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Area of the triangle

Expert replies
by gmattesttaker2 » Sun Dec 01, 2013 1:31 pm
Hello,

Can you please assist with this:

What is the area of the triangular region above if its sides have integer length?

(1) PQ + QR = 21
(2) PR = 15

OA: D

Thanks,
Sri
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Source: — Data Sufficiency |

by [email protected] » Sun Dec 01, 2013 7:50 pm
Hi Sri,

With a right triangle, and no restrictions, there would be limitless possibilities to consider. However, the prompt tells us that ALL SIDES are INTEGERS. There aren't that many possibilities for the Pythagorean Theorem in which all 3 numbers are integers. The two Facts deal with relatively small numbers, so I'm going to list out the simple possibilities for easy reference:

3/4/5 and the multiples
6/8/10
9/12/15
12/16/20
etc.

5/12/13 and the multiples
10/24/26
etc.

The prompt asks us for the area of the triangle, so we'll need the base and the height. We're told that all sides are integers.

Fact 1: PQ + QR = 21

We need a right triangle, with all integer sides, AND the two short sides add up to 21.

There's ONLY ONE possibility: 9/12/15
Fact 1 is SUFFICIENT

Fact 2: PR = 15

We need a triangle with a hypotenuse of 15 and the other 2 sides are also integers.

There's ONLY ONE possibility: 9/12/15
Fact 2 is SUFFICIENT

Final Answer: D

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by Matt@VeritasPrep » Sun Dec 01, 2013 11:12 pm
Suppose PQ = a, QR = b, and PR = c.

We know that a² + b² = c², and we know that a, b, and c are integers. We also know that there aren't very many Pythagorean triples (right triangles for which all the sides are integers), and the first three are the 3-4-5 (and its multiples), the 5-12-13 (and its multiples), and the 7-24-25 (and its multiples). (These seem to be the only three that the GMAT assumes that we know.)

S1:: a + b = 21

We can't have a 7-24-25 right triangle, as the sides are too long. We can't have a 5-12-13 (or its "first" multiple, a 10-24-26), as the legs don't sum to 21. So we must have a 9-12-15. SUFFICIENT.

S2:: c = 15

We have to have a multiple of 3-4-5, so our triangle is a 9-12-15. SUFFICIENT.
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