any other approach for this triangle??

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arpan307: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Tue Apr 26, 2011 3:28 am

by arpan307 » Thu Jun 23, 2011 6:56 pm

Q. A certain right triangle has sides of length x, y and z, where x<y<z. If the area is 1, which of the following indicates all possible values of y?

(A) y > root(2)
(B) root(3) / 2 < y < root(2)
(C) root(2) / 3 < y < root(3) / 2
(D) root(3) / 4 < y < root(2) / 3
(E) y< root(3) / 4

The OA is A. I understood the explanation, but its all about adjusting till u reach to some ans option. any other approach people?

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Source: Beat The GMAT — Problem Solving | Thu Jun 23, 2011 6:56 pm

Ashley@VeritasPrep: GMAT Instructor; Posts: 199; Joined: Tue May 17, 2011 6:06 amFollowed by: 121 members; Location: Cambridge, MA; Thanked: 192 times; GMAT Score:780

by Ashley@VeritasPrep » Thu Jun 23, 2011 7:59 pm

I don't know how different this will be from the explanation you've already seen, but here's how I'd think about it. Since this is a right triangle with x<y<z, we know that x and y must be the legs (and z the hypotenuse). z is therefore irrelevant to the area of the triangle -- I can use one of the legs as the base and the other as the height. So the area, 1, = (1/2)(xy), so xy = 2.

Now I apply the knowledge that x is less than y. Whenever you have two things you will multiply together to yield some product, and you know that one of them is less than the other, the lesser one MUST BE less than the square root of the product, and the greater one must be greater than that square root, because since sqrt(product) * sqrt(product) = product, the smaller one could only go down from there and the larger could only go up. Since here we have xy = 2, if x and y were allowed to be equal, they'd both be sqrt(2). But since y must be actively greater than x, we can conclude that y must be > sqrt(2).

Ashley Newman-Owens
GMAT Instructor
Veritas Prep

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