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Perimeter

by sarahw_gmat » Thu Dec 30, 2010 1:26 pm
Q. The perimeter of a certain isosceles right triangle is 16+16* root 2. What is the length of the hypotenues of the triangle?
Ans : 16

This seems to be a simple problem, but I am getting 8*root 2 as answer. Please help.

Sol: let x be the length.

2x + x root 2 = 16 + 16 root 2
x = 8
therefore, hypotenues = 8 root 2.
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by anshumishra » Thu Dec 30, 2010 1:32 pm
sarahw_gmat wrote:Q. The perimeter of a certain isosceles right triangle is 16+16* root 2. What is the length of the hypotenues of the triangle?
Ans : 16

This seems to be a simple problem, but I am getting 8*root 2 as answer. Please help.

Sol: let x be the length.

2x + x root 2 = 16 + 16 root 2
x = 8
therefore, hypotenues = 8 root 2.
Equal sides = a, hypotenues = b =a*sqrt2 = ?

2a+b = 16(1+sqrt 2)
=> 2a +a*sqrt2 = 16(1+sqrt2)
=>a*sqrt2(sqrt2+1) = 16*(1+sqrt2)
=> a*sqrt2 = 16
Thanks
Anshu

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by Geva@EconomistGMAT » Thu Dec 30, 2010 10:19 pm
sarahw_gmat wrote:Q. The perimeter of a certain isosceles right triangle is 16+16* root 2. What is the length of the hypotenues of the triangle?
Ans : 16

This seems to be a simple problem, but I am getting 8*root 2 as answer. Please help.

Sol: let x be the length.

2x + x root 2 = 16 + 16 root 2
x = 8
therefore, hypotenues = 8 root 2.
x is not equal to 8, and no amount of wishful thinking would make it so: 2x = 16 will make x=8, but that doesn't jive with xroot 2= 16 root 2.

i know the problem - conceptually, it is best solved by reverse Plug in the answer choices back into the question, keeping in mind that the answer choices represent the hypotenuse: the legs will each be hypo / root 2.
The answer is hypo=16:
each leg is 16/root 2:
multiply and divide by root 2 to get rid of the fraction in the denominator, without changing the fraction. You get:
16 root 2 / root 2 root 2
16 root 2 / 2
8 root 2.
Thus, if each leg is 8 root 2, then the perimeter is 8 root 2 + 8 root 2 + 16 (our chosen hypotenuse) - or 16 root 2 + 16.
Geva
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