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M16-22

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What is the perimeter of rectangle \(ABCD\)?


(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 \(\frac{1}{3}\)
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What is the perimeter of rectangle ABCD?

(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.

Answer: C
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New post 27 Nov 2014, 11:55
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.

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New post 01 Dec 2014, 06:09
JackSparr0w wrote:
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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New post 06 Feb 2016, 06:44
Hi guys,

can somebody explain why "a2+b2=d2" is "a2−32a−32=0"?! I cannot see where the 32 is coming from...

Thanks a lot!!!

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AnikaJu wrote:
Hi guys,

can somebody explain why "a2+b2=d2" is "a2−32a−32=0"?! I cannot see where the 32 is coming from...

Thanks a lot!!!


Substitute b = (a + 2)/3 and d = a + 2 into a^2 + b^2 = d^2 to get a^2 + ((a + 2)/3)^2 = (a + 2)^2 and simplify.
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New post 05 Jul 2016, 22:46
Hi Bunuel,

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?

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New post 06 Jul 2016, 07:33
amariappan wrote:
Hi Bunuel,

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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New post 06 Jul 2016, 07:55
Hi Bunuel,

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.

Please advice.

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New post 06 Jul 2016, 07:58
amariappan wrote:
Hi Bunuel,

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.

Please advice.

Thanks,
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I would suggest to read the whole thread: m16-184091.html#p1706107
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Equation for breath is x^2−12x+4 and the roots of this equation are 6 + 4*sqrt(2) = 11.656 & 6-4*sqrt(2) = 0.344 both of which are indeed positive

However when you substitute and solve for length ( = 3b - 2) , the second root will yield a negative value since ( 3* 0.344 - 2) = -0.968

Since length cannot be negative you can eliminate this value of b
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New post 30 Nov 2016, 12:46
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.

what would u personally suggest ?

Thank in advance

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

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New post 30 Nov 2016, 23:52
gmatdemolisher1234 wrote:
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.

what would u personally suggest ?

Thank in advance


No, we cannot assume that. So, I'd still solve.

Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm
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New post 21 Apr 2017, 10:37
Hi Bunnel,

Considering statements (1) & (2) separately,

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 ?

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New post 21 Apr 2017, 10:41
Kchaudhary wrote:
Hi Bunnel,

Considering statements (1) & (2) separately,

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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New post 21 Apr 2017, 10:48
As it is given the question ABCD is a rectangle and in a rectangle, the diagonal divides the angles into 45-degree.

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New post 21 Apr 2017, 11:01
Agreed, thank you.

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New post 09 Jul 2017, 12:00
Hi Bunuel

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.

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New post 21 Sep 2017, 18:29
Hi Bunuel

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.

Thanks in advance!

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Re: M16-22   [#permalink] 21 Sep 2017, 18:29

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