Similar figures and similar triangles – Enlargement and Similarity in 2D (Senior 3 Mathematics) | Ganzaa

Enlargement and Similarity in 2D

Lessons

Senior 3 · Mathematics

Enlargement and Similarity in 2D

5 lessons

Lesson 1 of 5

Similar figures and similar triangles

25 min

Paper that keeps its shape

Fold a sheet of A4 paper in half and you get two sheets of A5; fold one of those and you get A6. Each smaller sheet has exactly the same shape as the bigger one, only turned sideways. That is why a photocopier can shrink an A4 page onto A5 with nothing cut off. Two figures with the same shape but not necessarily the same size are called similar.

Definition

Two figures are similar if one is an enlargement of the other, possibly followed by a reflection, rotation or translation. For polygons this means both:

corresponding angles are equal, and
corresponding lengths are in the same ratio $k$ , the scale factor (or ratio of similarity).

Notation: $\triangle ABC \sim \triangle PQR$ means "is similar to", with the vertices listed in matching order: $A \leftrightarrow P$ , $B \leftrightarrow Q$ , $C \leftrightarrow R$ . Then $\frac{PQ}{AB} = \frac{QR}{BC} = \frac{RP}{CA} = k$ .

Example: any two squares are similar. A square and a long rectangle are not: their angles match, but their sides are not in one ratio.

Worked example 1: are they similar?

(a) A4 is $210 \times 297$ mm and A5 is $148.5 \times 210$ mm. (b) A photo is $10 \times 15$ cm and a frame is $20 \times 25$ cm.

Solution All angles are right angles, so only the side ratios need checking, long side to long side and short to short.

(a) $\frac{297}{210} \approx 1.414$ and $\frac{210}{148.5} \approx 1.414$ : the same ratio, so A4 and A5 are similar, with $k \approx 1.414$ (in fact $\sqrt{2}$ ).

(b) $\frac{20}{10} = 2$ but $\frac{25}{15} \approx 1.67$ : not similar. The photo will not fill the frame without cropping.

Important

Similar triangles: three shortcuts. Two triangles are similar if any one of these holds:

Condition	What you must show
AA	two angles of one equal two angles of the other (the third angles then match, by the angle sum)
SSS	all three pairs of sides are in the same ratio
SAS	two pairs of sides are in the same ratio and the angles between them are equal

Why AA is enough. Slide the smaller triangle onto the larger one so that the equal angles at $A$ and $P$ coincide. The equal angles at $B$ and $Q$ are then corresponding angles, so the third side is parallel to the larger triangle's side, and Thales' theorem (Senior 2) gives equal ratios on all three sides.

For polygons with more than three sides there is no shortcut: check all the angles and all the ratios.

Worked example 2: two angles are enough

In the figure, $\angle A = \angle P$ and $\angle B = \angle Q$ . $A B = 6$ cm, $B C = 8$ cm, $C A = 7$ cm and $P Q = 9$ cm. Find $Q R$ and $R P$ .

Solution $\triangle ABC \sim \triangle PQR$ (AA). The scale factor is $k = \frac{PQ}{AB} = \frac{9}{6} = 1.5$ .

$QR = 1.5 \times BC = 1.5 \times 8 = 12$ cm and $RP = 1.5 \times CA = 1.5 \times 7 = 10.5$ cm.

Worked example 3: checking three sides

Which of these triangles is similar to one with sides 4 cm, 6 cm and 7 cm: (a) 10 cm, 15 cm, 17.5 cm; (b) 8 cm, 12 cm, 15 cm?

Solution Compare shortest with shortest, middle with middle and longest with longest.

(a) $\frac{10}{4} = 2.5$ , $\frac{15}{6} = 2.5$ , $\frac{17.5}{7} = 2.5$ : similar (SSS), with $k = 2.5$ .

(b) $\frac{8}{4} = 2$ , $\frac{12}{6} = 2$ , but $\frac{15}{7} \approx 2.14$ : not similar.

Worked example 4: a parallel line inside a triangle

$DE \parallel BC$ , $A D = 5$ cm, $D B = 3$ cm and $D E = 6$ cm. Find $B C$ .

Solution $\angle ADE = \angle ABC$ (corresponding angles, $DE \parallel BC$ ), and $\angle A$ is common, so $\triangle ADE \sim \triangle ABC$ (AA).

