R12, G19 - Similarity ratios for lengths, areas and volumes

Use corresponding lengths to scale areas and volumes of similar figures, then work backwards from area or volume information. Link these ratios to capacity, mass and coating, and explain why equal areas alone do not guarantee the same shape.

Length, area and volume scale factors (Higher)

If a model's lengths are all doubled, will its paint and its capacity both double? A rectangle and a cube show why different measurements change by different factors.

You already know

Similar figures have equal corresponding angles and one common positive factor for all corresponding lengths. Match corresponding sides and keep the direction of the ratio consistent. A rectangle's area is base times height; a cuboid's volume is length times width times height.

Doubling a rectangle's base and height multiplies its area by $2\times2=4$ . Doubling a cuboid's three dimensions multiplies its volume by $2\times2\times2=8$ .

[DIAGRAM: asset_name: Length, area and volume scaling; asset_slug: 1ma1_m44_scale_powers; caption: Doubling every length multiplies area by four and volume by eight.; description: Similar rectangles of dimensions 2 by 1 and 4 by 2 have areas 2 and 8 square units. Similar cubes with edges 1 and 2 have volumes 1 and 8 cubic units, illustrating factors 2, 4 and 8.]

The same reasoning works with a general factor $k$ . A rectangle's new area is $(kb)(kh)=k^2bh$ . A triangle's area gains the same factor because $\frac12(kb)(kh)=k^2(\frac12bh)$ . Uniform enlargement scales both dimensions of each small piece of a shape, so the square relationship extends to other similar plane shapes. Each face of a solid also scales in two dimensions.

For a cuboid, $(kl)(kw)(kh)=k^3lwh$ . A uniform enlargement scales three dimensions of every small volume element, so the cube relationship extends to other similar solids.

Scaling similar figures

\begin{aligned} \text{length factor}&=k\\ \text{area or surface-area factor}&=k^2\\ \text{volume factor}&=k^3 \end{aligned}

All corresponding lengths must have the same positive factor $k$ . A length ratio $a:b$ gives area ratio $a^2:b^2$ and volume ratio $a^3:b^3$ , in the same order.

Not on the formula sheet: learn it.

For example, length ratio A:B $=2:3$ gives area ratio $4:9$ and volume ratio $8:27$ . From A to B, use factors $3/2$ , $9/4$ and $27/8$ respectively.

Worked example 1: an enlargement

Question. A solid has surface area $32\,\mathrm{cm}^2$ and volume $64\,\mathrm{cm}^3$ . A similar solid has every length doubled. Find its surface area and volume.

Plan. Square the length factor for surface area and cube it for volume.

\begin{aligned} k&=2 &&\text{given length factor}\\ S_{\text{new}}&=32(2^2)=128 &&\text{two-dimensional surface}\\ V_{\text{new}}&=64(2^3)=512 &&\text{three-dimensional space} \end{aligned}

Why use different powers? Surface area measures two dimensions; volume measures three.

Answer. $128\,\mathrm{cm}^2$ and $512\,\mathrm{cm}^3$ .

Check. The surface area quadruples and the volume increases eightfold, matching the diagram's factors.

Change only the length factor from $2$ to $1/2$ ; keep the starting measurements.

Worked example 2: a reduction

Question. A similar solid has every length half that of the solid with surface area $32\,\mathrm{cm}^2$ and volume $64\,\mathrm{cm}^3$ . Find its surface area and volume.

Plan. Apply the same powers to the reducing factor.

\begin{aligned} k&=\frac12 &&\text{halve every length}\\ S_{\text{new}}&=32\left(\frac12\right)^2=8 &&\text{area factor }1/4\\ V_{\text{new}}&=64\left(\frac12\right)^3=8 &&\text{volume factor }1/8 \end{aligned}

Why can the answers have the same number? They measure different quantities in different units and came from different starting values.

Answer. $8\,\mathrm{cm}^2$ and $8\,\mathrm{cm}^3$ .

Check. Reversing the reduction doubles lengths: $8(4)=32$ and $8(8)=64$ .

Similarity is essential. Doubling only a cuboid's height doubles its volume, but leaves the other two length factors at $1$ . There is no common factor $2$ , so the cubed similarity rule cannot be used.

Watch out

Wrong: a length ratio $2:3$ means the area ratio is $2:3$ . That applies a one-dimensional ratio to a two-dimensional measurement. Correct: square both ratio terms to get $4:9$ , keeping their order.

