$△$ ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. $△$ DEF is similar to $△$ ABC. If EF= 4 cm, then find the perimeter of $△$ DEF.

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Solution

Given that $△ A B C \sim △ D E F$
and AB = 3 cm, BC = 2 cm and CA = 2.5 cm, EF = 4 cm
$\frac{A B}{D E} = \frac{A C}{D F} = \frac{B C}{B F} = \frac{2}{4} = \frac{1}{2}$
DE = 2 $\times$ AB = 6 cm, DF = 2 $\times$ AC = 5 cm
$\Rightarrow$ Perimeter of $△$ DEF
= DE+ EF+ DF
=6 + 4 + 5 = 15 cm.

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△ ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. △ DEF is similar to △ABC. If EF= 4 cm, then find the perimeter of △DEF.

Given that △ABC∼△DEF and AB = 3 cm, BC = 2 cm and CA = 2.5 cm, EF = 4 cm ABDE=ACDF=BCBF=24=12 DE = 2 × AB = 6 cm, DF = 2 × AC = 5 cm ⇒ Perimeter of △DEF = DE+ EF+ DF =6 + 4 + 5 = 15 cm.

$△$ ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. $△$ DEF is similar to $△$ ABC. If EF= 4 cm, then find the perimeter of $△$ DEF.

Given that $△ A B C \sim △ D E F$
and AB = 3 cm, BC = 2 cm and CA = 2.5 cm, EF = 4 cm
$\frac{A B}{D E} = \frac{A C}{D F} = \frac{B C}{B F} = \frac{2}{4} = \frac{1}{2}$
DE = 2 $\times$ AB = 6 cm, DF = 2 $\times$ AC = 5 cm
$\Rightarrow$ Perimeter of $△$ DEF
= DE+ EF+ DF
=6 + 4 + 5 = 15 cm.