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.