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1. The longer side of the rectangle is 2 meters shorter than its diagonal 2. The ratio of the shorter side of the rectangle to its diagonal is \(\frac{1}{3}\)

Hey, Its all a matter of working organized. 1 - INS. we can say that D(diagonal) worth 20 and than L(length) = 18. Or D=1000, L=998. of course - not the same perimeter. 2 - INS. again - D=99, W(width) = 33. Or D=999, W=333

1+2 Lets say \(D=3x\). so from statement 1 we know that the longer side (L)=3x-2 from statement 2 - \(W=3x/3\)>>>>>\(w=x\)

Using Pythagoras - \(X^2+(3x-2)^2=(3x)^2\) Simplifying will give us - \(x^2-12x+4=0\)

So \(12+-squreroot128\) will always give us one positive and one negative solutions (128<144) and we know it will give us a solution.

hope I did no mistakes in the math. was doing it quickly. good luck.
_________________

1. The longer side of the rectangle is 2 meters shorter than its diagonal 2. The ratio of the shorter side of the rectangle to its diagonal is \(\frac{1}{3}\)

(1) Let a = longer side, b= shorter side a = d -2 d^2 = a^2 + b^2 (a-2)^2 = a^2 + b^2 b^2 = 4 - 4a Not sufficient

1. The longer side of the rectangle is 2 meters shorter than its diagonal 2. The ratio of the shorter side of the rectangle to its diagonal is \(\frac{1}{3}\)

Hey, Its all a matter of working organized. 1 - INS. we can say that D(diagonal) worth 20 and than L(length) = 18. Or D=1000, L=998. of course - not the same perimeter. 2 - INS. again - D=99, W(width) = 33. Or D=999, W=333

1+2 Lets say \(D=3x\). so from statement 1 we know that the longer side (L)=3x-2 from statement 2 - \(W=3x/3\)>>>>>\(w=x\)

Using Pythagoras - \(X^2+(3x-2)^2=(3x)^2\) Simplifying will give us - \(x^2-12x+4=0\)

So \(12+-squreroot128\) will always give us one positive and one negative solutions (128<144) and we know it will give us a solution.

hope I did no mistakes in the math. was doing it quickly. good luck.

Hi!

just did this Problem using the gmatclub test bank and i don't seem to understand it. the square root of 128 will always be lower than 12, hence, we should get two solutions and hence the answer should be E. where's my mistake?

just did this Problem using the gmatclub test bank and i don't seem to understand it. the square root of 128 will always be lower than 12, hence, we should get two solutions and hence the answer should be E. where's my mistake?

Length is always positive, that's the hidden clue. So, positive \sqrt{128} is unique.
_________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks!

1. The longer side of the rectangle is 2 meters shorter than its diagonal 2. The ratio of the shorter side of the rectangle to its diagonal is \(\frac{1}{3}\)

Hey, Its all a matter of working organized. 1 - INS. we can say that D(diagonal) worth 20 and than L(length) = 18. Or D=1000, L=998. of course - not the same perimeter. 2 - INS. again - D=99, W(width) = 33. Or D=999, W=333

1+2 Lets say \(D=3x\). so from statement 1 we know that the longer side (L)=3x-2 from statement 2 - \(W=3x/3\)>>>>>\(w=x\)

Using Pythagoras - \(X^2+(3x-2)^2=(3x)^2\) Simplifying will give us - \(x^2-12x+4=0\)

So \(12+-squreroot128\) will always give us one positive and one negative solutions (128<144) and we know it will give us a solution.

hope I did no mistakes in the math. was doing it quickly. good luck.

How are you jumping from: "Simplifying will give us - \(x^2-12x+4=0\)"

just did this Problem using the gmatclub test bank and i don't seem to understand it. the square root of 128 will always be lower than 12, hence, we should get two solutions and hence the answer should be E. where's my mistake?

Length is always positive, that's the hidden clue. So, positive \sqrt{128} is unique.

I know, but 12 plus or minus the square root will always be positive, hence we have two different solutions which are both positive... i'm confused.. could you elaborate please why you think there's only one solution?

ooohhh... got it :D :D because 3x-2 will become negative for one of the roots :D

(1) The longer side of the rectangle is 2 meters shorter than its diagonal --> \(a=d-2\), where \(a\) is the length of the longer side and \(d\) is the length of the diagonal. Not sufficient on its own to get the perimeter.

(2) The ratio of the shorter side of the rectangle to its diagonal is \(\frac{1}{3}\) --> \(\frac{b}{d}=\frac{1}{3}\) --> \(d=3b\), where \(b\) is the length of the shorter side. Not sufficient on its own to get the perimeter.

(1)+(2) Given that \(a=d-2\) and \(d=3b\) --> \(a=3b-2\). Also from Pythagorean theorem \(a^2+b^2=d^2\) --> \((3b-2)^2+b^2=(3b)^2\). We can get the value of \(b\) (we'll have two solutions for \(b\): one positive and another negative, hence not valid ) and therefore the value of \(a\), so the value of the perimeter too. Sufficient.

We're asked for the perimeter of rectangle ABCD. To answer this question, we'll need to figure out the Length (L) and Width (W) of the rectangle. This prompt involves a great 'System Algebra' shortcut that you can use to avoid doing a lot of math.

1) The longer side of the rectangle is 2 meters shorter than its diagonal

With the information in Fact 1, we can create 2 equations (with the diagonal represented by D): L = D - 2 L^2 + W^2 = D^2

Unfortunately, with 3 unknowns, but only 2 equations, we cannot solve for any of the variables, so we cannot figure out the perimeter. Fact 1 is INSUFFICIENT

2) The ratio of the shorter side to the diagonal is 1/3

With the information in Fact 2, we can create 2 equations (one of which is the same as we created in Fact 1): W/D = 1/3....... D = 3W L^2 + W^2 = D2

Again, with 3 unknowns, but only 2 equations, we cannot solve for any of the variables, so we cannot figure out the perimeter. Fact 2 is INSUFFICIENT

Combined, we know that we're dealing with a rectangle, so the variables can ONLY be POSITIVE numbers. As such, even though we're dealing with 'squared terms', the negative answers are not possible here. We have 3 variables and 3 unique equations, so we CAN solve for all 3 variables - and there will be just one solution. Combined, SUFFICIENT