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GMATPrep Math Question # 5

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GMATPrep Math Question # 5

by tpetel » Tue Feb 17, 2009 5:03 am
The perimeter of an isosceles triangle is equal to 16 + {16*((2)^.5))} ;

What is the length of the hypotenuse?

The answer is 16... I did not know how to attack this problem.

Any help?
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by DanaJ » Tue Feb 17, 2009 5:27 am
Now, you have an isosceles triangle. Say the length of the cathetus is a. Then, since it is a right triangle, you get that (hypotenuse)^2 = a^2 + a^2 = 2a^2. This in turn means that length of hypotenuse is sqrt(2a^2) = a*sqrt(2). This is an important property of isosceles triangles, and you should remember it, because it saves a lot of time on the real thing
Now you get that the perimeter of your right triangle is a + a + a*sqrt(2) = 2a + a*sqrt(2) = a[2 + sqrt(2)].
This thing is equal to what you provided, meaning that a[2 + sqrt(2)] = 16 + 16sqrt(2) = 8[2 + 2sqrt(2)]. In order for us to get 2 + sqrt(2) in the [...], we need to factor sqrt(2), therefore getting that a[2 + sqrt(2)] = 8sqrt(2)[sqrt(2) + 2]. This means that the length of the cathetus is 8sqrt(2) and the length of the hypotenuse will be sqrt(2)*8sqrt(2) = 8 * 2 = 16.
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