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Isosceles Triangles

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nkhalaieff
 Post subject: Isosceles Triangles
 Post Posted: Wed Feb 09, 2011 2:03 am 
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Students


Posts: 3
An isosceles triangle has a perimeter of 16+16(sqrt2). What is the length of the hypotenuse?

8
16
4(sqrt2)
8(sqrt2)
16(sqrt2)

I keep getting 8(sqrt2). The length of the sides are 8, then the hypotenuse must be 8(sqrt2), because of the 1,1, (sqrt2) relationship. anyone?
Correct answer is 16.
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narendra47
 Post subject: Re: Isosceles Triangles
 Post Posted: Wed Feb 09, 2011 9:03 am 
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Students


Posts: 1
The question is quite tricky...
you can proceed like this way

Let us consider the length of a equal side is X
then the length of hypotenuse would be X(sqrt(2))

Now, as given the perimeter is 16+16(sqrt(2))
i,e.
X+X+X(sqrt(2))=16+16(sqrt(2))
=> 2X+X(sqrt(2))=16[1+(sqrt(2))]
=> X(sqrt(2))[1+(sqrt(2))]=16[1+(sqrt(2))]
=> X=16/(sqrt(2))

Hence hypotenuse is 16.
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RonPurewal
 Post subject: Re: Isosceles Triangles
 Post Posted: Wed Feb 09, 2011 9:13 am 
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ManhattanGMAT Staff


Posts: 7146
please search the forum, people

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