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Source: — Data Sufficiency |
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by Night reader » Mon Jan 10, 2011 4:38 pm
GHong14 wrote:Image
triangle ABC - right, isosceles, find hypotenuse C-?

A=B, \_A=\_B=45`, hence C=sqr(2)A (or B)
A+B+C=16+16sqr(2) => form equation x+x+sqr(2)*x= 16+16sqr(2) <=> 2x+sqr(2)*x= 16+16sqr(2) <=>
sqr(2)*x *[sqr(2) + 1] = 16*[sqr(2) +1) <=> sqr(2)*x = 16 <=> x=16/sqr(2)
since hypotenuse = sqr(2)*A (or B) => [16/sqr(2)] * sqr(2) = 16

Answer B
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by anshumishra » Mon Jan 10, 2011 8:07 pm
GHong14 wrote:Image
.
Isosceles Right Angled triangle : Two sides = x, hypotenuse = x√2 = ?
Perimeter = x+x+x√2 = 2x+x√2 = 16+16√2 = 16(√2+1)
=> x√2(√2+1) = 16(√2+1)
=> x√2 = 16

B

I just realized night reader's solution is exactly the same. Basically it is the same.
Thanks
Anshu

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