Year 9 Maths Advanced Congruence and Similar Figures

Areas of Similar Figures

Q: How do you find the area of a similar figure?

Find the scale factor k from a pair of matching lengths. Multiply the known area by k squared.

Q: If the sides are doubled, what happens to the area?

The area is multiplied by 2 squared, which is 4. The volume of a solid would be multiplied by 2 cubed, which is 8.

Q: What is the ratio of volumes of similar solids?

If matching lengths are in the ratio a : b, volumes are in the ratio a cubed : b cubed.

20 practice questions 1 video lesson Theory + worked examples

If matching lengths of similar figures are in the ratio \(a : b\), then the ratio of their areas is \(a^2 : b^2\) and the ratio of their volumes is \(a^3 : b^3\). This NSW Year 9 Advanced maths subtopic uses these ratios (NESA outcome MA5‑GEO‑P‑01).

You will find area ratios from side ratios and side ratios from area ratios, then find unknown areas and lengths of triangles, circles, squares and rectangles. You will cost carpet for a similar room, find a paddock's real area from a map, price a larger pizza, cut a triangle into two equal areas, and find volumes and surface areas of similar solids. It is part of Year 9 Advanced Congruence and Similar Figures.

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Practice questions

Every question with a fully worked solution.

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Areas of Similar Figures - Video - Similar Shapes: Areas Watch

Theory

If two figures are similar with lengths in the ratio \(a : b\), their areas are in the ratio \(a^2 : b^2\) and the volumes of similar solids are in the ratio \(a^3 : b^3\). This Year 9 Advanced maths page shows how to move between lengths, areas and volumes. It comes up if you are heading for Extension maths.

When a figure is enlarged by a scale factor \(k\), every length is multiplied by \(k\): sides, heights, radii, diagonals and perimeters.

Area is length \(\times\) length, so every area is multiplied by \(k \times k = k^2\). For a triangle, \(\dfrac{1}{2}(kb)(kh) = k^2 \times \dfrac{1}{2}bh\). Surface areas of similar solids follow the same rule.

Volume is length \(\times\) length \(\times\) length, so every volume (and capacity) of a similar solid is multiplied by \(k^3\).

All circles, all squares and all equilateral triangles are similar, so the rules always apply to them. Anything that depends on area, such as the cost of carpet or paint, also scales by \(k^2\).

Scale factor \(k = 2\): lengths double, area is multiplied by \(2^2 = 4\).

Scale factor \(k = 2\): volume is multiplied by \(2^3 = 8\).

For similar figures with matching lengths in the ratio \(a : b\) (scale factor \(k = \dfrac{b}{a}\)):

\[\text{lengths } a : b \qquad \text{areas } a^2 : b^2 \qquad \text{volumes } a^3 : b^3\]

a^{2} : b^{2}

\[A_{\text{new}} = k^2 A \qquad V_{\text{new}} = k^3 V\]

A_{new} = k^{2} A

Going backwards:

\[k = \sqrt{\dfrac{A_{\text{new}}}{A}} \qquad k = \sqrt[3]{\dfrac{V_{\text{new}}}{V}}\]

k = \sqrt{\frac{A_{new}}{A}}

Percentages. A \(10\%\) increase in every length is \(k = 1.1\), so area is multiplied by \(1.21\): a \(21\%\) increase, not \(10\%\).

Maps. On a \(1 : n\) map, \(1\) cm\(^2\) stands for \(n^2\) cm\(^2\). Change to m\(^2\) (\(\div 10\,000\)) and hectares (\(\div 10\,000\) again).

How to solve an area or volume problem

Get the length ratio: from matching sides, perimeters, radii or heights. Simplify it.
Square for area, cube for volume.
Multiply the known area or volume by the ratio, the right way up: bigger figure, bigger answer.

Going the other way

Areas given: simplify the area ratio, then take square roots to get the length ratio.
Volumes given: take cube roots. Capacities (L, mL) count as volumes.
Area to volume: square root to get lengths, then cube.
Parts of a triangle: if \(DE \parallel BC\), compare \(\triangle ADE\) with the whole \(\triangle ABC\). The trapezium is the difference.
Exact answers: if the square root is not whole, simplify the surd, e.g. \(\sqrt{72} = 6\sqrt{2}\).

Example 1 — From sides to areas

Two similar pentagons have matching sides of \(6\) cm and \(15\) cm. The smaller pentagon has an area of \(36\) cm\(^2\). Find the area of the larger pentagon.

Solution

Simplify the ratio of sides, then square it.

