Two shapes are similar if one is an enlargement of the other. Everyday language uses "similar" to mean "a bit alike"; in maths it has a precise meaning.
โข all pairs of corresponding angles are equal, and
โข all pairs of corresponding sides are in the same ratio.
That common ratio is the scale factor, written $k$.
Every side of B is $2.5$ times the matching side of A, and the angles are unchanged.
- Identify which sides correspond (redraw the shapes the same way up if it helps).
- Find the scale factor using a pair of sides you know both of.
- Multiply (to go bigger) or divide (to go smaller) to find the missing length.
- Sense check: the bigger shape must have the bigger sides.
Triangle $ABC$ has $AB = 6$ cm and $BC = 9$ cm. Triangle $PQR$ is similar, with $PQ = 15$ cm corresponding to $AB$. Find $QR$.
Two similar rectangles have widths $12$ cm and $8$ cm. The larger has length $21$ cm. Find the length of the smaller.
Sense check: $14 < 21$ โ
This is the single most important idea on this page, and the one most often got wrong.
| Length factor $k$ | Area factor $k^2$ | Volume factor $k^3$ |
|---|---|---|
| $2$ | $4$ | $8$ |
| $3$ | $9$ | $27$ |
| $5$ | $25$ | $125$ |
| $1.5$ | $2.25$ | $3.375$ |
| $0.5$ | $0.25$ | $0.125$ |
Two similar triangles have corresponding sides $4$ cm and $10$ cm. The smaller has area $18\text{ cm}^2$. Find the area of the larger.
Two similar jugs have heights $12$ cm and $18$ cm. The smaller holds $640$ ml. How much does the larger hold?
Two similar shapes have areas $45\text{ cm}^2$ and $180\text{ cm}^2$. A side of the smaller is $6$ cm. Find the corresponding side of the larger.
Two similar cones have volumes $54\text{ cm}^3$ and $128\text{ cm}^3$. The smaller has slant height $9$ cm. Find the slant height of the larger.
Exam questions often present the information as a ratio rather than a scale factor.
Two similar solids have lengths in the ratio $2 : 5$. Write the ratio of their surface areas and the ratio of their volumes.
Trigonometry exists because of similar triangles. In any right-angled triangle with a given angle $\theta$, the ratio of two named sides is always the same, no matter how big the triangle is.
In triangle $ABC$, the point $D$ lies on $AB$ and $E$ on $AC$, with $DE$ parallel to $BC$. $AD = 4$ cm, $DB = 6$ cm and $DE = 5$ cm. Find $BC$.
Similar
Equal angles, all sides in the same ratio $k$.
Scale factor
$k = \dfrac{\text{new length}}{\text{old length}}$, using corresponding sides.
The big three
Length $\times k$, area $\times k^2$, volume $\times k^3$.
Backwards
From area: $k = \sqrt{\text{area factor}}$. From volume: $k = \sqrt[3]{\text{volume factor}}$.
Ratios
Lengths $a : b$ โ areas $a^2 : b^2$ โ volumes $a^3 : b^3$.
Mass
For solids of the same material, mass scales like volume, i.e. $\times k^3$.
Parallel lines
A line parallel to one side creates a similar triangle.
Trig link
$\sin$, $\cos$, $\tan$ are side ratios fixed by the angle alone.
Two similar rectangles have widths $5$ cm and $12.5$ cm. The smaller has length $8$ cm. Find the length of the larger.
โถ Show solution
$k = \dfrac{12.5}{5} = 2.5$
Length $= 8 \times 2.5 = 20$ cm
Two similar shapes have lengths in the ratio $3 : 7$. Write the ratio of (a) their areas, (b) their volumes.
โถ Show solution
(a) $3^2 : 7^2 = 9 : 49$
(b) $3^3 : 7^3 = 27 : 343$
A photograph measuring $10$ cm by $15$ cm is enlarged so that the longer side becomes $24$ cm. Find (a) the new shorter side, (b) the area of the enlargement.
