Similar Figures
Similar figures have the same shape but not always the same size. Matching angles are equal and matching sides are in the same ratio, the scale factor. In NSW Year 10 Advanced maths the scale factor is used to find lengths, and its square and cube give the ratios of areas and volumes (NESA outcome MA5‑GEO‑C‑01).
You will find unknown sides in similar quadrilaterals, trapeziums and triangles, check whether two boxes are similar, and work with scale drawings in a brochure. You will use area and volume ratios for drink bottles, chocolate rabbits, a model shed and a scale-model yacht, and work back from a volume ratio to a length. Harder questions solve equations for unknown sides, mount a photo in a similar frame, and prove that one rectangle can never be similar to another. It is part of Year 10 Advanced Geometric Figures.
Every question with a fully worked solution.
- Similar Figures - Video - Finding Missing Length on Similar Figures using Scale Factor Watch
Theory
Similar figures have the same shape: matching angles are equal and every length is multiplied by the same scale factor \(k\). This Year 10 Advanced maths page covers finding sides, testing for similarity, similarity with algebra, and how areas scale by \(k^2\) and volumes by \(k^3\), the part that comes up again on the Extension path.
Two figures are similar if one is an enlargement or reduction of the other, possibly also moved, turned or flipped. Matching angles are equal and matching sides are in the same ratio. The symbol is \(\sim\), with vertices written in matching order.
The scale factor is \(k = \dfrac{\text{image length}}{\text{original length}}\). \(k \gt 1\) is an enlargement and \(0 \lt k \lt 1\) a reduction. Perimeters scale by \(k\) too.
To test polygons for similarity, check that all matching sides give the same ratio and all matching angles are equal. For rectangles, compare long side \(\div\) short side.
Areas are lengths times lengths, so they scale by \(k^2\). Volumes, capacities and masses of the same material scale by \(k^3\).
| Given ratio | Length ratio \(k\) | Area ratio | Volume ratio |
|---|---|---|---|
| lengths \(2 : 3\) | \(2 : 3\) | \(4 : 9\) | \(8 : 27\) |
| areas \(16 : 49\) | \(4 : 7\) | \(16 : 49\) | \(64 : 343\) |
| volumes \(1 : 125\) | \(1 : 5\) | \(1 : 25\) | \(1 : 125\) |
How to solve a similar-figures problem
- Match the sides using the vertex order or the shapes' positions. Use the same units for both figures.
- Find \(k\) from one pair of matching lengths whose values are both known.
- Multiply by \(k\) to go from original to image; divide by \(k\) to go back.
- Areas and volumes: use \(k^2\) or \(k^3\). Given an area or volume ratio, take the square or cube root to get \(k\) first.
Other question types
- Similar or not: work out every ratio of matching sides; one different ratio means not similar.
- Algebra: write ratio = ratio and cross-multiply. With \(x\) on both sides of a pair, the equation may be quadratic; reject negative lengths.
- Percentage changes: \(+5\%\) on lengths is \(k = 1.05\), so volume \(\times 1.05^3 \approx 1.158\), an increase of about \(15.8\%\).
- Show never similar: assume the figures are similar, solve, and show the result contradicts the given conditions.
- Unknown in a ratio of perimeters: perimeters are in the length ratio, not the area ratio.
(i) Find \(LM\) and \(CD\).
(ii) Are rectangles of \(6\) cm by \(15\) cm and \(10\) cm by \(24\) cm similar?
(i) \(KL\) matches \(AB\), so the scale factor from \(ABCD\) to \(KLMN\) is:
| \(k\) | \(=\) | \(\dfrac{KL}{AB} = \dfrac{8}{6} = \dfrac{4}{3}\) |
| \(LM\) | \(=\) | \(9 \times \dfrac{4}{3} = \mathbf{12}\text{ cm}\) |
| \(CD\) | \(=\) | \(10 \div \dfrac{4}{3} = \mathbf{7.5}\text{ cm}\) |
Multiply to go from the first figure to the second; divide to go back.
(ii) Compare long side \(\div\) short side.
| \(\dfrac{15}{6}\) | \(=\) | \(2.5\) |
| \(\dfrac{24}{10}\) | \(=\) | \(2.4\) |
The ratios differ, so the rectangles are not similar, even though all their angles are \(90^\circ\).
