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triangle isosceles problem

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by Mr_T » Wed Jan 13, 2010 11:26 am
Hi guys,

Here's a problem from the Prep test that I'm having difficulty with. It seems easy, but I do not come up to the right answer.

The perimeter of a certain isosceles right triangle is 16 + 16 squareRoot(2) . What is the length of the hypotenuse of the triangle?

My strategy was to use the perimeter and write it as an algebraic equation and then use pythogoras equation.
So 2x + y = 16 + 16 squareRoot(2)
then x^2 - (y/2)^2 = h^2

x^2 - (y/2)^2 = h^2 => (x-y/2)(x+y/2) = h^2
2x + y = 16 + 16 squareRoot(2) => (x+y/2) = 8 + 8squareRoot(2)

So (8 + 8squareRoot(2)) (8 - 8squareRoot(2)) = h^2

then solving for h, but it doesn't give me the right answer, which is 16.

Help anyone?

Thanks
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by Brent@GMATPrepNow » Wed Jan 13, 2010 12:49 pm
Mr_T wrote:Hi guys,

Here's a problem from the Prep test that I'm having difficulty with. It seems easy, but I do not come up to the right answer.

The perimeter of a certain isosceles right triangle is 16 + 16 squareRoot(2) . What is the length of the hypotenuse of the triangle?

My strategy was to use the perimeter and write it as an algebraic equation and then use pythogoras equation.
So 2x + y = 16 + 16 squareRoot(2)
then x^2 - (y/2)^2 = h^2

x^2 - (y/2)^2 = h^2 => (x-y/2)(x+y/2) = h^2
2x + y = 16 + 16 squareRoot(2) => (x+y/2) = 8 + 8squareRoot(2)

So (8 + 8squareRoot(2)) (8 - 8squareRoot(2)) = h^2

then solving for h, but it doesn't give me the right answer, which is 16.

Help anyone?

Thanks
The important point here is that, in any isosceles right triangle, the sides have length x, x, and (root2)x for some positive value of x.

Note: (root2)x is the length of the hypotenuse (our goal here)

From here, we can see that the perimeter will be x+x+(root2)x

In your question the perimeter is 16 + 16(root2), so we can create the equation x+x+(root2)x = 16 + 16(root2),
We get: x+x+(root2)x = 16 + 16(root2) --> solve for x
Factor out the x to get: (2+ root2)x = 16 + 16(root2)
Solve for x: x = [16 + 16(root2)]/(2+ root2)

The hypotenuse will be (root2)x, so the hypotenuse = (root2) [16 + 16(root2)]/(2+ root2) - a bit of a mess, but not too hard to simplify.

Multiply to get [16(root2)+32]/(2+ root2)
Factor 16 from numerator: [16(root2 + 2)]/(2+ root2)
Simplify: 16
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by Mr_T » Wed Jan 13, 2010 1:24 pm
Cool thanks for the quick reply.

Yeah, I read the question too quickly. I was trying to find the height of a normal isosceles (non-right angled) triangle. Way off!!!

You're right the simplication at the end is a little tricky. I multiplied it by (2-root(2)) / (2-root(2)) to get to the answer.

Thanks,

mr T
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by Stuart@KaplanGMAT » Wed Jan 13, 2010 2:13 pm
Once you recognize that 2x + x(root2) = 16 + 16 squareRoot(2), you can find the hypotenuse very quickly (and with a lot simpler arithmetic) via backsolving.

Of course, we don't have the answer choices here, so it's hard to completely demonstrate how backsolving works (so please always post the choices in the future), but just looking at the correct answer:

If h=16, then x(root2) = 16
x = 16/root2

So the perimiter would be:

16 + 2(16/root2)
= 16 + 32/root2

We then rationalize the denominator of the second term by multiplying by (root2/root2):

= 16 + (32root2)/2
= 16 + 16 root2

Since we've generated the right perimiter, 16 is indeed the hypotenuse.

If we generated a perimiter too big, we'd eliminate 16 and all larger choices; if we generated a perimiter too small, we'd eliminate 16 and all smaller choices. As long as we started with choice B or D, we'd never have to check more than 2 choices.
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