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

1 Answer
Jan 6, 2016

The final answer is

Explanation:

PlatinumGMAT, 2015PlatinumGMAT, 2015
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