4.11 - Similar figures and length scales

Match the sides and angles of similar figures, use a common length scale, and test whether an enlargement keeps the same shape. The method works even when a figure is turned or reflected.

Corresponding sides and angles

Would a $12\,\mathrm{cm}$ by $8\,\mathrm{cm}$ rectangle be a faithful enlargement of an $8\,\mathrm{cm}$ by $5\,\mathrm{cm}$ rectangle? Looking similar is not enough: we need to match their features and compare their lengths.

You already know

A ratio compares quantities by multiplication. A triangle's angles sum to $180^\circ$ , and a rectangle has four right angles.

For a turned figure, the side at the bottom may correspond to a side drawn vertically. Match the features by their role in the shape, rather than their position on the page.

Similar figures

Similar figures have equal corresponding angles and corresponding lengths in one common ratio. Corresponding means matching: a vertex matches a vertex, and the side joining two vertices matches the side joining their partners. Congruent figures are similar with length factor $1$ .

A surprisingly small square and a very large square can still have exactly the same shape. A square of side $1\,\mathrm{cm}$ and a square of side $1\,\mathrm{km}=100\,000\,\mathrm{cm}$ have length factor $100\,000$ , but every corresponding angle is still $90^\circ$ . Similarity does not require equal size. These similarity properties are not printed on either Foundation or Higher formulae sheet: learn them.

An ordered statement names the matching vertices. If triangle $ABC$ is similar to triangle $DEF$ , read $A\leftrightarrow D$ , $B\leftrightarrow E$ , $C\leftrightarrow F$ . The endpoints then determine the matching sides.

Worked example 1: read the ordered match

Question. Triangles $ABC$ and $DEF$ are similar in that order. Angles $A$ , $B$ , $C$ are $40^\circ$ , $60^\circ$ , $80^\circ$ . State angle $D$ and the side corresponding to $AC$ .

Plan. Pair the vertices in order, then use the matched endpoints.

\begin{aligned} A&\leftrightarrow D,\quad C\leftrightarrow F &&\text{read first and third positions}\\ \angle D&=\angle A=40^\circ &&\text{corresponding angles agree}\\ AC&\leftrightarrow DF &&\text{join matched endpoints} \end{aligned}

Why this step? The ordering tells us which features match, even if the figures face different ways.

Answer. $40^\circ$ and side $DF$ .

Check. The other angles are $60^\circ$ and $80^\circ$ ; $40+60+80=180$ .

Change only the order of the second triangle's labels. Swapping two labels changes the correspondence; it does not change the rule about angles.

Worked example 2: a changed label order

Question. Triangles $ABC$ and $DFE$ are similar in that order. Angles $A$ , $B$ , $C$ are $40^\circ$ , $60^\circ$ , $80^\circ$ . State angle $D$ and the side corresponding to $AC$ .

Plan. Read the new order before transferring the angle or side.

\begin{aligned} A&\leftrightarrow D,\quad B\leftrightarrow F,\quad C\leftrightarrow E &&\text{new pairing}\\ \angle D&=40^\circ &&\text{still corresponds to A}\\ AC&\leftrightarrow DE &&\text{use the new third vertex} \end{aligned}

Why this step? Side $AC$ joins the first and third vertices, now $D$ and $E$ .

Answer. $40^\circ$ and side $DE$ .

Check. $\angle F=60^\circ$ and $\angle E=80^\circ$ , so the matched angles still agree.

The same idea applies to polygons: if $ABCD$ matches $WXYZ$ , then $AD$ matches $WZ$ . A corresponding $115^\circ$ angle remains $115^\circ$ ; there is no angle multiplier.

Watch out

If lengths double, the line $\angle D=2\times40^\circ=80^\circ$ is wrong. Similarity scales lengths, while corresponding angles stay equal: $\angle D=40^\circ$ .

Read a new ordering for this check; all needed correspondence information is in the question.

