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
DS Perimeter Question
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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.
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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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.
This is also true for right triangles.
For squares and circles, you only need one fact to know everything.
Greg Michnikov, Founder of GMAT Boost
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GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
View sample questions and tips without signing up, or sign up now for full access.
Also, check out the most useful GMAT Math blog on the internet here.