Finding Unknown Lengths in Similar Figures

Use a given similarity to write equal ratios of corresponding sides and find unknown lengths: figures side by side or turned, a triangle inside a triangle, triangles across parallel lines, and heights from shadows and mirrors.

Before you start

If $\triangle ABC$ is similar to $\triangle PQR$ , then $\angle A = \angle P$ , $\angle B = \angle Q$ , $\angle C = \angle R$ and $\frac{PQ}{AB} = \frac{QR}{BC} = \frac{RP}{CA}\text{,}$ the scale factor from $\triangle ABC$ to $\triangle PQR$ (lesson 8.2).
An equation with the unknown on top of a fraction is solved by one multiplication: $\frac{x}{6} = \frac{5}{4}$ gives $x = 6 \times \frac{5}{4} = 7.5$ .
Where a line crosses two parallel lines, corresponding angles are equal (corr. ∠s) and alternate angles are equal (alt. ∠s); where two lines cross, vertically opposite angles are equal (vert. opp. ∠s).

Equal Ratios of Corresponding Sides

When two figures are given as similar, the ratio of any pair of corresponding sides is the same. So one pair of corresponding sides whose lengths are both known gives the ratio, and every other unknown side follows from it.

Finding an unknown side

1. Match the vertices from the similarity, in order.

2. Write two equal ratios of corresponding sides: one with both lengths known, one with the unknown. Put the unknown on top and the matching figure's sides on top in both ratios.

3. Multiply to find the unknown, and check that its size makes sense: a side of the larger figure must come out longer.

Or find the scale factor $k$ first: a length of the new figure is $k$ times the corresponding length of the first, and a length of the first is the new length divided by $k$ .

Example 1: Two triangles side by side

$\triangle ABC$ is similar to $\triangle PQR$ . $AB = 6$ cm, $BC = 8$ cm, $CA = 7$ cm and $PQ = 9$ cm. Find $QR$ and $RP$ , marked $x$ and $y$ .

Solution

$A$ , $B$ , $C$ correspond to $P$ , $Q$ , $R$ . The known pair is $PQ$ and $AB$ .

$\frac{QR}{BC} = \frac{PQ}{AB}$ , so $\frac{x}{8} = \frac{9}{6}$ and $x = 8 \times \frac{9}{6} = 12$ .

$\frac{RP}{CA} = \frac{PQ}{AB}$ , so $\frac{y}{7} = \frac{9}{6}$ and $y = 7 \times \frac{9}{6} = 10.5$ .

Check with the scale factor: $k = \frac{9}{6} = 1.5$ , and $8 \times 1.5 = 12$ , $7 \times 1.5 = 10.5$ . $\triangle PQR$ is the larger, and both answers are longer than the sides they match.

$QR = 12$ cm and $RP = 10.5$ cm.

Example 2: A turned triangle, unknowns in both

$\triangle ABC$ is similar to $\triangle DEF$ . $AB = 12$ cm, $CA = 15$ cm, $DE = 8$ cm and $EF = 9$ cm. Find $DF$ and $BC$ .

Solution

$A$ , $B$ , $C$ correspond to $D$ , $E$ , $F$ . The known pair is $DE$ and $AB$ .

$DF$ is in $\triangle DEF$ , so put $\triangle DEF$ on top: $\frac{DF}{AC} = \frac{DE}{AB}$ , $\frac{DF}{15} = \frac{8}{12}$ , $DF = 15 \times \frac{8}{12} = 10\text{.}$

$BC$ is in $\triangle ABC$ , so put $\triangle ABC$ on top: $\frac{BC}{EF} = \frac{AB}{DE}$ , $\frac{BC}{9} = \frac{12}{8}$ , $BC = 9 \times \frac{12}{8} = 13.5\text{.}$

Check: $\triangle DEF$ is the smaller ( $8 < 12$ ), so $DF = 10$ is less than $AC = 15$ , and $BC = 13.5$ is more than $EF = 9$ .

$DF = 10$ cm and $BC = 13.5$ cm.

