area and volume GCSE Maths questions, the essential rule is that a length scale factor of $k$ produces an area scale factor of $k^2$ and a volume scale factor of $k^3$.

This Higher-tier revision guide explains how to identify the correct scale factor, move between length, area and volume ratios, handle reverse problems and solve exam-style questions. You can also use MathsGenie's GCSE Maths topics by exam board and tier to place this skill within your wider revision.

What are similar shapes?

Two shapes are mathematically similar when:

• corresponding angles are equal;
• corresponding lengths are in the same ratio.

One shape is therefore an enlargement of the other. Its orientation may be different, so the shapes do not have to be facing the same way.

For example, suppose one triangle has side lengths $3$ cm, $5$ cm and $7$ cm, while another has side lengths $6$ cm, $10$ cm and $14$ cm. Every length has been multiplied by $2$, so the triangles are similar with a length scale factor of $2$ from the smaller triangle to the larger triangle.

Congruent shapes are a special case of similar shapes. They have length scale factor $1$, so they have the same shape and the same size.

The length, area and volume scale-factor rules

If the length scale factor from shape A to shape B is $k$, the scale factors are:

This happens because an area has two dimensions. If both dimensions are multiplied by $k$, the area is multiplied by $k\times k=k^2$.

A volume has three dimensions. Multiplying each dimension by $k$ multiplies the volume by

$$k\times k\times k=k^3.$$

These rules apply to any pair of mathematically similar figures or solids. You do not need to know the specific area or volume formula if the relevant figures are similar and enough corresponding information is given.

Ratios of corresponding measurements

If corresponding lengths are in the ratio

$$a:b,$$

then corresponding areas are in the ratio

$$a^2:b^2,$$

and corresponding volumes are in the ratio

$$a^3:b^3.$$

For example, a length ratio of $2:5$ gives an area ratio of $4:25$ and a volume ratio of $8:125$.

Direction matters. If the scale factor from the smaller solid to the larger solid is $3$, then the scale factor from the larger solid to the smaller solid is $\frac{1}{3}$.

How to solve similar-shapes questions

Use the following method rather than deciding immediately whether to square or cube a number.

1. Identify the two similar figures and the direction of scaling.
2. Match corresponding lengths, areas or volumes.
3. Find the length scale factor $k$ if possible.
4. Choose $k$, $k^2$ or $k^3$ according to the measurement being scaled.
5. Multiply in the direction of the scale factor, or divide when moving in the opposite direction.
6. Check the size and units of the answer.

If the target figure is larger, its corresponding positive measurements should be larger. This simple check catches many reversed scale factors.

Similar shapes worked examples

Worked example 1: finding a corresponding length

Two similar quadrilaterals have corresponding sides of $8$ cm and $14$ cm. Another side of the smaller quadrilateral is $12$ cm. Find the corresponding side of the larger quadrilateral.

The scale factor from smaller to larger is

$$k=\frac{14}{8}=\frac{7}{4}.$$

Lengths scale by $k$, so the required side is

$$12\times \frac{7}{4}=21.$$

The corresponding side is $21$ cm.

Only the length scale factor is needed. Squaring the scale factor would be incorrect because the question asks for a length, not an area.

Worked example 2: finding an area

Two mathematically similar shapes have a length scale factor of $3$ from the smaller shape to the larger shape. The smaller area is $18$ cm${}^2$. Find the larger area.

The area scale factor is

$$3^2=9.$$

Therefore,

$$18\times 9=162.$$

The larger area is $162$ cm${}^2$.

Worked example 3: finding a volume

Two similar containers have corresponding heights of $15$ cm and $24$ cm. The smaller container has a volume of $500$ cm${}^3$. Find the volume of the larger container.

The length scale factor is

$$k=\frac{24}{15}=\frac{8}{5}.$$

The volume scale factor is

$$k^3={\left(\frac{8}{5}\right)}^3=\frac{512}{125}.$$

The larger volume is

$$500\times \frac{512}{125}=2048.$$

The volume of the larger container is $2048$ cm${}^3$.

Keeping the scale factor as a fraction gives an exact calculation and often makes cancellation easier.

Worked example 4: finding a length from two areas

Two similar shapes have areas of $80$ cm${}^2$ and $245$ cm${}^2$. A corresponding length on the smaller shape is $12$ cm. Find the matching length on the larger shape.

First find the area scale factor:

$$\frac{245}{80}=\frac{49}{16}.$$

The length scale factor is the positive square root:

$$k=\sqrt{\frac{49}{16}}=\frac{7}{4}.$$

The corresponding length is

$$12\times \frac{7}{4}=21.$$

The larger length is $21$ cm.

The square root is required because an area scale factor is the square of the length scale factor.

Worked example 5: finding a length from two volumes

Two similar solids have volumes of $250$ cm${}^3$ and $686$ cm${}^3$. A corresponding length on the smaller solid is $20$ cm. Find the corresponding larger length.

The volume scale factor is

$$\frac{686}{250}=\frac{343}{125}.$$

Take the positive cube root to obtain the length scale factor:

$$k=\sqrt[3]{\frac{343}{125}}=\frac{7}{5}.$$

Scale the length:

$$20\times \frac{7}{5}=28.$$

The larger length is $28$ cm.

Moving backwards from area or volume

Higher-tier questions often provide an area or volume relationship and ask for a length. This reverses the usual process.

For ratios, apply the root to both parts. If the area ratio is $25:81$, the length ratio is $5:9$. If the volume ratio is $64:343$, the length ratio is $4:7$.

