Question

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

Answer 1

The final answer is

Explanation:

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

`:.xsqrt(2)= 16`