Example 3: Similar quadrilaterals, one flipped

$ABCD$ is similar to $PQRS$ . $AB = 10$ cm, $BC = 6$ cm, $DA = 7.5$ cm, $PQ = 6$ cm and $RS = 3.3$ cm. Find $QR$ , $SP$ and $CD$ .

Solution

$A$ , $B$ , $C$ , $D$ correspond to $P$ , $Q$ , $R$ , $S$ . Scale factor from $ABCD$ to $PQRS$ : $k = \frac{PQ}{AB} = \frac{6}{10} = 0.6\text{.}$

$QR = 0.6 \times BC = 0.6 \times 6 = 3.6$ cm and $SP = 0.6 \times DA = 0.6 \times 7.5 = 4.5$ cm.

$CD$ is in the larger figure: $\frac{CD}{RS} = \frac{AB}{PQ}$ , so $CD = 3.3 \times \frac{10}{6} = 5.5$ cm. (Or $CD = 3.3 \div 0.6 = 5.5$ cm.)

$QR = 3.6$ cm, $SP = 4.5$ cm and $CD = 5.5$ cm.

A Triangle Inside a Triangle

In $\triangle ABC$ , the point $D$ is on $AB$ and $E$ is on $AC$ , with $DE$ parallel to $BC$ . The small triangle $ADE$ and the large triangle $ABC$ share the angle at $A$ , and $\angle ADE = \angle ABC$ (corr. ∠s, $DE \parallel BC$ ). In such questions you are told that $\triangle ADE$ is similar to $\triangle ABC$ .

The corresponding sides are $AD$ and $AB$ , $AE$ and $AC$ , $DE$ and $BC$ . A side of the large triangle is a whole side, such as $AB = AD + DB$ , never the piece $DB$ alone.

Example 4: Finding the base

In the figure, $DE \parallel BC$ and $\triangle ADE$ is similar to $\triangle ABC$ . $AD = 4$ cm, $DB = 6$ cm and $DE = 5$ cm. Find $BC$ , marked $x$ .

Solution

$A$ , $D$ , $E$ correspond to $A$ , $B$ , $C$ . The whole side $AB = AD + DB = 4 + 6 = 10$ cm.

$\frac{BC}{DE} = \frac{AB}{AD}$ , so $\frac{x}{5} = \frac{10}{4}$ and $x = 5 \times \frac{10}{4} = 12.5\text{.}$

Check: $BC$ is a side of the larger triangle, and $12.5 > 5$ .

$BC = 12.5$ cm.

Example 5: Finding a piece of a side

In the figure, $DE \parallel BC$ and $\triangle ADE$ is similar to $\triangle ABC$ . $AD = 10$ cm, $AE = 8$ cm, $EC = 4$ cm and $DE = 12$ cm. Find $AB$ , $DB$ and $BC$ .

Solution

The whole side $AC = AE + EC = 8 + 4 = 12$ cm. The pair with both lengths known is $AC$ and $AE$ .

$\frac{AB}{AD} = \frac{AC}{AE}$ , so $\frac{AB}{10} = \frac{12}{8}$ and $AB = 10 \times \frac{12}{8} = 15$ cm.

$DB = AB - AD = 15 - 10 = 5$ cm.

$\frac{BC}{DE} = \frac{AC}{AE}$ , so $\frac{BC}{12} = \frac{12}{8}$ and $BC = 12 \times \frac{12}{8} = 18$ cm.

$AB = 15$ cm, $DB = 5$ cm and $BC = 18$ cm.

Two Triangles Across Parallel Lines

Two straight lines $AE$ and $BD$ cross at $C$ , and $AB \parallel DE$ . The triangles $ABC$ and $EDC$ sit on opposite sides of $C$ , one turned upside down. $\angle ACB = \angle ECD$ (vert. opp. ∠s) and $\angle BAC = \angle DEC$ (alt. ∠s, $AB \parallel DE$ ), and you are told that $\triangle ABC$ is similar to $\triangle EDC$ .

Read the corresponding sides from the order of the letters, not from the picture: $A$ matches $E$ , $B$ matches $D$ and $C$ matches $C$ .

