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.
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
RedeemGMAT Prep Question - Hypotenuse of a right isosles triangle
This topic has expert replies
Source: Beat The GMAT — Problem Solving |
-
scoobydooby
- Legendary Member
- Posts: 1035
- Joined: Wed Aug 27, 2008 10:56 pm
- Thanked: 104 times
- Followed by:1 members
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
-
Vemuri
- Legendary Member
- Posts: 682
- Joined: Fri Jan 16, 2009 2:40 am
- Thanked: 32 times
- Followed by:1 members
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