In $ \triangle \mathrm{ABC}, \mathrm{AB}=\mathrm{AC} $ and $ \mathrm{AM} $ is an altitude. If $ A M=15 $ and the perimeter of $ \triangle A B C $ is 50 , find the area of $ \triangle \mathrm{ABC} $.

Given:

In \( \triangle \mathrm{ABC}, \mathrm{AB}=\mathrm{AC} \) and \( \mathrm{AM} \) is an altitude.

\( A M=15 \) and the perimeter of \( \triangle A B C \) is 50.

To do:

We have to find the area of \( \triangle \mathrm{ABC} \).

Solution:

Let $AB=AC=x$

In $\triangle ABC$,

By Pythagoras theorem,

$BC^2 =AB^2+AC^2$

$=x^2+x^2$

$=2x^2$

$BC=\sqrt{2x^2}$

$BC=x\sqrt2$

$AB+BC+CA=50$

$x+x+x\sqrt2=50$

$x(2+\sqrt2)=50$

$x=\frac{50}{2+\sqrt2}$

$x=\frac{50(2-\sqrt2)}{(2+\sqrt2)(2-\sqrt2)}$

$x=\frac{50(2-\sqrt2)}{4-2}$

$x=25(2-\sqrt{2})$

Area of the triangle $=\frac{1}{2}\times [25(2-\sqrt2)]^2$

$=\frac{625(4+2-4\sqrt2)}{2}$

$=625(3-2\sqrt2)$