Example 6: Crossing lines

In the figure, $AB \parallel DE$ , and $AE$ and $BD$ meet at $C$ . $\triangle ABC$ is similar to $\triangle EDC$ . $AB = 8$ cm, $BC = 7$ cm, $CA = 6$ cm and $CE = 9$ cm. Find $DE$ and $CD$ .

Solution

$A$ , $B$ , $C$ correspond to $E$ , $D$ , $C$ . The known pair is $CE$ and $CA$ : $k = \frac{CE}{CA} = \frac{9}{6} = 1.5\text{.}$

$DE$ corresponds to $BA$ : $\frac{DE}{AB} = \frac{CE}{CA}$ , so $DE = 8 \times \frac{9}{6} = 12$ cm.

$CD$ corresponds to $CB$ : $\frac{CD}{CB} = \frac{CE}{CA}$ , so $CD = 7 \times \frac{9}{6} = 10.5$ cm.

$DE = 12$ cm and $CD = 10.5$ cm.

Heights from Shadows and Mirrors

The height of something too tall to measure, such as a tree, a flagpole or a building, can be found from similar triangles.

Shadows. At one moment the sun's rays fall at the same slant everywhere. A person and a flagpole standing upright on level ground, with their shadows, make two right-angled triangles with the same angles. (In Secondary 3 you will learn why two triangles with the same angles must be similar.)

A mirror. Light leaves a mirror on the ground at the same angle as it arrives. A person who sees the top of a tree in a mirror on the ground makes, with the tree, two right-angled triangles with the same angles that meet at the mirror.

In each question below the triangles are given as similar.

Example 7: A flagpole's shadow

At 4 p.m., Hui Min, who is 1.6 m tall, casts a shadow 2 m long on the school field. At the same time the flagpole casts a shadow 10 m long. The two triangles in the figure are similar. Find the height of the flagpole, $h$ m.

Solution

$\triangle HFS$ is similar to $\triangle TBU$ : $H$ , $F$ , $S$ correspond to $T$ , $B$ , $U$ (top, foot, end of shadow).

$\frac{TB}{HF} = \frac{BU}{FS}$ , so $\frac{h}{1.6} = \frac{10}{2}$ and $h = 1.6 \times \frac{10}{2} = 8$ .

Check: the flagpole's shadow is 5 times Hui Min's, so the flagpole is 5 times her height: $5 \times 1.6 = 8$ .

The flagpole is $8$ m tall.

Example 8: A mirror on the ground

Ravi puts a small mirror $M$ on level ground, 6 m from the foot $B$ of a tree. He stands 2 m from the mirror and sees the top $T$ of the tree in it. His eyes $E$ are 1.5 m above the ground. $\triangle EFM$ is similar to $\triangle TBM$ . Find the height of the tree, $h$ m.

Solution

$E$ , $F$ , $M$ correspond to $T$ , $B$ , $M$ .

$\frac{TB}{EF} = \frac{BM}{FM}$ , so $\frac{h}{1.5} = \frac{6}{2}$ and $h = 1.5 \times \frac{6}{2} = 4.5\text{.}$

The tree is $4.5$ m tall.

Common Mistakes

Using a piece of a side instead of the whole side: in a triangle inside a triangle, $AB = AD + DB$ matches $AD$ ; $DB$ matches nothing.
Mixing the figures in a ratio: in $\frac{x}{8} = \frac{9}{6}$ both tops come from one figure and both bottoms from the other.
Pairing sides by their places on the page. In a turned or flipped figure, read the pairs from the order of the letters.
An answer of the wrong size: a side of the larger figure must be longer than the side it matches. Check every answer this way.

Practice Questions

Try each question before you open its solution. Write the equal ratios with the unknown on top, and check the size of each answer.

Question 1

$\triangle PQR$ is similar to $\triangle XYZ$ . $PQ = 5$ cm, $QR = 7$ cm, $RP = 6$ cm and $XY = 7.5$ cm. Find $YZ$ and $ZX$ , marked $x$ and $y$ .