(i) A rectangle measures \((x + 3)\) cm by \((3x - 2)\) cm. It is similar to a \(2\) cm by \(5\) cm rectangle, with the short sides matching. Find \(x\).
(ii) A \(24\) cm by \(16\) cm print has a \(2\) cm border above and below it and an \(x\) cm border at each end. The outer edge is similar to the print. Find \(x\).
(i) Matching sides are in the same ratio.
| \(\dfrac{3x - 2}{x + 3}\) | \(=\) | \(\dfrac{5}{2}\) |
| \(2(3x - 2)\) | \(=\) | \(5(x + 3)\) |
| \(6x - 4\) | \(=\) | \(5x + 15\) |
| \(x\) | \(=\) | \(\mathbf{19}\) |
Check: \(22\) cm by \(55\) cm, and \(\dfrac{55}{22} = 2.5 = \dfrac{5}{2}\).
(ii) The outer height is \(16 + 2 + 2 = 20\) cm. Scale factor \(\dfrac{20}{16} = 1.25\), so the outer length is \(24 \times 1.25 = 30\) cm.
| \(24 + 2x\) | \(=\) | \(30\) |
| \(x\) | \(=\) | \(\mathbf{3}\) |
A border of equal width all round would not give a similar rectangle: adding the same amount to both sides changes their ratio.
(i) The smaller one has a surface area of \(200\) cm\(^2\). Find the surface area of the larger one.
(ii) The larger one holds \(2.7\) L. How much does the smaller one hold?
Find the length scale factor first, then square it for area or cube it for volume.
| \(k\) | \(=\) | \(\dfrac{18}{12} = 1.5\) |
| \(k^2\) | \(=\) | \(2.25\) |
| \(k^3\) | \(=\) | \(3.375\) |
| \(\text{(i) area}\) | \(=\) | \(200 \times 2.25 = \mathbf{450}\text{ cm}^2\) |
| \(\text{(ii) capacity}\) | \(=\) | \(2.7 \div 3.375 = \mathbf{0.8}\text{ L}\) |
Going from large to small, divide. Dividing by \(1.5\) instead of \(1.5^3\) is a common slip. Check: \(0.8 \times 3.375 = 2.7\).
(i) Every length of a photo is enlarged by \(20\%\). By what percentage does its area increase?
(ii) Two similar statues are made of the same metal and have masses of \(8\) kg and \(27\) kg. The smaller one is \(60\) cm tall. How tall is the larger one?
(i) A \(20\%\) increase in length is a scale factor of \(1.2\).
| \(k^2\) | \(=\) | \(1.2^2 = 1.44\) |
The area becomes \(144\%\) of the original: an increase of \(\mathbf{44\%}\), not \(40\%\).
(ii) Mass of the same material is proportional to volume, so the mass ratio is \(k^3\). Take the cube root to get \(k\).
| \(k^3\) | \(=\) | \(\dfrac{27}{8}\) |
| \(k\) | \(=\) | \(\sqrt[3]{\dfrac{27}{8}} = \dfrac{3}{2}\) |
| \(\text{height}\) | \(=\) | \(60 \times \dfrac{3}{2} = \mathbf{90}\text{ cm}\) |
From an area ratio, take the square root instead: areas in the ratio \(4 : 9\) mean lengths in the ratio \(2 : 3\).
Common pitfalls
Frequently asked questions
What makes two shapes similar?
They have the same shape: all matching angles are equal and all matching sides are in the same ratio, called the scale factor. One is an enlargement or reduction of the other.
How do you find the scale factor of similar figures?
Divide a length on the new figure by the matching length on the original. If a \(4\) cm side becomes \(6\) cm, the scale factor is \(1.5\).
If lengths double, what happens to area and volume?
Area is multiplied by \(2^2 = 4\) and volume by \(2^3 = 8\). In general area scales by \(k^2\) and volume by \(k^3\).
How do you find the length ratio from an area ratio?
Take the square root of each part. Areas in the ratio \(9 : 25\) give lengths in the ratio \(3 : 5\).
Are all squares similar?
Yes. Every square has four right angles and four equal sides, so any two squares have all matching sides in the same ratio. All circles are similar too.