GMAT Prep Question - Hypotenuse of a right isosles triangle

[Vemuri]

Hi,

I am sure this question was posted earlier. None of search queries resulted in the question showing up. So, I am posting it. Please explain your answers.

[Feep]

I find it hard to believe this hasn't been answered before, but...

We know that an isosceles right triangle is, by definition, a 45-45-90 right triangle. Therefore, the hypotenuse will be sqrt(2) times the length of either of the sides. If the length of a side is s, the perimeter is 2s + s*sqrt(2).

2s + s*sqrt2 = 16 + 16*sqrt(2). Note that the "obvious" answer, s = 16, does not fit this equation. Instead, we see that the hypotenuse is actually 16, making the sides 16 / sqrt(2) = 8*sqrt(2), which DOES fit the equation. (Alternatively, you just could just solve the equation with algebra if you don't "see" this, but it'll take longer.)

So, the hypotenuse is 16.

[scoobydooby]

let the 2 equal sides be a and the hypotanuse be rt 2a

perimeter: 2a+rt 2a=16+16*rt 2
=> rt 2a (rt 2+1)=16(1+rt 2)
=> rt 2a=16
=>a=16/rt 2

hypotanuse= rt 2a=rt 2*(16/rt 2)=16

[Vemuri]

let the 2 equal sides be a and the hypotanuse be rt 2a

perimeter: 2a+rt 2a=16+16*rt 2
=> rt 2a (rt 2+1)=16(1+rt 2)
=> rt 2a=16
=>a=16/rt 2

hypotanuse= rt 2a=rt 2*(16/rt 2)=16

Thanks scoobydooby. I see where I went wrong. I answered for the side of the isoseles triangle instead of its hypotenuse.