โถ Show solution
(a) $k = \dfrac{24}{15} = 1.6$; shorter side $= 10 \times 1.6 = 16$ cm.
(b) Area $= 16 \times 24 = 384\text{ cm}^2$.
Or via the area factor: original area $= 150\text{ cm}^2$; $k^2 = 2.56$; $150 \times 2.56 = 384\text{ cm}^2$ โ
Two similar triangles have corresponding sides $6$ cm and $9$ cm. The larger has area $81\text{ cm}^2$. Find the area of the smaller.
โถ Show solution
Going large โ small: $k = \dfrac{6}{9} = \dfrac{2}{3}$.
Area factor $= \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}$.
Area $= 81 \times \dfrac{4}{9} = 36\text{ cm}^2$.
Two similar bottles have heights $15$ cm and $25$ cm. The taller holds $1.25$ litres. Find the capacity of the shorter, in millilitres.
โถ Show solution
$k = \dfrac{15}{25} = 0.6$
Volume factor $= 0.6^3 = 0.216$
$1.25$ litres $= 1250$ ml
Capacity $= 1250 \times 0.216 = 270$ ml
Two similar shapes have areas $28\text{ cm}^2$ and $175\text{ cm}^2$. A length on the smaller is $4$ cm. Find the corresponding length on the larger.
โถ Show solution
Area factor $= \dfrac{175}{28} = 6.25$
$k = \sqrt{6.25} = 2.5$
Length $= 4 \times 2.5 = 10$ cm
Two similar solid statues are made of the same bronze. The smaller is $30$ cm tall and has mass $5.4$ kg. The larger is $50$ cm tall. Find its mass.
โถ Show solution
$k = \dfrac{50}{30} = \dfrac{5}{3}$
Mass scales like volume, so the factor is $\left(\dfrac{5}{3}\right)^3 = \dfrac{125}{27}$.
Mass $= 5.4 \times \dfrac{125}{27} = 25$ kg
In triangle $PQR$, $S$ lies on $PQ$ and $T$ on $PR$, with $ST$ parallel to $QR$. $PS = 6$ cm, $SQ = 9$ cm and $QR = 20$ cm. Find $ST$.
โถ Show solution
Triangles $PST$ and $PQR$ are similar.
$PQ = 6 + 9 = 15$ cm.
Scale factor from large to small $= \dfrac{6}{15} = 0.4$.
$ST = 20 \times 0.4 = 8$ cm.
Two similar cones have volumes $135\text{ cm}^3$ and $320\text{ cm}^3$. The surface area of the smaller is $63\text{ cm}^2$. Find the surface area of the larger.
โถ Show solution
Volume factor $= \dfrac{320}{135} = \dfrac{64}{27}$
$k = \sqrt[3]{\dfrac{64}{27}} = \dfrac{4}{3}$
Area factor $= k^2 = \dfrac{16}{9}$
Surface area $= 63 \times \dfrac{16}{9} = 112\text{ cm}^2$
A cone of height $18$ cm has a smaller cone cut from its top by a cut parallel to the base. The small cone has height $6$ cm. The volume of the whole cone is $972\text{ cm}^3$.
(a) Find the volume of the small cone. (b) Find the volume of the frustum (the piece left behind). (c) Write the ratio small cone : frustum in its simplest form.
โถ Show solution
(a) The small cone is similar to the whole cone with $k = \dfrac{6}{18} = \dfrac{1}{3}$.
Volume factor $= \left(\dfrac{1}{3}\right)^3 = \dfrac{1}{27}$.
Small cone $= 972 \times \dfrac{1}{27} = 36\text{ cm}^3$.
(b) Frustum $= 972 - 36 = 936\text{ cm}^3$.
(c) $36 : 936$. Divide both by $36$: $\;\mathbf{1 : 26}$.