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