\(\text{sides}\)	\(=\)	\(6 : 15 = 2 : 5\)
\(\text{areas}\)	\(=\)	\(2^2 : 5^2 = 4 : 25\)
\(A\)	\(=\)	\(36 \times \dfrac{25}{4} = \mathbf{225}\text{ cm}^2\)

The larger figure has the larger area, so multiply by \(\dfrac{25}{4}\), not \(\dfrac{4}{25}\).

A = 36 \times \frac{25}{4} = 225

Example 2 — From areas back to sides

Two similar triangles have areas of \(32\) cm\(^2\) and \(50\) cm\(^2\). A side of the larger triangle is \(12\) cm. Find the matching side of the smaller triangle.

Solution

Simplify the ratio of areas, then take square roots.

\(\text{areas}\)	\(=\)	\(32 : 50 = 16 : 25\)
\(\text{sides}\)	\(=\)	\(\sqrt{16} : \sqrt{25} = 4 : 5\)
\(\text{side}\)	\(=\)	\(12 \times \dfrac{4}{5} = \mathbf{9.6}\text{ cm}\)

Check: \(\left(\dfrac{9.6}{12}\right)^2 = 0.64 = \dfrac{32}{50}\) \(\checkmark\)

\sqrt{\frac{32}{50}} = \frac{4}{5}

Example 3 — Volumes and surface areas

Two similar jars are \(8\) cm and \(12\) cm tall. The larger jar holds \(810\) mL.
(i) How much does the smaller jar hold?
(ii) The label on the smaller jar has an area of \(96\) cm\(^2\). Find the area of the matching label on the larger jar.

Solution

Lengths: \(8 : 12 = 2 : 3\). Volumes use the cube, areas use the square.

\(\text{(i) volume}\)	\(=\)	\(810 \times \left(\dfrac{2}{3}\right)^3 = 810 \times \dfrac{8}{27}\)
\(\)	\(=\)	\(\mathbf{240}\text{ mL}\)
\(\text{(ii) area}\)	\(=\)	\(96 \times \left(\dfrac{3}{2}\right)^2 = 96 \times \dfrac{9}{4}\)
\(\)	\(=\)	\(\mathbf{216}\text{ cm}^2\)

810 \times {(\frac{2}{3})}^{3} = 240

Example 4 — Areas of the parts

In \(\triangle ABC\), \(DE \parallel BC\). The area of \(\triangle ADE\) and the area of the trapezium \(DBCE\) are in the ratio \(4 : 5\). \(AB = 15\) cm. Find \(AD\).

Solution

\(\triangle ADE \sim \triangle ABC\) (common angle, corresponding angles). Compare the small triangle with the whole triangle, not with the trapezium.

\(\triangle ADE : \triangle ABC\)	\(=\)	\(4 : (4 + 5) = 4 : 9\)
\(AD : AB\)	\(=\)	\(\sqrt{4} : \sqrt{9} = 2 : 3\)
\(AD\)	\(=\)	\(15 \times \dfrac{2}{3} = \mathbf{10}\text{ cm}\)

\frac{A D}{A B} = \sqrt{\frac{4}{9}} = \frac{2}{3}

Common pitfalls

Areas do not scale by \(k\). Doubling the sides of a shape multiplies its area by \(4\), not \(2\).

Square root before using a side. If areas are \(9 : 16\), the sides are \(3 : 4\). Using \(\dfrac{16}{9}\) on a length is a common slip.

Compare with the whole triangle. With a line parallel to the base, the trapezium is not similar to anything. Use the small triangle and the whole triangle.

Cube for volume. Capacity, mass of solid objects made of the same material, and volume all scale by \(k^3\).

Percent increase in area. A \(30\%\) increase in lengths gives \(1.3^2 = 1.69\), a \(69\%\) increase in area.

Frequently asked questions

How do you find the area of a similar figure?

Find the scale factor \(k\) from a pair of matching lengths. Multiply the known area by \(k^2\).

If the sides are doubled, what happens to the area?

The area is multiplied by \(2^2 = 4\). The volume of a solid would be multiplied by \(2^3 = 8\).

How do you find the ratio of sides from the ratio of areas?

Simplify the ratio of areas, then take the square root of each part. For example, areas \(25 : 49\) give sides \(5 : 7\).

What is the ratio of volumes of similar solids?

If matching lengths are in the ratio \(a : b\), volumes are in the ratio \(a^3 : b^3\).

Are all circles similar?

Yes. Every circle is an enlargement of every other circle, so areas of circles are in the ratio of the squares of their radii.

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Tests for Similar Triangles

Areas of Similar Figures

📖 Theory

How to solve an area or volume problem

Going the other way

Common pitfalls

Frequently asked questions

Theory