ShowHide solution

$k = \frac{XY}{PQ} = \frac{7.5}{5} = 1.5\text{.}$

$x = 7 \times 1.5 = 10.5$ and $y = 6 \times 1.5 = 9$ .

$YZ = 10.5$ cm and $ZX = 9$ cm.

Question 2

$\triangle ABC$ is similar to $\triangle LMN$ . $AB = 14$ cm, $BC = 10$ cm, $LM = 7$ cm and $LN = 6$ cm. Find $MN$ and $AC$ .

ShowHide solution

$\frac{MN}{BC} = \frac{LM}{AB}$ : $MN = 10 \times \frac{7}{14} = 5$ cm.

$\frac{AC}{LN} = \frac{AB}{LM}$ : $AC = 6 \times \frac{14}{7} = 12$ cm.

Check: $\triangle LMN$ is the smaller ( $7 < 14$ ); $MN = 5 < 10$ and $AC = 12 > 6$ .

$MN = 5$ cm and $AC = 12$ cm.

Question 3

A rectangle 15 cm by 9 cm is similar to a rectangle 10 cm by $x$ cm, the 15 cm side matching the 10 cm side. Find $x$ .

ShowHide solution

$\frac{x}{9} = \frac{10}{15}$ , so $x = 9 \times \frac{10}{15} = 6$ .

$x = 6$

Question 4

Quadrilateral $ABCD$ is similar to quadrilateral $PQRS$ . $AB = 8$ cm, $BC = 5$ cm, $CD = 6$ cm, $DA = 4$ cm and $PQ = 12$ cm. Find $QR$ , $RS$ and $SP$ .

ShowHide solution

$k = \frac{PQ}{AB} = \frac{12}{8} = 1.5\text{.}$

$QR = 5 \times 1.5 = 7.5$ cm, $RS = 6 \times 1.5 = 9$ cm, $SP = 4 \times 1.5 = 6$ cm.

$QR = 7.5$ cm, $RS = 9$ cm and $SP = 6$ cm.

Question 5

In the figure, $DE \parallel BC$ and $\triangle ADE$ is similar to $\triangle ABC$ . $AD = 3$ cm, $DB = 6$ cm and $DE = 4$ cm. Find $BC$ .

ShowHide solution

$AB = 3 + 6 = 9$ cm. $\frac{BC}{DE} = \frac{AB}{AD}$ , so $BC = 4 \times \frac{9}{3} = 12$ cm.

$BC = 12$ cm.

Question 6

In the figure, $PQ \parallel BC$ and $\triangle APQ$ is similar to $\triangle ABC$ . $AP = 6$ cm, $PB = 4$ cm, $AQ = 4.5$ cm and $PQ = 7.2$ cm. Find $AC$ , $QC$ and $BC$ .

ShowHide solution

$AB = 6 + 4 = 10$ cm, and $\frac{AB}{AP} = \frac{10}{6}$ .

$\frac{AC}{AQ} = \frac{AB}{AP}$ : $AC = 4.5 \times \frac{10}{6} = 7.5$ cm, so $QC = 7.5 - 4.5 = 3$ cm.

$\frac{BC}{PQ} = \frac{AB}{AP}$ : $BC = 7.2 \times \frac{10}{6} = 12$ cm.

$AC = 7.5$ cm, $QC = 3$ cm and $BC = 12$ cm.

Question 7

In the figure, $AB \parallel CD$ , and $AD$ and $BC$ meet at $O$ . $\triangle AOB$ is similar to $\triangle DOC$ . $AO = 4$ cm, $OD = 6$ cm, $AB = 5$ cm and $OC = 7.2$ cm. Find $DC$ and $OB$ .

ShowHide solution

$A$ , $O$ , $B$ correspond to $D$ , $O$ , $C$ . $k = \frac{OD}{OA} = \frac{6}{4} = 1.5\text{.}$

$DC = 5 \times 1.5 = 7.5$ cm.

$OB$ is in the smaller triangle: $\frac{OB}{OC} = \frac{OA}{OD}$ , so $OB = 7.2 \times \frac{4}{6} = 4.8$ cm.

$DC = 7.5$ cm and $OB = 4.8$ cm.

