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M20-24

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M20-24 [#permalink]

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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 \(AD\)
[Reveal] Spoiler: OA

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Re M20-24 [#permalink]

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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.


Answer: C
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Re: M20-24 [#permalink]

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New post 26 Jul 2016, 11:21
Hi Bunuel,

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

Thanks
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Re: M20-24 [#permalink]

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New post 26 Jul 2016, 11:39
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


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Re: M20-24 [#permalink]

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New post 16 Aug 2017, 06:04
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.


Answer: C[/quote]

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

Thanks!
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Re: M20-24 [#permalink]

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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.


Answer: C

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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Re: M20-24 [#permalink]

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New post 16 Oct 2017, 05:09
+1 for option C.
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Re: M20-24 [#permalink]

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New post 08 Nov 2017, 10:47
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?
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Re: M20-24 [#permalink]

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New post 08 Nov 2017, 22:17
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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M20-24 [#permalink]

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New post 09 Nov 2017, 10:10
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?

Thank you for your help!
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Re: M20-24 [#permalink]

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New post 16 Jan 2018, 03:37
Hi,

Could you please explain why the answer is not E,

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

Please to explain

Regards,
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Re: M20-24 [#permalink]

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New post 16 Jan 2018, 04:29
krucaix2 wrote:
Hi,

Could you please explain why the answer is not E,

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

Please to explain

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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M20-24   [#permalink] 16 Jan 2018, 04:29
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