Question

Help me please with the question.​

Answer 1

Answer:

$\rm \sqrt{32}$

Step-by-step explanation:

In an isosceles right triangle both base and height are equal,

so, area of the triangle is,

$\rm \dfrac{1}{2}×s×s$

where s is equal sides,

$\rm 8=\dfrac{s^2}{2}$

$\rm s^2=8×2$

$\rm s = \sqrt{16}$

$\rm s = 4$

by Pythagoras theorem,

$\rm hypotenuse= \sqrt{4^2+4^2}$

$\rm = \sqrt{32}$