Question

The perimeter of an isosceles right angled triangle is $2p$. Its area is

• A. $\frac{p^2}{(2+\sqrt{2})}$
• B. $\frac{p^2}{(2-\sqrt{2})}$
• C. $\frac{p^2}{(3-2\sqrt{2})}$
• D. $\frac{p^2}{(3+2\sqrt{2})}$

Answer 1

The correct option is D $\frac{p^2}{(3+2\sqrt{2})}$

The ratio of the sides of an isosceles right triangle is$1:1:\sqrt{2}$ .By the Pythogoras Theorem,The perimeter (P) is,
$P=1x+1x+\left(\sqrt{2}\right)x$
$=(2+\sqrt{2})x$
$=2p$
$x=\frac{2p}{(2+\sqrt{2})}$
Here x is the scale factor and it is also the length of each of the two identical sides $1x$.
Therefore,
The area is,
$A=\frac{1}{2}\times b\times h$
$=\frac{1}{2}\times \left(1x\right)\times \left(1x\right)$
$=\frac{x^2}{2}$
$=\left(\frac{(\frac{2p}{(2+\sqrt{2})}{)}^2}{2}\right)$
$=\frac{2p^2}{(2+\sqrt{2})}$
$=\frac{2p^2}{(4+4\sqrt{2}+2)}$
$=\frac{2p^2}{(6+4\sqrt{2})}$
$=\frac{p^2}{(3+2\sqrt{2})}$