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Geometry

by nkaur » Tue May 24, 2011 6:26 am
The perimeter of a certain isoceles triange is 16+ 16 sqrt 2. What is the length of the hypothenuse?

(1) 8
(2) 16
(3) 4 sqrt 2
(4) 8 sqrt 2
(5) 16 sqrt 2

I understand that the answer is 16 sqrt 2. But couldnt it be 16 too? Is there a certain rule for why it should be answer choice E?

Thanks
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by Frankenstein » Tue May 24, 2011 6:38 am
nkaur wrote:The perimeter of a certain isoceles triange is 16+ 16 sqrt 2. What is the length of the hypothenuse?

(1) 8
(2) 16
(3) 4 sqrt 2
(4) 8 sqrt 2
(5) 16 sqrt 2

I understand that the answer is 16 sqrt 2. But couldnt it be 16 too? Is there a certain rule for why it should be answer choice E?

Thanks
Hi,
As you have stated that the triangle is isosceles and asked about hypotenuse, I can assume that it is an isosceles right-angled triangle
.
Let the sides be a,a,asqrt2. Perimeter is a(2+sqrt2) = 16(2+sqrt2) => a=8sqrt2.
So hypotenuse is asqrt2 = 16.
Answer B Can you check the OA again?

Cheers!
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by amar66 » Thu May 26, 2011 10:11 am
Frankenstein is correct. Even, I got the same result.
Answer must be 16.

Can you pls point the source of this question?
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by Brian@VeritasPrep » Fri May 27, 2011 3:08 pm
Great question - and the answer is definitely 16.

This may very well be my FAVORITE GMAT geometry device. The authors of the test know that you have these side ratio rules memorized cold:

45-45-90
x - x - x(sqrt2)

30-60-90
x - x(sqrt3) - 2x

And they know how you think. You want to see that radical sign on the hypotenuse of the 45-45-90 triangle and on the 60-degree side of the 30-60-90. So what do they do? They love to make x = to a multiple of the square root of 2 (or of 3) so that the radical shows up where you'd least expect it. So please take note - they can put the radical wherever they want...it's not that the square root of 2 must go on the hypotenuse side; it's just that the hypotenuse must be equal to x * sqrt 2.

There are only two distinct terms here: an integer term (16) and a square root term (16 sqrt 2). And one of them has to be divided by 2 to account for the two sides that are x each. If we divide the 16, it doesn't work:

8, 8, 16 sqrt 2 --> the third term should be 8 sqrt 2

8sqrt 2, 8 sqrt 2, 16 --> this works; the third side would be x * sqrt 2, or 8(sqrt 2)(sqrt 2), which gives you 8*2 = 16.

So be forewarned - the RATIO must be satisfied on these triangle side ratios, but that doesn't necessarily mean that the radical will correspond to the side on which the ratio tells you there's a radical! If x has a radical, then it will multiply out to an integer on the hypotenuse side.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
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