Similarity Calculations

Congruent and similar figures

Two congruent figures have the same shape and the same size. One may have been translated, rotated or reflected, but every corresponding length and angle remains equal. Their length scale factor is $1$.

Two similar figures have the same shape but need not have the same size. Their corresponding angles are equal, while every corresponding length is multiplied by the same number. This common multiplier is the length scale factor, usually written $k$. Congruent figures can therefore be regarded as the special case of similar figures for which $k=1$.

For example, rectangles measuring $3\text{ cm}$ by $5\text{ cm}$ and $6\text{ cm}$ by $10\text{ cm}$ are similar: both dimensions have been multiplied by $2$. A $3\text{ cm}$ by $5\text{ cm}$ rectangle and a $6\text{ cm}$ by $9\text{ cm}$ rectangle are not similar because the two multipliers would be different.

Matching corresponding features

A calculation begins by matching features that occupy the same position in the two figures. Shapes may be turned over or drawn in different orientations, so sides that merely look horizontal or longest do not necessarily correspond. Equal angles and the order of labelled vertices can help identify the correct pairs.

Once one pair of corresponding lengths is known, the scale factor from an old figure to a new figure is

$k=\frac{\text{new corresponding length}}{\text{old corresponding length}}.$

The direction matters. If the scale factor from A to B is $2.5$, the factor from B back to A is $\frac{1}{2.5}=0.4$.

Suppose similar triangles have corresponding sides of $6\text{ cm}$ and $15\text{ cm}$. From the first triangle to the second,

$k=\frac{15}{6}=2.5.$

A different side of length $9\text{ cm}$ on the first triangle therefore corresponds to a side of length

$9\times2.5=22.5\text{ cm}.$

Why area and volume use powers of the scale factor

A length changes in one direction, so it is multiplied by $k$. Area measures a two-dimensional region. If both relevant dimensions are multiplied by $k$, the area is multiplied by

$k\times k=k^2.$

Volume measures a three-dimensional region. When all three linear dimensions are multiplied by $k$, the volume is multiplied by

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

This gives the three central relationships:

$\text{length factor}=k,$

$\text{area factor}=k^2,$

$\text{volume factor}=k^3.$

The area rule applies to all corresponding areas, including the surface areas of similar solids. The volume rule also applies to capacity because capacity measures an amount of three-dimensional space.

If every length doubles, $k=2$: corresponding areas become $2^2=4$ times as large and corresponding volumes become $2^3=8$ times as large. If every length halves, $k=0.5$: areas are multiplied by $0.25$ and volumes by $0.125$. The rules therefore work for enlargements and reductions.

Finding an area or volume

First find the length scale factor from a pair of corresponding lengths. Then square or cube it according to the quantity being scaled.

For example, two similar triangles have corresponding sides of $4\text{ cm}$ and $10\text{ cm}$. The smaller triangle has area $18\text{ cm}^2$. The length factor from smaller to larger is

$k=\frac{10}{4}=2.5,$

so the area factor is

$k^2=2.5^2=6.25.$

The larger area is therefore

$18\times6.25=112.5\text{ cm}^2.$

For similar solids with heights $12\text{ cm}$ and $18\text{ cm}$, the length factor from smaller to larger is $1.5$. If the smaller volume is $640\text{ cm}^3$, the larger volume is

$640\times1.5^3=2160\text{ cm}^3.$

Working backwards from area or volume

Sometimes the area factor or volume factor is given instead of the length factor. Reverse the power operation:

$k=\sqrt{\text{area factor}}$

and

$k=\sqrt[3]{\text{volume factor}}.$

If two similar figures have areas $45\text{ cm}^2$ and $180\text{ cm}^2$, the area factor from smaller to larger is

$\frac{180}{45}=4.$

The length factor is therefore $\sqrt4=2$. A $6\text{ cm}$ side on the smaller figure corresponds to a $12\text{ cm}$ side on the larger one.

The same approach can connect volume and surface area. Suppose the volumes of two similar cones are in the ratio $27:64$. Their length ratio is

$\sqrt[3]{27}:\sqrt[3]{64}=3:4,$

so their surface-area ratio is

$3^2:4^2=9:16.$

Using ratio notation

If corresponding lengths are in the ratio $a:b$, then

$\text{length ratio}=a:b,$

$\text{area ratio}=a^2:b^2,$

$\text{volume ratio}=a^3:b^3.$

Thus a length ratio of $2:5$ gives an area ratio of $4:25$ and a volume ratio of $8:125$. The order must stay consistent: if the length ratio is written as first figure to second figure, the area and volume ratios must follow that same order.

Similar triangles within larger diagrams

Similarity may need to be established before any calculation is made. For example, a line parallel to one side of a triangle creates a smaller triangle with matching angles, so the smaller and larger triangles are similar.

If a smaller triangle has a side of $4\text{ cm}$ and the corresponding whole side of the larger triangle is $4+6=10\text{ cm}$, the scale factor from small to large is $\frac{10}{4}=2.5$. It would be incorrect to use $\frac{6}{4}$ because the $6\text{ cm}$ section is not the whole corresponding side of the larger triangle.

At Foundation tier, the specified similarity content includes relationships between lengths. Relationships involving areas and volumes of similar figures are Higher-tier content.

Core facts

For similar figures, corresponding angles are equal and corresponding lengths have one constant ratio. Congruent figures have the special scale factor $k=1$.

From one figure to another:

$k=\frac{\text{new corresponding length}}{\text{old corresponding length}}$

$\text{lengths}\times k,\qquad \text{areas}\times k^2,\qquad \text{volumes}\times k^3.$

Surface area follows the area rule; capacity follows the volume rule.

Ratios

If the length ratio is $a:b$:

$\text{area ratio}=a^2:b^2,$

$\text{volume ratio}=a^3:b^3.$

Keep the figures in the same order in every ratio.

Reverse scaling

$k=\sqrt{\frac{\text{new area}}{\text{old area}}}$

$k=\sqrt[3]{\frac{\text{new volume}}{\text{old volume}}}.$

Then use $k$, $k^2$ or $k^3$ for the required quantity.

Method

1. Match corresponding features.
2. State the direction of scaling.
3. Find the appropriate factor.
4. Apply the correct power of $k$.
5. Check size, ratio order and units.

Length relationships are Additional Foundation content; area and volume relationships in similar figures are Higher-tier content.