Change only the length factor from Worked example 2.

Recovering length scales from area and volume (Higher)

Work backwards through the power: take a positive square root for a length factor from areas, or a positive cube root from volumes.

A length factor $3$ makes an area factor $9$ and a volume factor $27$ . Reversing those steps requires roots, not division by $2$ or $3$ . This root method is not on the formula sheet: derive it by reversing the similarity scale-factor rules.

Worked example 3: an area ratio gives a length

Question. Two similar triangles have areas $27\,\mathrm{cm}^2$ and $75\,\mathrm{cm}^2$ . A side of the smaller triangle is $12\,\mathrm{cm}$ . Find the corresponding larger side.

Plan. Find the area factor, take its square root, then scale the side.

\begin{aligned} k^2&=\frac{75}{27}=\frac{25}{9} &&\text{larger divided by smaller}\\ k&=\sqrt{\frac{25}{9}}=\frac53 &&\text{positive length factor}\\ \text{larger side}&=12\left(\frac53\right)=20 &&\text{scale a length} \end{aligned}

Why a square root? Area gained two copies of the length factor; the root recovers one copy.

Answer. $20\,\mathrm{cm}$ .

Check. $(20/12)^2=25/9=75/27$ .

Change only the larger area from $75$ to $108\,\mathrm{cm}^2$ .

Worked example 4: a different area factor

Question. Similar triangles have areas $27\,\mathrm{cm}^2$ and $108\,\mathrm{cm}^2$ . The corresponding smaller side is $12\,\mathrm{cm}$ . Find the larger side.

Plan. Recover the length factor before applying it to the side.

\begin{aligned} k^2&=108/27=4 &&\text{area factor}\\ k&=\sqrt4=2 &&\text{length factor}\\ \text{larger side}&=12(2)=24 &&\text{scale the given side} \end{aligned}

Why not multiply the side by $4$ ? That is the area factor, rather than the length factor.

Answer. $24\,\mathrm{cm}$ .

Check. Doubling lengths quadruples area: $27(4)=108$ .

Not every root is a whole number. Areas $20\,\mathrm{cm}^2$ and $50\,\mathrm{cm}^2$ give length factor $\sqrt{50/20}=\sqrt{2.5}$ . Keep the root exact until a final approximation is requested.

Worked example 5: a volume ratio gives a surface-area ratio

Question. Similar solids A and B have volumes $64\,\mathrm{cm}^3$ and $216\,\mathrm{cm}^3$ . Find their length ratio and surface-area ratio A:B.

Plan. Take cube roots to recover lengths, then square the length-ratio terms.

\begin{aligned} V_A:V_B&=64:216=8:27 &&\text{simplify the given ratio}\\ L_A:L_B&=\sqrt[3]8:\sqrt[3]{27}=2:3 &&\text{recover lengths}\\ S_A:S_B&=2^2:3^2=4:9 &&\text{scale surfaces} \end{aligned}

Why pass through lengths? Volume and area use different powers of the common length factor.

Answer. Length ratio $2:3$ ; surface-area ratio $4:9$ .

Check. Cubing $2:3$ returns $8:27$ , the simplified volume ratio.

Change only B's volume from $216$ to $512\,\mathrm{cm}^3$ .

Worked example 6: a different volume ratio

Question. Similar solids A and B have volumes $64\,\mathrm{cm}^3$ and $512\,\mathrm{cm}^3$ . Find their length ratio and surface-area ratio A:B.

Plan. Simplify the volume ratio, recover lengths by cube roots, then square.

\begin{aligned} V_A:V_B&=64:512=1:8 &&\text{divide both terms by }64\\ L_A:L_B&=1:2 &&\text{take cube roots}\\ S_A:S_B&=1:4 &&\text{square the length terms} \end{aligned}

Why are the ratios different? One compares three-dimensional volumes and the other compares two-dimensional surfaces.

Answer. Length ratio $1:2$ ; surface-area ratio $1:4$ .

Check. A factor $2$ in length gives factors $4$ in surface area and $8$ in volume.

Root-taking uses known similarity; it cannot establish similarity from areas or volumes alone.

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R12, G19 - Similarity ratios for lengths, areas and volumes

Length, area and volume scale factors (Higher)

Recovering length scales from area and volume (Higher)

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