# What is the value of the hypotenuse of an isosceles triangle with a perimeter equal to 16 + 16sqrt2?

Jan 6, 2016

#### Explanation:

From the photo, we can see that the perimeter of an isosceles right triangle is $2 x + x \sqrt{2}$, so we can construct the following equation:
$2 x + x \sqrt{2} = 16 + 16 \sqrt{2}$
$x \left(2 + \sqrt{2}\right) = 16 + 16 \sqrt{2}$
$x = \frac{16 + 16 \sqrt{2}}{2 + \sqrt{2}}$
And since we're looking for the * hypotenuse $x \sqrt{2}$ * we can multiply both sides of the equation by $\sqrt{2}$
$x \sqrt{2} = \frac{16 + 16 \sqrt{2}}{2 + \sqrt{2}} \cdot \sqrt{2}$
xsqrt(2)=(16sqrt(2)+32)/(2+sqrt(2)
$x \sqrt{2} = \frac{16 \left(\sqrt{2} + 2\right)}{2 + \sqrt{2}}$

$\therefore x \sqrt{2} = 16$