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Senior Manager
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GMAT Prep [#permalink] New post 02 Jun 2009, 22:25
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I have answered this Q but want to understand the way to solve this.
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Senior Manager
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Re: GMAT Prep [#permalink] New post 03 Jun 2009, 01:25
The Question clearly says that the traingle is a right angles isosceles triangle..

hence the sides wud be x,x,root2*x

2x+root2*x = 16+16root2
root2*x*(root2 + 1) = 16root2(root2 + 1)

X = 16

but the question asks hypotenuse.. hence the answer will be 16root2.
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Re: GMAT Prep [#permalink] New post 03 Jun 2009, 09:22
As we know the right angle triangle with two equal side (isosceles) has side ratio in 1:1:root2
so according to question , the perimeter is 16+16root2
so the sides are 8root2:8root2:16
so the hypotenuse is 16
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Re: GMAT Prep [#permalink] New post 03 Jun 2009, 09:28
hence the sides wud be x,x,root2*x

2x+root2*x = 16+16root2
root2*x*(root2 + 1) = 16root2(root2 + 1)

X = 16

but the question asks hypotenuse.. hence the answer will be 16root2.

Neochronic , please check the detail before posting.
you said 2x+root2*x = 16+16root2
in first part 2x is equal to 16, that means x = 8
and in second part root2*x is equal to 16root2, that means x = 16
hows that possible?
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Re: GMAT Prep [#permalink] New post 03 Jun 2009, 13:58
vaivish1723 wrote:
I have answered this Q but want to understand the way to solve this.

Isosceles triangle has :

Base = perpendicular = ab = bc
Hypoteneous = ac

ac^2 = ab^2 + bc^2
ac^2 = ab^2 + ab^2
ac^2 = 2ab^2
ac = ab Sqrt(2)
Perimeter of the triangle = ab+bc+ac
16+16(sqrt2) = ab+ ab + ab Sqrt (2)
16 + 8 (sqrt2) + 8 (sqrt2) = ab Sqrt (2) + ab + ab
ab = bc = 8 (sqrt2)
ac = 16
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Re: GMAT Prep [#permalink] New post 04 Jun 2009, 01:10
my bad..

i was so wrong here..

Thats why i loose so many points uncessarily..

will defnitely try to improve..

2x+root2*x = 16+16root2
root2*x*(root2 + 1) = 16root2(root2 + 1) ---> wrong deduction .. it shud be 16(1+root2)
Re: GMAT Prep   [#permalink] 04 Jun 2009, 01:10
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