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(1) The longer side of the rectangle is 2 meters shorter than its diagonal. Given: \(a=d-2\), 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 \(\frac{1}{3}\). Given: \(\frac{b}{d}=\frac{1}{3}\), so \(d=3b\), where \(b\) is the length of the shorter side. Not sufficient on its own to get the perimeter.

(1)+(2) We have that \(a=d-2\) and \(d=3b\), so \(a=3b-2\). Also from Pythagorean theorem \(a^2+b^2=d^2\), hence \(a^2 -32a-32=0\). We can get the value of \(a\) (we'll have two solutions for \(a\): one positive and another negative, hence not valid ) and therefore the value of \(b\), so the value of the perimeter too. Sufficient.

When I factor the above I get (b^2)-12b+4=0; is this correct? If so wouldn't we get (b+neg)(b+neg), yielding two positive roots?

If not can you please provide the steps to factor?

Thanks.

Yes. But the point is that for one of the positive values of b, a, which is 3b - 2, would become negative, hence not valid. Edited the solution.
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Can you please validate my logic. Let breath be "X" then the diagonal is 3X and the Longest side is 3X-2. Now X^2 + (3X-2)^2 = 9(X^2)

Solving this equation we get X^2 - 12X+ 4 =0; using quadratic equation formula for roots ( -b+- (b^2-4ac) )/ 2a . Now we get the two roots as -6 (+/-) 4(sqrt 2) . But in both cases the roots are negative. Where am I making a mistake?

Can you please validate my logic. Let breath be "X" then the diagonal is 3X and the Longest side is 3X-2. Now X^2 + (3X-2)^2 = 9(X^2)

Solving this equation we get X^2 - 12X+ 4 =0; using quadratic equation formula for roots ( -b+- (b^2-4ac) )/ 2a . Now we get the two roots as -6 (+/-) 4(sqrt 2) . But in both cases the roots are negative. Where am I making a mistake?

Thanks, Arun

You should get \(6+4\sqrt{2}\) and \(6-4\sqrt{2}\) as the roots.
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Thanks for the prompt reply. I beg to differ with the solution. Even if the the roots are 6+4√2 and 6−4√2, then we have 2 positive roots meaning 2 values are possible for side x. Then we cannot determine the perimeter of the rectangle.

Thanks for the prompt reply. I beg to differ with the solution. Even if the the roots are 6+4√2 and 6−4√2, then we have 2 positive roots meaning 2 values are possible for side x. Then we cannot determine the perimeter of the rectangle.

when we get an equation like a^2 − 32a −32=0 , would you recommend solving the equation during the official test.. isnt it really time consuming.. Since GMAT isnt a calulation based exam.. Can we assume that one value will be positive and the other neg.

what would u personally suggest ?

Thank in advance

Last edited by gmatdemolisher1234 on 01 Dec 2016, 03:30, edited 1 time in total.

Bunuel when we get an equation like a^2 − 32a −32=0 , would you recommend solving the equation during the official test.. isnt it really time consuming.. Since GMAT isnt a calulation based exam.. Can we assume that one value will be positive and the other neg.

Can't be both the statements will be sufficient on its own if I consider diagonal and one side of a rectangle as a right angle triangle with angles 45-90-45 ?

Can't be both the statements will be sufficient on its own if I consider diagonal and one side of a rectangle as a right angle triangle with angles 45-90-45 ?

Why would you assume that? We don't know whether it's true.
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Is there a way to determine the sign of roots of a quadratic equation without solving the equation? By reading the above thread, it seems that there is not a simple way to determine the sign. However, I just wanted to confirm. Please reply.

Since you say it's better to solve (when we get an equation like a^2 − 32a −32=0, as we cannot assume that one value will be positive and the other neg.) Would you mind posting a complete resolution to this problem? i.e. including your calculations to get to the positive and negative (invalid) solution.

I am asking this as I think you might be doing the calculus in a much more straightforward way than I am.