Identify congruent and similar figures, apply the conditions for congruence and similarity, and use scale factors for lengths, areas and volumes

What this dot point is asking

SEAB wants you to recognise when two figures are congruent (identical) or similar (same shape, different size), to justify each with the correct conditions, and to use scale factors to find missing lengths, areas and volumes. The squared and cubed scale-factor relationships are a frequent exam focus.

The answer

Congruence

Two figures are congruent if they are identical in shape and size, so one can be placed exactly on the other. For triangles, the standard congruence conditions are SSS (three sides), SAS (two sides and the included angle), ASA or AAS (two angles and a corresponding side), and RHS (right angle, hypotenuse and one side).

Similarity

Two figures are similar if they have the same shape but possibly different size: corresponding angles are equal and corresponding sides are in the same ratio. For triangles, showing two pairs of equal angles (AA) is enough, because the third angle then matches automatically.

The linear scale factor

The linear scale factor $k$ is the ratio of a length in one figure to the corresponding length in the other:

Every corresponding length is multiplied by $k$, so missing lengths follow at once.

Area and volume scale factors

When lengths scale by $k$:

• areas scale by $k^2$,
• volumes scale by $k^3$.

So doubling every length quadruples the area and multiplies the volume by eight. This is the most tested idea in similarity questions.

Proving two triangles similar in a diagram

A common E-Maths task is to justify similarity before using it, and the cleanest argument lists two pairs of equal angles (AA). Equal angles often come from shared angles, vertically opposite angles, or the equal corresponding and alternate angles created by parallel lines. In a figure where a line is drawn parallel to one side of a triangle, the smaller triangle it cuts off is similar to the whole, because the parallel line produces two pairs of equal angles. Writing out the reason for each equal angle, then concluding "two pairs of equal angles, so the triangles are similar (AA)", is the structured justification markers expect.

Working backwards from an area or volume ratio

Because area scales by $k^2$ and volume by $k^3$, you can recover the linear scale factor by taking a square root or cube root. If two similar shapes have areas in the ratio $9 : 16$, the linear scale factor is $\sqrt{9} : \sqrt{16} = 3 : 4$; if two similar solids have volumes in the ratio $8 : 27$, the lengths are in the ratio $\sqrt[3]{8} : \sqrt[3]{27} = 2 : 3$. This reverse direction is just as examinable as the forward one, so being comfortable taking roots of the area or volume ratio to get back to lengths is essential.

Examples in context

Example 1. Scale models. A model car at $1 : 20$ has every length one twentieth of the real car, its surface area $\dfrac{1}{20^2}$ and its volume (and so its mass for the same material) $\dfrac{1}{20^3}$. The cube law explains why small models are surprisingly light.

Example 2. Map areas. On a map at scale $1 : 25\,000$, a region's area on the ground is $25\,000^2$ times its area on the map. Converting map area to real area always uses the squared scale factor.

Try this

Q1. Two similar rectangles have lengths $4\ \text{cm}$ and $10\ \text{cm}$. State the linear scale factor from the smaller to the larger. [1 mark]

• Cue. $\dfrac{10}{4} = 2.5$.

Q2. The linear scale factor between two similar shapes is $3$. State the area scale factor. [1 mark]

• Cue. $3^2 = 9$.

Q3. Two similar solids have volumes in the ratio $8 : 27$. Find the ratio of their corresponding lengths. [2 marks]

• Cue. Take cube roots: $\sqrt[3]{8} : \sqrt[3]{27} = 2 : 3$.