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 $18\ cm$.
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