Question 8

In the figure, $DE \parallel BC$ and $\triangle ADE$ is similar to $\triangle ABC$ . $AD = x$ cm, $DB = 4$ cm, $DE = 6$ cm and $BC = 9$ cm. Find $x$ .

ShowHide solution

$AB = x + 4$ . $\frac{AD}{AB} = \frac{DE}{BC}$ , so $\frac{x}{x + 4} = \frac{6}{9}$ .

Multiply both sides by $9(x + 4)$ : $9x = 6(x + 4) = 6x + 24\text{,}$ so $3x = 24$ and $x = 8$ .

Check: $AB = 12$ , and $\frac{8}{12} = \frac{6}{9}$ , both $\frac{2}{3}$ .

$x = 8$

Question 9

At one moment, Daniel, who is 1.5 m tall, casts a shadow 1.2 m long, and an HDB block casts a shadow 36 m long on level ground. The triangles formed by each with its shadow are similar. Find the height of the block.

ShowHide solution

$\frac{\text{height of block}}{1.5} = \frac{36}{1.2}\text{,}$ so the height is $1.5 \times \frac{36}{1.2} = 1.5 \times 30 = 45$ m.

The block is $45$ m tall.

Question 10

A mirror lies on level ground 30 m from the foot of a building. Aisha stands 1.2 m from the mirror, on the other side, and sees the top of the building in it. Her eyes are 1.6 m above the ground. The two right-angled triangles that meet at the mirror are similar. Find the height of the building.

ShowHide solution

The building and its distance from the mirror match Aisha's eye height and her distance from the mirror.

$\frac{\text{height}}{1.6} = \frac{30}{1.2}\text{,}$ so the height is $1.6 \times \frac{30}{1.2} = 1.6 \times 25 = 40$ m.

The building is $40$ m tall.

Question 11

In $\triangle ABC$ , $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$ , and $\triangle ADE$ is similar to $\triangle ABC$ . $AD = 5$ cm, $DB = 3$ cm and $DE = 10$ cm. Raju writes $\frac{BC}{DE} = \frac{DB}{AD}$ and gets $BC = 6$ cm. Explain his mistake and find $BC$ .

ShowHide solution

$DB$ is only a piece of $AB$ ; the side matching $AD$ is the whole side $AB = 5 + 3 = 8$ cm.

His answer cannot be right: $BC$ is the base of the larger triangle, so it must be longer than $DE = 10$ cm.

$\frac{BC}{DE} = \frac{AB}{AD}$ , so $BC = 10 \times \frac{8}{5} = 16$ cm.

He used $DB$ instead of $AB$ ; $BC = 16$ cm.

Question 12

The frame of a roof is the triangle $ABC$ , with a horizontal beam $DE$ parallel to the base $BC$ , $D$ on $AB$ and $E$ on $AC$ . $\triangle ADE$ is similar to $\triangle ABC$ . $AD = 3$ m, $DB = 2$ m and $BC = 7.5$ m. Find the length of the beam $DE$ .

ShowHide solution

$AB = 3 + 2 = 5$ m. $\frac{DE}{BC} = \frac{AD}{AB}$ , so $DE = 7.5 \times \frac{3}{5} = 4.5$ m.

Check: the beam is shorter than the base, as it must be.

The beam is $4.5$ m long.

Key Points

In similar figures, every pair of corresponding sides is in the same ratio, the scale factor.
Write two equal ratios, one with both lengths known and one with the unknown, with the unknown on top and the same figure on top in both.
Read corresponding sides from the order of the letters, especially when a figure is turned or flipped.
In a triangle inside a triangle ( $DE \parallel BC$ ), $AD$ matches the whole side $AB = AD + DB$ .
For triangles across parallel lines meeting at $C$ , $\triangle ABC$ is similar to $\triangle EDC$ : $A$ matches $E$ and $B$ matches $D$ .
Heights from shadows and mirrors: object's height ÷ person's height = object's shadow ÷ person's shadow, or, with a mirror, object's distance from the mirror ÷ person's distance from the mirror.
Check every answer: a side of the larger figure is longer than the side it matches.