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Triangle problem

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Junior | Next Rank: 30 Posts
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Triangle problem

by gmatme » Sun Mar 18, 2007 7:52 am
Can someone explain why the answer is B?

The perimeter of a certain isoscles right triangle is 16 + 16sqrt(2). What is the lenght of the hypotenuse of the triangle?

A) 8
b) 16
c) 4sqrt(2)
d) 8sqrt(2)
e) 16sqrt(2)

Isn't that answer should be d? 1:1:sqrt(2) so 8:8:8sqrt(2)?
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by jaskaran_d » Sun Mar 18, 2007 8:06 am
Perimeter of an isosceles right angle triangle is x(sqrt(2) + 2) where x is the lenght of the equal side.

therefore, x(sqrt(2) + 2)= 16 + 16sqrt(2) and x = (16 + 16sqrt(2))/(sqrt(2) + 2)

Lenght of hypotenues is x* sqrt(2)

Hence, length of hypo= {(16 + 16sqrt(2))* sqrt(2)}/(sqrt(2) + 2)

Ans is 16
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Junior | Next Rank: 30 Posts
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by gmatme » Sun Mar 18, 2007 8:21 am
Got it thanks!
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by Cybermusings » Wed Mar 28, 2007 6:55 am
The perimeter of a certain isoscles right triangle is 16 + 16sqrt(2). What is the lenght of the hypotenuse of the triangle?

A) 8
b) 16
c) 4sqrt(2)
d) 8sqrt(2)
e) 16sqrt(2)

Great Solution...Thanks Jaskaran!!!!
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