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Math Expert V
Joined: 02 Sep 2009
Posts: 55623

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Simba9 wrote:
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.

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Manager  B
Joined: 23 Jun 2016
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Can someone correct where am i going wrong?

let diameter = d
long side: d-2
short side: d /3

using Pythagorean theorem
d^2 = (d-2)^2 + (d/3)^2 or
d^2 = d^2 + 4 - 4d + d^2/9 subtracting d^2 on both sides
0 = 4 - 4d + d^2/9
multiplying both sides by 9
d^2 - 36d - 36 = 0

this is a different equation that derived in the original expert ans. What am I missing here?
Math Expert V
Joined: 02 Sep 2009
Posts: 55623

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sevenplusplus wrote:
Can someone correct where am i going wrong?

let diameter = d
long side: d-2
short side: d /3

using Pythagorean theorem
d^2 = (d-2)^2 + (d/3)^2 or
d^2 = d^2 + 4 - 4d + d^2/9 subtracting d^2 on both sides
0 = 4 - 4d + d^2/9
multiplying both sides by 9
d^2 - 36d - 36 = 0

this is a different equation that derived in the original expert ans. What am I missing here?

You have: d^2 = (d-2)^2 + (d/3)^2
Original solution: a^2 -32a-32=0

Do you see the difference? You have d, while the OE has a (longer side).
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Intern  B
Joined: 23 Jan 2018
Posts: 3

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We know the ratio of the length of the right triangle (1: \sqrt{3}: 2) and since the larger side is 2 meters less than the diagonal.
Can't we just assume that d is equal to 2x = x\sqrt{3} + 2 ?
And so we just need statement A to figure out the perimeter ?

Thank you
Math Expert V
Joined: 02 Sep 2009
Posts: 55623

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alvarezy wrote:
We know the ratio of the length of the right triangle (1: \sqrt{3}: 2) and since the larger side is 2 meters less than the diagonal.
Can't we just assume that d is equal to 2x = x\sqrt{3} + 2 ?
And so we just need statement A to figure out the perimeter ?

Thank you

Do we know that the diagonal will divide the rectangle into 30-60-90 triangles? No. Why it cannot be say 1-89-90? Or 7.5-82.5-90?
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Intern  B
Joined: 23 Jan 2018
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Understood. Very clear. Thank you for the quick reply.

Sent from my iPhone using GMAT Club Forum
Manager  G
Joined: 01 Aug 2017
Posts: 190
Location: India
GMAT 1: 500 Q47 V15 GPA: 3.4
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Simba9 wrote:
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.

Here you go!! I know it's a bit late but it might be useful for others as well.

$$a^2 -32a-32=0$$. You can't solve this using factorization. So we should use other method

$$D = \sqrt{b^2-4*a*c}$$

$$D = \sqrt{(-32)^2-4*1*(-32)}$$

$$D = \sqrt{(32 * 32) + (4*32)}$$

$$D = \sqrt{32 * (32 + 4)}$$

$$D = \sqrt{32 * 36}$$

$$D = 24 \sqrt{2}$$

$$D = 34$$ approx.

$$x = \frac{(-b + D)}{(2*a)}$$

$$a =\frac{(- - 32 + 34)}{(2*1)}$$

$$a = \frac{(32 + 34)}{2}$$

$$a = 33$$

$$x = \frac{(-b - D)}{(2* a)}$$
$$a = \frac{(- - 32 - 34)}{(2*1)}$$
$$a = \frac{(32 - 34)}{2}$$
$$a = \frac{-2}{2}$$ = -1 Not Applicable.

so a = 33, we can calculate value of b. And Perimeter of Rectangle as well.
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Sudhanshu
Intern  B
Joined: 17 Apr 2018
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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. If longer side has two value ,then shorter side will have two value then how will we come to solution the value of perimeter?
Math Expert V
Joined: 02 Sep 2009
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hellogmat123999 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. If longer side has two value ,then shorter side will have two value then how will we come to solution the value of perimeter?

I'd suggest to solve a^2 -32a-32=0 and see what you get. I think this will clear your doubts.
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if shorter side is 2, diagonal is 6 and longer side is 4 (Meet conditions of both statements)
then perimeter is=2(2+4)=12

if shorter side is 10, diagonal is 30 and longer side is 28 (Meet conditions of both statements)
then perimeter is=2(10+28)=76

two different answers so should't the correct option be E?
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Math Expert V
Joined: 02 Sep 2009
Posts: 55623

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HasnainAfxal wrote:
if shorter side is 2, diagonal is 6 and longer side is 4 (Meet conditions of both statements)
then perimeter is=2(2+4)=12

if shorter side is 10, diagonal is 30 and longer side is 28 (Meet conditions of both statements)
then perimeter is=2(10+28)=76

two different answers so should't the correct option be E?

In a rectangle two adjacent sides and the diagonal form a right triangle, so the lengths of the sides must satisfy side^2 + side^2 = diagonal^2. Do your examples satisfy this?
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Manager  G
Joined: 06 Sep 2018
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Location: Pakistan
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Bunuel wrote:
HasnainAfxal wrote:
if shorter side is 2, diagonal is 6 and longer side is 4 (Meet conditions of both statements)
then perimeter is=2(2+4)=12

if shorter side is 10, diagonal is 30 and longer side is 28 (Meet conditions of both statements)
then perimeter is=2(10+28)=76

two different answers so should't the correct option be E?

In a rectangle two adjacent sides and the diagonal form a right triangle, so the lengths of the sides must satisfy side^2 + side^2 = diagonal^2. Do your examples satisfy this?

Thanks got it Posted from my mobile device
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Manager  G
Joined: 22 Jun 2017
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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. It would be better if there is a rectangle showing what's each letter written on the solution.
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Manager  B
Joined: 26 Jan 2015
Posts: 76

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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 ?

Generally, in such situations, I prefer to quickly use the concept of product of roots. In a quadratic equation, ax^2 + bx + c, product of roots is c/a. If the product of roots is negative, one of the roots is negative else both are positive.
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