$\triangle \mathrm{ABC} \sim \triangle \mathrm{ZYX} .$ 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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Updated on: 10-Oct-2022

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