Properties of Similar Figures
In similar figures every length is multiplied by the same scale factor, so one pair of matching sides is enough to find the rest. This NSW Year 9 Advanced maths subtopic finds unknown sides in similar figures (NESA outcome MA5‑GEO‑C‑01).
You will find unknown sides in similar triangles, parallelograms, trapeziums and pentagons, including a triangle cut off by a parallel line. You will enlarge and reduce photos and rectangles, find a flagpole's height from its shadow and the length of a roof beam, find unknowns from algebraic sides, and use a street lamp to find the length of a shadow. It is part of Year 9 Advanced Congruence and Similar Figures.
Every question with a fully worked solution.
- Properties of Similar Figures - Video - Similar Polygons Watch
Theory
In similar figures every length is multiplied by the same scale factor. Find it from one matching pair of sides, then multiply (or divide) to get any other length, including perimeters. This Year 9 Advanced maths page also covers triangles inside triangles, hourglass shapes and real-life problems.
If two figures are similar, the ratio of any pair of matching sides is the same. That ratio is the scale factor \(k\).
Because every length is multiplied by \(k\), perimeters are multiplied by \(k\) too. Heights, diagonals and any other matching lengths follow the same rule.
Two common similar-triangle shapes come from parallel lines. A line parallel to one side of a triangle cuts off a small triangle similar to the whole one. Two segments crossing between parallel lines make an hourglass of two similar triangles.
Real objects make similar triangles too: shadows cast at the same moment, ladders, ramps and roof trusses. Look for two triangles with two pairs of equal angles.
Scale factor and unknown lengths:
As a ratio equation (unknown on top):
How to find an unknown length in similar figures
- Match the figures: use the similarity statement, equal angles, or sizes (shortest with shortest). Redraw nested or hourglass triangles separately if it helps.
- Find a matching pair with both lengths known and work out \(k\). Keep it as a fraction if the decimal does not end.
- Multiply or divide: multiply by \(k\) to go from the original to the image; divide by \(k\) to go back.
- Check that the answer is sensible: a bigger figure has longer sides.
Harder set-ups
- Pieces of a side: if \(AB = AD + DB\), write the whole side before forming the ratio, e.g. \(x + 4\).
- Unknown in two places: write a ratio equation, cross-multiply and solve.
- Perimeter given: find the other figure's perimeter, divide to get \(k\), then scale each side.
- Shadows and lamps: the object and its shadow, and the larger object and its shadow (or the lamp post and the whole shadow), make similar right-angled triangles.
\(PQ\) and \(AB\) are a matching pair with both lengths known.
| \(k\) | \(=\) | \(\dfrac{PQ}{AB} = \dfrac{4}{10} = 0.4\) |
\(QR\) matches \(BC\): multiply by \(k\). \(PR\) matches \(AC\): go back by dividing by \(k\).
| \(y\) | \(=\) | \(14 \times 0.4 = \mathbf{5.6}\) |
| \(x\) | \(=\) | \(7.2 \div 0.4 = \mathbf{18}\) |
The matching side for \(AD\) is the whole of \(AB\), not \(DB\): \(AB = 5 + 3 = 8\).
| \(k\) | \(=\) | \(\dfrac{AD}{AB} = \dfrac{5}{8}\) |
| \(x\) | \(=\) | \(12 \times \dfrac{5}{8} = \mathbf{7.5}\) |
\(AE\) matches \(AC = y + 4.2\).
| \(\dfrac{y}{y + 4.2}\) | \(=\) | \(\dfrac{5}{8}\) |
| \(8y\) | \(=\) | \(5y + 21\) |
| \(y\) | \(=\) | \(\mathbf{7}\) |
(i) Explain why \(\triangle ABC \sim \triangle EDC\).
(ii) Find \(x = CE\) and \(y = BC\).
(i) \(\angle ACB = \angle ECD\) (vertically opposite) and \(\angle BAC = \angle DEC\) (alternate angles, \(AB \parallel DE\)). Two pairs of equal angles make the triangles similar.
(ii) From the statement, \(A \leftrightarrow E\), \(B \leftrightarrow D\), \(C \leftrightarrow C\). So \(CE\) matches \(CA\) and \(CD\) matches \(CB\).
| \(k\) | \(=\) | \(\dfrac{ED}{AB} = \dfrac{9}{6} = 1.5\) |
| \(x\) | \(=\) | \(4 \times 1.5 = \mathbf{6}\) |
| \(y\) | \(=\) | \(7.5 \div 1.5 = \mathbf{5}\) |
The small triangle at the top (wall above the point, horizontal \(d\), ladder) is similar to the whole triangle: they share the angle at the top and both have a right angle.
Its height is \(6 - 1.5 = 4.5\) m, matching the full height \(6\) m. Its base \(d\) matches \(2.4\) m.
| \(\dfrac{d}{2.4}\) | \(=\) | \(\dfrac{4.5}{6}\) |
| \(d\) | \(=\) | \(2.4 \times 0.75 = \mathbf{1.8}\text{ m}\) |
Common pitfalls
Frequently asked questions
How do you find a missing side in similar figures?
Find the scale factor by dividing a known side in one figure by its matching side in the other. Then multiply the matching side of the missing length by the scale factor, or divide to go the other way.
Do similar figures have the same perimeter?
No. Their perimeters are in the same ratio as their sides. If the scale factor is \(3\), the perimeter is \(3\) times as long.
Why does a line parallel to a side make similar triangles?
The small triangle and the large triangle share one angle, and the parallel line makes equal corresponding angles. Two pairs of equal angles means the triangles are similar.
How do you solve shadow problems with similar triangles?
At the same moment the sun's rays hit at the same angle, so each object and its shadow form similar right-angled triangles. Match height with height and shadow with shadow, then use the scale factor.
What is an hourglass shape in similar triangles?
Two segments crossing between two parallel lines form two triangles that meet at a point. Vertically opposite angles and alternate angles are equal, so the triangles are similar.