What is the value of the hypotenuse of an isosceles triangle with a perimeter equal to $16 + 16 \sqrt{2}$ $16 + 16sqrt2$ ?

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Answer 1 · 2015-11-14T16:43:32

16

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

$2 p = t + t + t \sqrt{2} = 16 + 16 \sqrt{2}$ $2p = t + t + t sqrt2 = 16 + 16sqrt 2$

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

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

What is the value of the hypotenuse of an isosceles triangle with a perimeter equal to 16+16√216 + 16sqrt2?