In \( \triangle \mathrm{PQR}, \quad \angle \mathrm{P}=\angle \mathrm{Q}+\angle \mathrm{R}, \mathrm{PQ}=7 \) and \( \mathrm{QR}=25 \). Find the perimeter of \( \triangle \mathrm{PQR} \).
Given:
In \( \triangle \mathrm{PQR}, \angle \mathrm{P}=\angle \mathrm{Q}+\angle \mathrm{R}, \mathrm{PQ}=7 \) and \( \mathrm{QR}=25 \).
To do:
We have to find the perimeter of \( \triangle \mathrm{PQR} \).
Solution:
We know that,
Sum of the angles in a triangle is $180^o$.
Therefore,
$\angle P+\angle Q+\angle R=180^o$
$\angle P+\angle P=180^o$
$\angle P=\frac{180^o}{2}=90^o$
 This implies,
\( \triangle \mathrm{PQR} \) is a right-angled triangle.
Therefore,
$QR^2=PQ^2+PR^2$
$25^2=7^2+PR^2$
$PR^2=625-49$
$\Rightarrow PR=\sqrt{576}=24$
The perimeter of \( \triangle \mathrm{PQR}=PQ+QR+PR \)
$=7+25+24$
$=56$
The perimeter of \( \triangle \mathrm{PQR} \) is 56.
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