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DS Perimeter Question

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DS Perimeter Question

by tonebeeze » Wed Aug 10, 2011 4:44 pm
If m is the perimeter of rectangle N, what is the value of m?

(1) The area of rectangle N is 60.

(2) Each diagonal of rectangle N has length 13.

I got the problem correct, but can someone write out the algebra. I am trying to tighten my skills in this area. thanks.

OA = C
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by goalevan » Wed Aug 10, 2011 6:56 pm
This has a pretty tough algebraic solution, but see the shortcut at the bottom:

m = 2(L + W), "what is m?"

Statement 1) LW = 60, so L = 60/W and W = 60/L

but it could be that (L,W) is (6,10) and 2(L + W) = 2(16) = 32
or it could be that (L,W) is (5,12) and 2(L + W) = 2(17) = 34
or even that (L,W) is (100000,0.0006) and 2(L + W) = 2(100000.0006) > 200000

Insufficient.

Statement 2) through the pythagorean theorem, L^2 + W^2 = 13^2 = 169, but it could be that (L,W) is (0,13) and 2(L + W) is 2(13) = 26, or that (L,W) is (12,5) and 2(L + W) is 2(17) = 34.

Insufficient.

Combined) we know that L = 60/W and that L^2 + W^2 = 169, so substituting we get (60/W)^2 + W^2 = 169.

So multiplying both sides by W^2 we have (60^2) + W^4 = 169W^2 and W^4 - 169W^2 + 60^2 = 0 or x^2 - 169x + 60^2 = 0 where x = W^2.

We can factor this as (W^2 - 144)(W^2 - 25) = 0, so W^2 = 144 or W^2 = 25 and W = {-12,-5,5,12}. The positive values that make sense in this problem are W = 5 and W = 12.

With a W of 5 or 12, the perimeter would remain the same at 2[W + (60/W)] = 2[5 + (60/5)] = 34 = 2[12 + (60/12)] = 34

Thus the statements combined are sufficient. The hard part is breaking down the quadratic that results from combining the statements from W^4 - 169W^2 + 60^2 to (W^2 - 144)(W^2 - 25) = 0.


You could also solve this with a shortcut:

m^2 = [2(L + W)]^2 = 4(L^2 + W^2 + 2LW).

With statement 1 we are given that LW = 60, so 2LW = 120.
With statement 2 we are given that L^2 + W^2 = 169.

So m^2 = 4(169 + 120) = 4(289) = (2^2)(17^2) = (34)^2, and m = {-34,34}, so the perimeter can be positive 34.
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by gmatboost » Thu Aug 11, 2011 7:56 am
In general, once you know two facts about a rectangle, you know everything about it. When you only know one fact, you don't know anything else.

This is also true for right triangles.

For squares and circles, you only need one fact to know everything.
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