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# M20-24

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Math Expert
Joined: 02 Sep 2009
Posts: 49303

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16 Sep 2014, 01:09
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Difficulty:

35% (medium)

Question Stats:

66% (00:55) correct 34% (00:58) wrong based on 155 sessions

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What is the area of rectangle $$ABCD$$ ?

(1) Diagonal $$AC$$ is twice as long as side $$CD$$

(2) Diagonal $$AC$$ is 0.1 meters longer than side $$AD$$

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Math Expert
Joined: 02 Sep 2009
Posts: 49303

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16 Sep 2014, 01:09
Official Solution:

Statement (1) by itself is insufficient. S1 defines the shape of the rectangle but does not specify its size.

Statement (2) by itself is insufficient. The area of the rectangle can be made arbitrarily small (as the length of side $$AD$$ goes to 0).

Statements (1) and (2) combined are sufficient. Because $$AC^2 = CD^2 + AD^2$$ and $$AC = 2CD$$, $$4CD^2 = CD^2 + AD^2$$ which simplifies to $$3CD^2 = AD^2$$. It follows from here that $$AD = \frac{\sqrt{3}}{2} AC$$.

This is one of the equations, the other one is $$AC = AD + 0.1$$. This system of linear equations is enough to determine both sides of the rectangle.

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Joined: 23 Jan 2014
Posts: 27
Schools: Sloan '19 (D)

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26 Jul 2016, 12:21
Hi Bunuel,

Wouldn't (1) work with the triangle type x:xsqrt3:2x?

Thanks
Senior Manager
Joined: 11 Nov 2014
Posts: 336
Location: India
WE: Project Management (Telecommunications)

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26 Jul 2016, 12:39
dauren88 wrote:
Hi Bunuel,

Wouldn't (1) work with the triangle type x:xsqrt3:2x?

Thanks

It will
But area of the said rectangle will change I.e directly proportional to x
Hence in suff

Intern
Joined: 20 May 2017
Posts: 3

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16 Aug 2017, 07:04
Statements (1) and (2) combined are sufficient. Because $$AC^2 = CD^2 + AD^2$$ and $$AC = 2CD$$, $$4CD^2 = CD^2 + AD^2$$ which simplifies to $$3CD^2 = AD^2$$. [color=#fff200]It follows from here that $$AD = \frac{\sqrt{3}}{2} AC$$.[/color]

This is one of the equations, the other one is $$AC = AD + 0.1$$. This system of linear equations is enough to determine both sides of the rectangle.

Hello Bunuel,
Can you please let me know how have you arrived at the highlighted text?

Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 49303

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16 Aug 2017, 07:12
1
anirudhasopa wrote:
Statements (1) and (2) combined are sufficient. Because $$AC^2 = CD^2 + AD^2$$ and $$AC = 2CD$$, $$4CD^2 = CD^2 + AD^2$$ which simplifies to $$3CD^2 = AD^2$$. [color=#fff200]It follows from here that $$AD = \frac{\sqrt{3}}{2} AC$$.[/color]

This is one of the equations, the other one is $$AC = AD + 0.1$$. This system of linear equations is enough to determine both sides of the rectangle.

Hello Bunuel,
Can you please let me know how have you arrived at the highlighted text?

Thanks!

$$3CD^2 = AD^2$$

Take the square root from both sides: $$\sqrt{3}CD = AD$$.

We know that $$AC = 2CD$$, so $$\frac{AC}{2} = CD$$.

Thus, $$\sqrt{3}*\frac{AC}{2} = AD$$.

Hope it's clear.
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GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
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16 Oct 2017, 06:09
+1 for option C.
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" The few , the fearless "

Intern
Joined: 06 Jun 2016
Posts: 9

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08 Nov 2017, 11:47
To all,

Combining both statements, is it safe to assume that the diagonal splits the rectangle into two, 30-60-90 triangles and that with information from statement 2 we can determine the area of each triangle?
Math Expert
Joined: 02 Sep 2009
Posts: 49303

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08 Nov 2017, 23:17
mrosale2 wrote:
To all,

Combining both statements, is it safe to assume that the diagonal splits the rectangle into two, 30-60-90 triangles and that with information from statement 2 we can determine the area of each triangle?

No. Why would you assume that? A diagonal of a rectangle could make any angle from 0 to 90, not inclusive with adjacent side. Why should it be 30 (or 60) specifically?
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Joined: 06 Jun 2016
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09 Nov 2017, 11:10
Bunuel wrote:
mrosale2 wrote:
To all,

Combining both statements, is it safe to assume that the diagonal splits the rectangle into two, 30-60-90 triangles and that with information from statement 2 we can determine the area of each triangle?

No. Why would you assume that? A diagonal of a rectangle could make any angle from 0 to 90, not inclusive with adjacent side. Why should it be 30 (or 60) specifically?

Hi Bunuel,

I misread statement 2 as "Diagonal AC is 0.1 meters longer than side CD.

Obviously I understand now that my reasoning is wrong, but for the sake of learning, would it be correct to assume that, if taking both statements into consideration, and if statement 2 did read how I misread it, my reasoning would be correct?

In this situation, the diagonal AC must be 90 degrees, and because it is twice the size of CD, then the triangle must be 30-60-90. AC is 2x, CD is x, AD is x sqrt3

Or maybe I need to brush up on my fundamentals?

Intern
Joined: 16 Oct 2017
Posts: 1

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16 Jan 2018, 04:37
Hi,

the value of the area is still undefined due to no value can be input into the equation" 3√2/2AC-AC = -0.1"

we still need to know the value of AC to know the exact area of ABCD

Regards,
Math Expert
Joined: 02 Sep 2009
Posts: 49303

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16 Jan 2018, 05:29
krucaix2 wrote:
Hi,

the value of the area is still undefined due to no value can be input into the equation" 3√2/2AC-AC = -0.1"

we still need to know the value of AC to know the exact area of ABCD

Regards,

$$AD = \frac{\sqrt{3}}{2} AC$$ and $$AC = AD + 0.1$$ has unique solution: $$AD \approx 0.64641$$ and $$AC \approx 0.74641$$
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Re: M20-24 &nbs [#permalink] 16 Jan 2018, 05:29
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# M20-24

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