$ \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} . \quad $ If $ \quad \mathrm{AB}+\mathrm{BC}=12 \mathrm{~cm} $ $ \mathrm{PQ}+\mathrm{QR}=15 \mathrm{~cm} $ and $ \mathrm{AC}=8 \mathrm{~cm} $, find $ \mathrm{PR} $.
Given:
\( \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} .\)
\( \mathrm{AB}+\mathrm{BC}=12 \mathrm{~cm} \) \( \mathrm{PQ}+\mathrm{QR}=15 \mathrm{~cm} \) and \( \mathrm{AC}=8 \mathrm{~cm} \).
To do:
We have to find \( \mathrm{PR} \).
Solution:
\( \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \)
The ratio of the perimeters of two similar triangles is same as the ratio of their corresponding sides.
Therefore,
$\frac{Perimeter\ of\ \triangle ABC}{Primeter\ of\ \triangle PQR}=\frac{AC}{PR}$
This implies,
$\frac{AB+BC+CA}{PQ+QR+RP}=\frac{8}{PR}$
$PR(12+8)=8(15+PR)$
$20PR=120+8PR$
$(20-8)PR=120$
$12PR=120$
$PR=10\ cm$
Hence, \( \mathrm{PR}=10\ cm \).
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