If $ \Delta \mathrm{ABC} \sim \Delta \mathrm{DEF}, \mathrm{AB}=4 \mathrm{~cm}, \mathrm{DE}=6 \mathrm{~cm}, \mathrm{EF}=9 \mathrm{~cm} $ and $ \mathrm{FD}=12 \mathrm{~cm} $, find the perimeter of $ \triangle \mathrm{ABC} $.

Given:

\( \Delta \mathrm{ABC} \sim \Delta \mathrm{DEF}, \mathrm{AB}=4 \mathrm{~cm}, \mathrm{DE}=6 \mathrm{~cm}, \mathrm{EF}=9 \mathrm{~cm} \) and \( \mathrm{FD}=12 \mathrm{~cm} \)

To do:

We have to find the perimeter of \( \triangle \mathrm{ABC} \).

Solution:

$\Delta A B C  \sim \Delta D E F$

This implies,

$\frac{A B}{E D} =\frac{B C}{E F}=\frac{A C}{D F}$

$\frac{4}{6} =\frac{B C}{9}=\frac{A C}{12}$

$\frac{4}{6} =\frac{B C}{9}$

$B C=\frac{4 \times 9}{6}$

$BC=6 \mathrm{~cm}$

$A C=\frac{6 \times 12}{9}$

$AC=8 \mathrm{~cm}$

Therefore,

Perimeter of  $\Delta A B C =A B+B C+A C$

$=4+6+8$

$=18 \mathrm{~cm}$

The perimeter of \( \triangle \mathrm{ABC} \) is $18\ cm$.