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.
GMAT Prep Question - Hypotenuse of a right isosles triangle
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scoobydooby
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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
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
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Vemuri
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Thanks scoobydooby. I see where I went wrong. I answered for the side of the isoseles triangle instead of its hypotenuse.scoobydooby wrote: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
