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right anngle isosceles

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Re: right anngle isosceles

by Brent@GMATPrepNow » Sun Feb 22, 2009 8:54 am
AJ2009 wrote:The perimeter of a certain isosceles right triangle is 16+16*sqrroot2. What is the hypotenuse of the triangle?
a. 8
b. 16
c. 4*sqroot2
d. 8*sqroot2
e. 16*sqroot2

help please!
thanks
Here's one approach:

As soon as you see the sqrt2 and the two 16's, you should be thinking that this triangle might be in the 45-45-90 family or right triangles.

The ratio of the sides of a 45-45-90 triangle is 1:1:sqrt2
The perimeter 16+16sqrt2 has an integer component and a root component.
To get 16sqrt2, we could multiply each side of the base 1:1:sqrt2 triangle by 16, but this would make the sides 16:16:16sqrt2 and the perimeter 32+16sqrt (nope)

At this point, it helps to create a second "base" triangle by multiplying each side of the 1:1:sqrt2 triangle by sqrt 2.
We get sides sqrt2:sqrt2:2 (this is a useful ratio to keep in your pocket)

Now we might see that if we multiply each side by 8 we get the desired dimensions. We get 8sqrt2:8sqrt2:16.
The perimeter is 16+16sqrt2, and so the hypotenuse is 16
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by bmlaud » Sun Feb 22, 2009 9:13 am
For a isosceles right triangle with sides a and b

2a+b = 16+16sqrt2 ------1

2a^2 = b^2 or b = a sqrt2, putting the value in 1

2a + a sqrt2 = 16(1+sqrt2)
or a sqrt2(1+ sqrt2) = 16 (1+sqrt2)

or a = 16/sqrt2 = 8 sqrt2 ..D
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by bravotalks » Sun Feb 22, 2009 11:06 am
bmlaud, your approach seems to be correct until the last step.

You deduced a = 8 sqrt2
But the question is asking for the hypotenuse, which in your case is variable 'b' (and not variable 'a')

i.e., b = a * sqrt2 = 8 * sqrt2 * sqrt 2 = 16.
i.e., Ans B

OA?
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by daimaomee » Sun Feb 22, 2009 7:52 pm
Let x be the length of the isosceles sides, therefore the hypotenuse = x.sqrt(2)

Since 2x + x.sqrt(2) = 16 + 16sqrt(2)

If we compare the rational(2x) with the rational (16) and the irrational (x.sqrt(2)) with the irrational (16sqrt(2)), we cannot get a solution. Therefore we try to compare the rationals with the irrationals, i.e.

2x = 16sqrt(2), and
x.sqrt(2) = 16

From both equations, we obtain x = 8sqrt(2). Since the hypotenuse is x.sqrt(2) = 16, therefore the answer should be B.
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by AJ2009 » Mon Feb 23, 2009 12:17 am
OA: B
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