Question

In figure if ΔABC ~ΔDEF and their sides are of lengths (in cm) as marked along them ,then find the lengths of the sides of each triangle. ​

Answer 1

Understanding the Question

A similar triangle is basically a magnified/shrunk triangle.

This is done by multiplying one ratio to the sides.

Solution

The same ratio is multiplied to the sides. The ratios of pairs of corresponding sides are hence equal.

$\overline{AB}:\overline{DE}=\overline{BC}:\overline{EF}=\overline{CA}:\overline{FD}$

$\Leftrightarrow (2x-1):18=(2x+2):(3x+9)=3x:6x$

Hence,

$(2x+1):18=1:2$

$\Leftrightarrow 2(2x+1)=18$

$\therefore x=4$ cm

But,

$(2x+2):(3x+9)=1:2$

$\Leftrightarrow 2(2x+2)=1(3x+9)$

$\Leftrightarrow 4x+4=3x+9$

$\therefore x=5$ cm

There is no such triangle following the given condition.

Answer 2

Given :-

• ΔABC ~ ΔDEF

• AB = 2x - 1

• BC = 2x + 2

• CA = 3x

• DE = 18

• EF = 3x + 9

• FD = 6x

To Find :-

• Length of sides of ΔABC and ΔDEF

CALCULATION :-

Since,

Its given that

$\bf \: \triangle \: ABC \: \sim \: \triangle \: DEF$

$\rm :\implies\:\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$

$\rm :\implies\:\dfrac{2x - 1}{18} = \dfrac{2x + 2}{3x + 9} = \dfrac{3x}{6x}$

$\rm :\implies\:\dfrac{2x - 1}{18} = \dfrac{2x + 2}{3x + 9} = \dfrac{1}{2}$

$\rm :\implies\:\dfrac{2x - 1}{18} = \dfrac{1}{2}$

$\rm :\implies\:4x - 2 = 18$

$\rm :\implies\:4x = 20$

$\rm :\implies\:x = 5$

Hence,

Sides of triangle are

• AB = 2x - 1 = 2×5 - 1 = 9 cm.

• BC = 2x + 2 = 2 × 5 + 2 = 12 cm.

• CA = 3x = 3 × 5 = 15 cm.

• DE = 18 cm.

• EF = 3x + 9 = 3 × 5 + 9 = 15 + 9 = 24 cm.

• FD = 6x = 6 × 5 = 30 cm.