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geometry: triangle question

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by deec4 » Mon Jul 20, 2009 2:57 pm
this one is killing me because it sounds so easy, but I don't get it...can someone help?

The perimeter of a certain isosceles right triangle is 16 + 16 radical 2. What is the hypothenuse of the triangle.

Answer: 16

would any one care to explain?

Thanks much!
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by truplayer256 » Mon Jul 20, 2009 3:06 pm
The 3 sides of the isosceles right triangle are s, s, and s*sqrt(2).

2s+s*sqrt(2)=16+16*sqrt(2)

s(2+sqrt(2))=16+16*sqrt(2)

s=16+16*sqrt(2)/2+sqrt(2)

Multiply the numerator and denominator by 2-sqrt(2) to get:

s=16*sqrt(2)/2=8*sqrt(2)

Hypotenuse= s*sqrt(2)=8*sqrt(2)*sqrt(2)=8*2=16.
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by bfman » Mon Jul 20, 2009 6:16 pm
Can you please explain how you assumed hypotenuse to be s*sqrt(2)?

Bcuz your equation is 2s + s*sqrt(2).
So, 2s would be the 2 equal sides of an Isosceles triangle, and s*sqrt(2) be the hypotenuse, correct? So why s*sqrt(2)?

Thanks.
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by mehravikas » Mon Jul 20, 2009 11:07 pm
Assume two equal sides of the triangle to be x and hypotenuse to be y

y^2 = X^2 + X^2

therefore, y = x sqrt(2)
bfman wrote:Can you please explain how you assumed hypotenuse to be s*sqrt(2)?

Bcuz your equation is 2s + s*sqrt(2).
So, 2s would be the 2 equal sides of an Isosceles triangle, and s*sqrt(2) be the hypotenuse, correct? So why s*sqrt(2)?

Thanks.
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