$k = \frac{AB}{AD} = \frac{5 + 3}{5} = 1.6$ , so $BC = 1.6 \times 6 = 9.6$ cm. Note that the ratio uses the whole side $A B = 8$ , not the part $D B$ .

Worked example 5: the width of a river

A tree $T$ stands on the far bank of a river. From $B$ , straight across from the tree, a pupil walks 40 m along the bank to $C$ and 10 m more to $D$ , then turns at right angles away from the river and walks until $C$ hides the tree: this is $E$ , with $D E = 12$ m. How wide is the river, $T B$ ?

Solution $\angle TBC = \angle EDC = 90^\circ$ , and $\angle TCB = \angle ECD$ (vertically opposite angles), so $\triangle TBC \sim \triangle EDC$ (AA).

\frac{TB}{ED} = \frac{BC}{DC} \quad\Longrightarrow\quad TB = 12 \times \frac{40}{10} = 48 \text{ m}.

Warning

Match the vertices before writing a ratio. The side of one triangle corresponds to the side opposite the equal angle in the other, not to the side that happens to be drawn in the same place. Writing the similarity with matching letters, such as $\triangle TBC \sim \triangle EDC$ , prevents the mistake.

Tip

Put the scale factor in words: "big over small". Find it once from a pair of sides you know, then multiply every small length by it (or divide every big length by it).

Corbettmaths: Similar shapes, missing sides. Watch how the scale factor is found from one pair of matching sides and then used for the others, as in Worked example 2.

✦

Practice

The last two are harder.

Are rectangles of $4 \times 6$ cm and $10 \times 15$ cm similar? If so, give the scale factor.
Are rectangles of $5 \times 8$ cm and $10 \times 13$ cm similar?
One triangle has angles $35^\circ$ and $75^\circ$ ; another has angles $70^\circ$ and $35^\circ$ . Are they similar?
$\triangle ABC \sim \triangle DEF$ , with $A B = 4$ , $B C = 5$ , $C A = 6$ and $D E = 10$ (in cm). Find $E F$ and $F D$ .
Are triangles with sides 5, 7, 8 and 7.5, 10.5, 12 (cm) similar?
In triangle $A B C$ , $DE \parallel BC$ with $D$ on $A B$ and $E$ on $A C$ . $A D = 4$ cm, $A B = 10$ cm and $B C = 15$ cm. Find $D E$ .
$AB \parallel CD$ , and the segments $A D$ and $B C$ cross at $X$ . $A X = 4$ cm, $X D = 10$ cm, $B X = 5$ cm and $A B = 6$ cm. Find $C D$ and $X C$ .
Explain why all squares are similar, but not all rhombuses are.

Show answersHide answers

$\frac{10}{4} = \frac{15}{6} = 2.5$ : yes, $k = 2.5$ .
$\frac{10}{5} = 2$ but $\frac{13}{8} = 1.625$ : no.
The third angles are $70^\circ$ and $75^\circ$ , so both triangles have angles $35^\circ$ , $70^\circ$ , $75^\circ$ : similar (AA).
$k = \frac{10}{4} = 2.5$ : $E F = 12.5$ cm, $F D = 15$ cm.
$\frac{7.5}{5} = \frac{10.5}{7} = \frac{12}{8} = 1.5$ : yes (SSS).
$\triangle ADE \sim \triangle ABC$ with $k = \frac{4}{10} = 0.4$ , so $DE = 0.4 \times 15 = 6$ cm.
$\triangle XAB \sim \triangle XDC$ (alternate angles, vertically opposite angles), $k = \frac{XD}{XA} = \frac{10}{4} = 2.5$ : $CD = 2.5 \times 6 = 15$ cm and $XC = 2.5 \times 5 = 12.5$ cm.
Every square has four right angles and four equal sides, so any two squares have equal angles and sides in one ratio. Two rhombuses always have sides in one ratio, but their angles can differ (for example $60^\circ$ and $120^\circ$ against $90^\circ$ ), so they need not be similar.

Sources and credits

Written by Ganzaa.
Video: Similar Shapes - Missing Sides, Corbettmaths, embedded from YouTube.

The diagrams, worked examples and practice items were written by Ganzaa. The A-series sizes are the ISO 216 paper sizes (A5 is usually rounded to 148 × 210 mm); the river measurements are illustrative.

1Paper that keeps its shape

Paper that keeps its shape