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, 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#