Area in terms of perimeter:Isosceles Triangle.

Geometry Level 3

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

\displaystyle (1-2\sqrt { 2 } ){ p }^{ 2 }

\displaystyle (3-2\sqrt { 2 } ){ p }^{ 2 }

\displaystyle (3+2\sqrt { 2 } ){ p }^{ 2 }

\displaystyle (2-\sqrt { 2 } ){ p }^{ 2 }

1 solution

Omkar Kulkarni
Jan 16, 2015

Let both the sides of the triangle, other than the hypotenuse, be $n$ . By Pythagoras theorem, the hypotenuse comes to be $\sqrt{2}n$ .

$\therefore 2n + \sqrt{2}n = 2p$

$(\sqrt{2}+1)n=\sqrt{2}p$

$n=\frac {\sqrt{2}p}{\sqrt{2}+1}=\sqrt{2}p(\sqrt{2}-1)$

Now, the area of the triangle is $\frac {n^2}{2}$

$= \frac {(\sqrt{2}p(\sqrt{2}-1))^{2}}{2}$

$= \boxed {(3-2\sqrt{2})p^{2}}$

@okmar nice solution

Mardokay Mosazghi - 6 years, 4 months ago