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heshamelaziry · Mon Oct 12, 2009 11:49 pm

The perimeter of a certain Isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle ?

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xcusemeplz2009 · Tue Oct 13, 2009 1:22 am

h is 16

h^2=2*a^2=>h=asqrt(2)

2a+h=16+16sqrt(2)

2a+sqrt(2)=16+16sqrt(2)

solving a=8*sqrt(2)
h=8*sqrt2*sqrt2=16

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NikolayZ · Tue Oct 13, 2009 1:23 am

Hey there !

Since you have an isosceles triangle, its perimeter = 2x+y. ( 2 sides are x, and the hypotenuse is y)
so the perimeter is
16+16sqrt(2)=2x+y.
Also the hypotenuse equals xsqrt(2), since it is 45 degree isosceles right triangle.

16+16sqrt(2)=2x+xsqrt(2)

x= (16+16*sqrt(2)/(2+sqrt(2))
Then , hypotenuse will be x*sqrt(2)= (16*sqrt(2)+32)/(sqrt(2)+2)=16.

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heshamelaziry · Tue Oct 13, 2009 9:45 am

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NikolayZ · Tue Oct 13, 2009 9:51 am

Sorry mate. "typo happens" (c) =) corrected.

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heshamelaziry · Tue Oct 13, 2009 9:56 am

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NikolayZ · Tue Oct 13, 2009 10:13 am

heshamelaziry, truly there is a flaw in my solvent, because it is not consistent =(

so, right answer to my mind is

2x+x*sqrt(2)=16+16*sqrt(2)
x(2+sqrt(2)=16+16*sqrt(2)
x=(16+16*sqrt(2))/2+sqrt(2)
x*sqrt(2)=(16*sqrt(2)+16*sqrt(2)*sqrt(2))/2+sqrt(2)
==> since hte x*sqrt(2) is the hypotenuse (h)
then
h=(16*sqrt(2)+32)/2+sqrt(2)
factorizing 16 from the numerator we'll get
16(sqrt(2)+2)/(sqrt(2)+2)=16.