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

Nov 14, 2015

16

#### Explanation:

That triangle is half a square. $\left(t , t , t \sqrt{2}\right)$

$2 p = t + t + t \sqrt{2} = 16 + 16 \sqrt{2}$

$t = 16 \cdot \frac{1 + \sqrt{2}}{2 + \sqrt{2}}$ // times $\sqrt{2}$

Hypotenuse = $t \sqrt{2} = 16 \cdot \frac{\sqrt{2} + 2}{2 + \sqrt{2}}$