Use the positive root because the lengths of geometric figures are positive.

Surface area and perimeter

Perimeter is made by adding lengths, so it follows the length scale factor $k$. It does not follow the area scale factor.

Surface area is an area measurement, even though it belongs to a three-dimensional solid. It therefore follows the area scale factor $k^2$, not the volume scale factor $k^3$.

For two similar solids with length scale factor $4$:

• the perimeter of any corresponding face scales by $4$;
• the surface area scales by $4^2=16$;
• the volume scales by $4^3=64$.

Common mistakes in similar-shapes exam questions

Using the length scale factor for every measurement

A scale factor of $5$ does not multiply an area by $5$ or a volume by $5$. It multiplies area by $25$ and volume by $125$.

Write $k$, $k^2$ and $k^3$ before substituting values. This makes the required relationship explicit.

Reversing the scale factor

If you are moving from a length of $6$ cm to a corresponding length of $15$ cm, the scale factor is

$$\frac{15}{6}=\frac{5}{2},$$

not $\frac{2}{5}$. Label the direction, such as “small to large”, beside the fraction.

Squaring or cubing the measurements themselves

You square or cube the length scale factor, not the stated area or volume. If $k=3$ and an area is $20$ cm${}^2$, calculate $20\times 3^2$, not $20^2$.

Forgetting to take a root

When areas are given and a length is required, take a square root. When volumes are given and a length is required, take a cube root.

Matching non-corresponding sides

Sides in the same visual position may not correspond if a shape has been rotated or reflected. Use equal angles, relative positions and the overall arrangement of sides to identify genuine corresponding lengths.

Using incorrect units

Length is measured in units such as cm, area in cm${}^2$, and volume in cm${}^3$. The power on the unit helps identify which scale-factor rule is required.

Rounding too early

Keep fractions, surds or full calculator values during the calculation. Round only the final answer and follow any accuracy instruction in the question.

Embedded exam-style question

Two solid ornaments are mathematically similar. The smaller ornament has a surface area of $150$ cm${}^2$ and a volume of $375$ cm${}^3$. The larger ornament has a surface area of $294$ cm${}^2$.

Calculate the volume of the larger ornament.

First find the area scale factor:

$$\frac{294}{150}=\frac{49}{25}.$$

The length scale factor is

$$k=\sqrt{\frac{49}{25}}=\frac{7}{5}.$$

The volume scale factor is therefore

$$k^3={\left(\frac{7}{5}\right)}^3=\frac{343}{125}.$$

The larger volume is

$$375\times \frac{343}{125}=1029.$$

The volume of the larger ornament is $1029$ cm${}^3$.

Try 4 more similar-shapes practice questions.

More similar shapes exam questions

Question 1

Two similar pentagons have corresponding sides of $9$ cm and $15$ cm. The area of the smaller pentagon is $54$ cm${}^2$. Find the area of the larger pentagon.

Question 2

Two mathematically similar solids have volumes in the ratio $27:125$. A corresponding length on the smaller solid is $7.2$ cm. Find the corresponding length on the larger solid.

Question 3

The surface areas of two similar cylinders are $144\pi$ cm${}^2$ and $225\pi$ cm${}^2$. The smaller cylinder has a volume of $320\pi$ cm${}^3$. Find the volume of the larger cylinder.

Question 4

Two similar shapes have areas of $98$ cm${}^2$ and $200$ cm${}^2$. A side on the larger shape is $35$ cm. Find the corresponding side on the smaller shape.

Answers

1. The length scale factor is $\frac{15}{9}=\frac{5}{3}$, so the area scale factor is $\frac{25}{9}$. The area is $54\times \frac{25}{9}=150$ cm${}^2$.
2. The length ratio is $\sqrt[3]{27}:\sqrt[3]{125}=3:5$. The larger length is $7.2\times \frac{5}{3}=12$ cm.
3. The area ratio is $144:225=16:25$, so the length ratio is $4:5$. The volume scale factor is ${\left(\frac{5}{4}\right)}^3=\frac{125}{64}$. The volume is $320\pi \times \frac{125}{64}=625\pi$ cm${}^3$.
4. The area ratio from smaller to larger is $98:200=49:100$, so the length ratio is $7:10$. The smaller side is $35\times \frac{7}{10}=24.5$ cm.

Exam technique for Higher-tier questions

Show the scale-factor relationship clearly, even when using a calculator. A strong solution normally identifies the length scale factor and then shows why it is squared or cubed.

For a problem that moves from surface area to volume, write a chain such as

$$area\; scale\; factor\to length\; scale\; factor\to volume\; scale\; factor.$$

Do not assume that shapes are similar only because they look similar in a diagram. In an exam, similarity will normally be stated or established using sufficient geometrical information. Diagrams may also not be drawn accurately, so calculate from the given values rather than measuring the page.

Conclusion

For similar figures, a length scale factor of $k$ gives an area scale factor of $k^2$ and a volume scale factor of $k^3$. Perimeter follows $k$, while surface area follows $k^2$.

When working backwards, take a square root of an area scale factor or a cube root of a volume scale factor. Label the direction, match corresponding measurements and retain exact values until the final step. MathsGenie practice questions and mark schemes can then help you apply these rules across full Higher-tier exam problems.

Sources and referenced URLs

• https://www.mathsgenie.co.uk/blog/gcse-maths-topics-by-exam-board-and-tier
• https://www.mathsgenie.co.uk/resources/5-similar-shapes.pdf