$ \triangle \mathrm{PQR} \sim \triangle \mathrm{ZYX} . \quad $ If $ \mathrm{PQ}: \mathrm{ZY}=5: 3 $ and $ \mathrm{PR}=10 \mathrm{~cm} $, find $ \mathrm{ZX} $.

Given:

\( \triangle \mathrm{PQR} \sim \triangle \mathrm{ZYX} \).

\( \mathrm{PQ}: \mathrm{ZY}=5: 3 \) and \( \mathrm{PR}=10 \mathrm{~cm} \).

To do:

We have to find \( \mathrm{ZX} \).

Solution:

\( \triangle \mathrm{PQR} \sim \triangle \mathrm{ZYX} \)

When two triangles are similar their corresponding angles are equal and corresponding angles are equal and corresponding sides are in proportion.

Therefore,

$\frac{PQ}{ZY}=\frac{QR}{YX}=\frac{PR}{ZX}$

$\frac{PQ}{ZY}=\frac{PR}{ZX}$

$\frac{5}{3}=\frac{10}{ZX}$

$ZX=\frac{10\times3}{5}$

$ZX=6\ cm$

Hence, \( \mathrm{ZX}=6\ cm \).