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# M19-21

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Math Expert
Joined: 02 Sep 2009
Posts: 47157

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16 Sep 2014, 01:06
00:00

Difficulty:

85% (hard)

Question Stats:

51% (01:26) correct 49% (01:45) wrong based on 182 sessions

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If the area of a square equals the area of a circle, which of the following is closest to the ratio of the diagonal of the square to the diameter of the circle?

A. 0.95
B. 1.26
C. 1.40
D. 1.57
E. 2.51

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 47157

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16 Sep 2014, 01:06
1
Official Solution:

If the area of a square equals the area of a circle, which of the following is closest to the ratio of the diagonal of the square to the diameter of the circle?

A. 0.95
B. 1.26
C. 1.40
D. 1.57
E. 2.51

If $$x$$ is the side of the square, then the area of the square is $$x^2$$ and the diagonal of the square is $$x\sqrt{2}$$. If $$d$$ is the diameter of the circle then the area of the circle is $$\pi(\frac{d}{2})^2$$. Because $$x^2 = \pi(\frac{d}{2})^2$$, $$d = \frac{2x}{\sqrt{\pi}}$$. The required ratio $$=\frac{x\sqrt{2}}{\frac{2x}{\sqrt{\pi}}} = \frac{\sqrt{\pi}}{\sqrt{2}} = \sqrt{\frac{\pi}{2}}$$ or approximately $$\sqrt{1.57}$$. This is slightly smaller than $$\sqrt{1.69} = 1.3$$. The best answer is therefore B.

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Joined: 13 Jan 2011
Posts: 22

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16 Sep 2015, 04:58
Most likely a stupid question but why does

x sqrt(2) / [(2x/sqrt(pi)] = sqrt(pi)/sqrt(2)?

Doesnt it simplify to sqrt(2) * sqrt(pi) / 2?

Bunuel wrote:
Official Solution:

If the area of a square equals the area of a circle, which of the following is closest to the ratio of the diagonal of the square to the diameter of the circle?

A. 0.95
B. 1.26
C. 1.40
D. 1.57
E. 2.51

If $$x$$ is the side of the square, then the area of the square is $$x^2$$ and the diagonal of the square is $$x\sqrt{2}$$. If $$d$$ is the diameter of the circle then the area of the circle is $$\pi(\frac{d}{2})^2$$. Because $$x^2 = \pi(\frac{d}{2})^2$$, $$d = \frac{2x}{\sqrt{\pi}}$$. The required ratio $$=\frac{x\sqrt{2}}{\frac{2x}{\sqrt{\pi}}} = \frac{\sqrt{\pi}}{\sqrt{2}} = \sqrt{\frac{\pi}{2}}$$ or approximately $$\sqrt{1.57}$$. This is slightly smaller than $$\sqrt{1.69} = 1.3$$. The best answer is therefore B.

Math Expert
Joined: 02 Sep 2009
Posts: 47157

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16 Sep 2015, 06:38
dor1209 wrote:
Most likely a stupid question but why does

x sqrt(2) / [(2x/sqrt(pi)] = sqrt(pi)/sqrt(2)?

Doesnt it simplify to sqrt(2) * sqrt(pi) / 2?

Bunuel wrote:
Official Solution:

If the area of a square equals the area of a circle, which of the following is closest to the ratio of the diagonal of the square to the diameter of the circle?

A. 0.95
B. 1.26
C. 1.40
D. 1.57
E. 2.51

If $$x$$ is the side of the square, then the area of the square is $$x^2$$ and the diagonal of the square is $$x\sqrt{2}$$. If $$d$$ is the diameter of the circle then the area of the circle is $$\pi(\frac{d}{2})^2$$. Because $$x^2 = \pi(\frac{d}{2})^2$$, $$d = \frac{2x}{\sqrt{\pi}}$$. The required ratio $$=\frac{x\sqrt{2}}{\frac{2x}{\sqrt{\pi}}} = \frac{\sqrt{\pi}}{\sqrt{2}} = \sqrt{\frac{\pi}{2}}$$ or approximately $$\sqrt{1.57}$$. This is slightly smaller than $$\sqrt{1.69} = 1.3$$. The best answer is therefore B.

Check below:
$$\frac{x\sqrt{2}}{(\frac{2x}{\sqrt{\pi}})} =x\sqrt{2}* \frac{\sqrt{\pi}}{2x}=\sqrt{2}* \frac{\sqrt{\pi}}{2}=\sqrt{2}* \frac{\sqrt{\pi}}{\sqrt{2}*\sqrt{2}}=\frac{\sqrt{\pi}}{\sqrt{2}} = \sqrt{\frac{\pi}{2}}$$.

Hope it helps.
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Intern
Joined: 07 Feb 2016
Posts: 21
GMAT 1: 650 Q47 V34
GMAT 2: 710 Q48 V39

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15 Apr 2017, 23:16
1
I used smart numbers to calculate the ratio and find it even faster than the algebrical approach.

$$A(square)=1$$ --> $$a=1$$
$$d(square)=1*\sqrt{2}≈1.4$$

$$A(circle)=1$$
$$r=\sqrt{\frac{1}{pi}}≈1.1$$

$$Ratio=\frac{1.4}{1.1}≈1.3$$
Manager
Joined: 09 Nov 2016
Posts: 56

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23 Aug 2017, 08:12
This is really a good Question..

Answer to this one is B.
Manager
Joined: 19 Jul 2017
Posts: 85
Location: India
Concentration: General Management, Strategy
GPA: 3.5

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26 Aug 2017, 03:53
22/7 * r^2 = a^2
a/r =sqrt(22/7)
sqrt(2) * a/2r = sqrt(22/14)
=sqrt(1.50..)
approximately equal to 1.26
Re: M19-21 &nbs [#permalink] 26 Aug 2017, 03:53
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# M19-21

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