The perimeter or a certain isosceles gmat prep

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by bajwa2307 » Tue Sep 27, 2011 3:45 pm
Verbal is testing my patience

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by Anurag@Gurome » Tue Sep 27, 2011 7:47 pm
Taniuca wrote:the perimeter of a certain isosceles right triangle is 16+16sqr2. What is the leght of the hypotenuse of the triangle?
Let a be the side of the isosceles right triangle.
16 + 16√2 = 2a + a√2
16 + 16√2 = a(2 + √2)
a = (16 + 16√2)/(2 + √2)
Rationalize the denominator by multiplying by its conjugate (2 - √2), we get
a = (16 + 16√2)(2 - √2)/(2 + √2)(2 - √2) = (16 + 16√2)(2 - √2)/2 = (32 - 16√2 + 32√2 - 32)/2
= 16√2/2
= 8√2
Therefore, hypotenuse = 8√2 x √2 = 8 x 2 = 16
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by Taniuca » Wed Sep 28, 2011 4:38 am
Anarug,
I'm not sure which part of the concept of triangles I'm missing. I do not understand how can a triagle that is iscoceles can have an hypotenuse of asqr2, this should be applicable if you are saying that a triagle has 2 sides that are 45 degrees, while an isosceles could have 30's,60's , and so on...
In advance thanks for ALL your help!