The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2). What is the length of the hypotenuse of the triangle?
A) 8
B) 16
C) 4*2^(1/2)
D) 8*2^(1/2)
E) 16*2^(1/2)
The correct answer is B.
I'm having trouble solving this problem. The perimeter of a right isosceles triangle is base+height+hypotenuse = x+x+x*2^(1/2) = 16 + 16*2^(1/2)
I square both sides to get rid of the square root and end up with (x^2)+(x^2)+(x^2)*2 = 256+256*2
Factoring that out I get x^2(1+1+1*2) = 768
4*x^2=768
x^2=192
x = approximately 14
(2^1/2) = approximately 1.4
14*1.4 = 19.6
(note: 2^1/2 is square root two)
What am I doing wrong?