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 | Wrong Answer in GMAT Official Guide 11? |  |
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avishal
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Posted: Wed Nov 07, 2007 7:04 am |
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Hi
I came across the following DS question in the 11th OG (i think last question of the initial DS diagnostics test). To paraphrase, the question read as follows:
what's the perimeter of the rectangle?
1) the diagonal is 10 inches.
2) the product of two sides is 48.
I believe the answer should be A, but the OG says it's C. Here's my logic:
If the diagonal is 10, it means the other two sides (of the right triangle) necessarily have to be 6 and 8 (6-8-10 right triangle), and that implies the perimeter is 2*6+2*8=12+16=28.
The OG says that we can't determine the sides of the triangle simply by the diagonal size. It then goes on to use the factors of 48 (from choice 2 of the question) to determine that the sides are 6 and 8.
My question really is - is it possible to have two right triangles with same diagonal lengths but different set of other dimensions? While it's possible to have 100 = 10 + 90, since 10 or 90 aren't perfect squares, only 36 + 64 is the right choice (i.e. sides are 6 and 8).
Am I right? If so, is there a mathematical proof to back this statement?
Thanks,
vishal
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 | Re: Wrong Answer in GMAT Official Guide 11? |  |
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shaji
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Posted: Wed Nov 07, 2007 11:33 pm |
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The OG is correct. The sides need not be perfect squares. Real numbers too are possibilities.
| avishal wrote: |
Hi
I came across the following DS question in the 11th OG (i think last question of the initial DS diagnostics test). To paraphrase, the question read as follows:
what's the perimeter of the rectangle?
1) the diagonal is 10 inches.
2) the product of two sides is 48.
I believe the answer should be A, but the OG says it's C. Here's my logic:
If the diagonal is 10, it means the other two sides (of the right triangle) necessarily have to be 6 and 8 (6-8-10 right triangle), and that implies the perimeter is 2*6+2*8=12+16=28.
The OG says that we can't determine the sides of the triangle simply by the diagonal size. It then goes on to use the factors of 48 (from choice 2 of the question) to determine that the sides are 6 and 8.
My question really is - is it possible to have two right triangles with same diagonal lengths but different set of other dimensions? While it's possible to have 100 = 10 + 90, since 10 or 90 aren't perfect squares, only 36 + 64 is the right choice (i.e. sides are 6 and 8).
Am I right? If so, is there a mathematical proof to back this statement?
Thanks,
vishal |
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 | you're right shaji |  |
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avishal
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Posted: Thu Nov 08, 2007 1:21 am |
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You are correct actually, I verified this [may sound dumb] :)
created two right triangles with same diagonal length (obviously, the other two angles are not fixed).
moral of the story (i feel so stupid that i didn't see this before) - if two sides of a right triangle follow 3-4-5 rule, you can get the third side but you can't assume the 3 sides to follow 3-4-5 based on one side alone (be it the diagonal)
thanks,
vishal
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srikant
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Posted: Sat Nov 10, 2007 5:20 pm |
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You can also think of it as a circle with diagonal of rectangle as its diameter.
The angle formed by diameter at the circumference is always right angle but the possibilities (different side lengths) are countless!
Therefore there is no unique solution, to answer your question.
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Stacey Koprince
MGMAT STAFF
| Joined: 06 Mar 2007 |
| Posts: 2644 |
Location: San Francisco
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Posted: Mon Nov 12, 2007 8:32 pm |
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I'm glad that you checked it out yourself, vishal, and figured it out - we always remember lessons better that way. And now you know why it works the way it does! (Or, technically, why it doesn't work... :)
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