$ \triangle \mathrm{ABC} \sim \triangle \mathrm{ZYX} . $ If $ \mathrm{AB}=3 \mathrm{~cm}, \quad \mathrm{BC}=5 \mathrm{~cm} $, $ \mathrm{CA}=6 \mathrm{~cm} $ and $ \mathrm{XY}=6 \mathrm{~cm} $, find the perimeter of $ \Delta \mathrm{XYZ} $.
Given:
\( \triangle \mathrm{ABC} \sim \triangle \mathrm{ZYX} . \)
\( \mathrm{AB}=3 \mathrm{~cm}, \mathrm{BC}=5 \mathrm{~cm} \), \( \mathrm{CA}=6 \mathrm{~cm} \) and \( \mathrm{XY}=6 \mathrm{~cm} \)
To do:
We have to find the perimeter of \( \Delta \mathrm{XYZ} \).
Solution:
\( \triangle \mathrm{ABC} \sim \triangle \mathrm{ZYX} \)
When two triangles are similar their corresponding angles are equal and corresponding angles are equal and corresponding sides are in proportion.
Therefore,
$\frac{AB}{ZY}=\frac{BC}{YX}=\frac{AC}{ZX}$
This implies,
$\frac{AB}{ZY}=\frac{BC}{YX}$
$\frac{3}{ZY}=\frac{5}{6}$
$ZY=\frac{3\times6}{5}$
$YZ=\frac{18}{5}\ cm$
$\frac{BC}{YX}=\frac{AC}{ZX}$
$\frac{5}{6}=\frac{6}{ZX}$
$ZX=\frac{6\times6}{5}$
$ZX=\frac{36}{5}\ cm$
The perimeter of \( \Delta \mathrm{XYZ}=6+\frac{18}{5}+\frac{36}{5} \)
$=\frac{6\times5+18+36}{5}\ cm$
$=\frac{84}{5}\ cm$
Hence, the perimeter of \( \Delta \mathrm{XYZ} \) is $\frac{84}{5}\ cm$.
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