One common length scale

One known pair of corresponding lengths supplies the multiplier for every other corresponding length. Multiplication preserves proportions; adding the same amount usually does not.

For example, $4\,\mathrm{cm}$ becoming $10\,\mathrm{cm}$ means multiplying by $2.5$ . A corresponding $3\,\mathrm{cm}$ length must also be multiplied by $2.5$ .

Directed length scale factor

k=\frac{\text{length in the new figure}}{\text{corresponding length in the original figure}},\qquad \text{new length}=k\times\text{original length}

Use the same unit for the two lengths in the ratio. A positive factor above $1$ increases size; one below $1$ reduces it. Neither result is printed on the Foundation or Higher sheet. Not in the formulae sheet: learn it.

The larger triangle below is reflected. Its labels, rather than its orientation, show that $AB$ matches $DE$ , $AC$ matches $DF$ , and $BC$ matches $EF$ .

[DIAGRAM: asset_slug: m080_similar_triangles; caption: Reflected similar triangles still have matching sides in one ratio; description: Similar right triangles ABC and DEF with A corresponding to D, B to E and C to F. AB is 4 cm, AC 3 cm, BC 5 cm, DE 10 cm, DF x cm and EF y cm. The larger triangle is a reflected 2.5-times copy; no measurement from the image is needed.; file: m080_similar_triangles.png; recommended_method: matplotlib]
Reflected similar triangles still have matching sides in one ratio

Worked example 3: two missing sides

Question. Similar triangles $ABC$ and $DEF$ have matching sides $AB\leftrightarrow DE$ , $AC\leftrightarrow DF$ , $BC\leftrightarrow EF$ . Given $AB=4$ , $AC=3$ , $BC=5$ and $DE=10$ , all in centimetres, find $DF=x$ and $EF=y$ .

Plan. Find the factor from $ABC$ to $DEF$ , then multiply both missing sides by it.

\begin{aligned} k&=\frac{10}{4}=2.5 &&\text{new divided by original}\\ x&=3\times2.5=7.5\,\mathrm{cm} &&\text{corresponds to AC}\\ y&=5\times2.5=12.5\,\mathrm{cm} &&\text{corresponds to BC} \end{aligned}

Why this step? Similarity gives one multiplier for all matching lengths, even after reflection.

Answer. $DF=7.5\,\mathrm{cm}$ , $EF=12.5\,\mathrm{cm}$ .

Check. $10/4=7.5/3=12.5/5=2.5$ .

Change only the given length $DE$ , from $10$ to $12\,\mathrm{cm}$ . The original triangle and all side matches stay fixed.

Worked example 4: a different target size

Question. Similar triangles have the same correspondences as above. The original sides are $AB=4$ , $AC=3$ , $BC=5$ , and now $DE=12$ , all in centimetres. Find $DF$ and $EF$ .

Plan. Recalculate the common factor using the changed target length.

\begin{aligned} k&=\frac{12}{4}=3 &&\text{directed length factor}\\ DF&=3\times3=9\,\mathrm{cm} &&\text{scale AC}\\ EF&=5\times3=15\,\mathrm{cm} &&\text{scale BC} \end{aligned}

Why this step? A new target size requires a new common multiplier; the correspondence is unchanged.

Answer. $DF=9\,\mathrm{cm}$ , $EF=15\,\mathrm{cm}$ .

Check. $12/4=9/3=15/5=3$ .

A proportion gives the same method. For instance, $DF/3=12/4$ says the unknown-to-original ratio equals the known ratio.

Watch out

$DF=3+(10-4)=9$ is wrong for Worked example 3. Adding $6$ does not preserve the ratio. Correct it to $DF=3\times(10/4)=7.5\,\mathrm{cm}$ .

For your turn, change only $DE$ again; use the supplied matching sides.

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Number

Algebra

Shape, space and measure

Handling data

4.11 - Similar figures and length scales

Corresponding sides and angles

One common length scale

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