hypotenuse = a each side = a/root(2) Perimeter of rt isosceles triangle = a + 2a/root(2) = a+a*root(2) = 16+16*root(2) so hypotenuse = 16 Option B

The easiest way is to plug the answer choices [ remember to start with smart choices ] 2a^2 = b^2 (b is the hypotenuse) or, a = b/sqrt(2) = b*sqrt(2)/2 -- THIS IS OUR MAIN EXPRESSION. By smart choices I mean -- look at the perimeter.. It has root2 in it. So start with choices where you don't have ROOT(2) because our expression above has already got one. Put the choices and see whether the value satisfies this equation .. 2a+b = 16+16*sqrt(2). this will take 45-60 secs. hth IMO [spoiler](16) [/spoiler] the correct answer.

2a^2=h^2 a=h/sqroot2 2*h/sqroot2+h=16+16*sqroot2 sqroot2*h+h=16+16*sqroot2 h*(sqroot2+1)=16(sqroot2+1) h=16

fbgeoff wrote: IMO D WHTS the OA?

Let the length of the perpendicular sides be a, Hypotenuse = a*Square root(2) = ? 16 + 16*(Square root(2)) = a+a+(a*Square root(2)) 16*(1+(Square root(2)) = a*Square root(2)* (1+(Square root(2)) 16 = a*Square root(2) = Hypotenuse ! IMO : B vaibhavgupta - I reckon you calculated the side, and not the Hypotenuse

Correct answer according to the GMATPrep is B

neelgandham wrote:Let the length of the perpendicular sides be a, Hypotenuse = a*Square root(2) = ? 16 + 16*(Square root(2)) = a+a+(a*Square root(2)) 16*(1+(Square root(2)) = a*Square root(2)* (1+(Square root(2)) 16 = a*Square root(2) = Hypotenuse ! IMO : B vaibhavgupta - I reckon you calculated the side, and not the Hypotenuse Hmm.. will keep this in mind, thanks