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Two circles with equal radius pass through centre of each other as sho

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New post 13 Sep 2014, 07:40
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66% (03:00) correct 34% (02:23) wrong based on 109 sessions

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Two circles with equal radius pass through centre of each other as shown in figure. If radius = r, the area of the overlapping region would be:
Attachment:
Circle.jpg
Circle.jpg [ 14.68 KiB | Viewed 21607 times ]


A: \(2\pi r^2 - \frac{\sqrt{3}}{2}\)

B: \(2\pi r^2 - \frac{\pi}{8}\)

C: \(r^2(\frac{4\pi}{3} - \frac{\sqrt{3}}{4})\)

D: \(r^2(\frac{\pi}{3} - \sqrt{2})\)

E: \(r^2(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})\)
[Reveal] Spoiler: OA

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Re: Two circles with equal radius pass through centre of each other as sho [#permalink]

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New post 13 Sep 2014, 09:31
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PareshGmat wrote:
Two circles with equal radius pass through centre of each other as shown in figure. If radius = r, the area of the overlapping region would be:
Attachment:
The attachment Circle.jpg is no longer available


A: \(2\pi r^2 - \frac{\sqrt{3}}{2}\)

B: \(2\pi r^2 - \frac{\pi}{8}\)

C: \(r^2(\frac{4\pi}{3} - \frac{\sqrt{3}}{4})\)

D: \(r^2(\frac{\pi}{3} - \sqrt{2})\)

E: \(r^2(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})\)

Attachment:
1.jpg
1.jpg [ 19.56 KiB | Viewed 21645 times ]

In the diagram above, we need to recognize that \(AB\) is the radius for both the circles. Since the line \(AB\) is the radius, \(A\) is the center of the circle on the left and \(B\) is the center of the circle on the right. So, \(AC, BC, AD\) and \(BD\) are all radii of the circles and \(AB = BC = AC = AD = BD = r\)

Which makes \(\triangle ABC\) and \(\triangle ABD\) equilateral. So, angles \(ABC, BAC, BAD\) and \(ABD\) all measure \(60^o\).

So, angle \(CAD = CBD = 120^o\)

Now the area required will be Area of sector \(CBD\) (Area enclosed by line segments \(AC\) & \(AD\) and arc \(CBD\))+ Area of sector CAD (Area enclosed by line segments \(BC\) & \(BD\) and arc \(CAD\)) - Area of rhombus CADB

Since, the two circles are the same, the areas of the two sectors will be the same. Also, the rhombus can be seen as two equilateral triangles with equal areas. So, our equation reduces to
Required Area = \(2 *\) Area of sector \(CBD\) - \(2 *\) Area of \(\triangle CAB\)

\(= 2* \frac{\pi}{3} * r^2 - 2 * \frac{\sqrt3}{4} * r^2\)

\(= r^2(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})\)

Hence the answer is E.

Hope that helps.
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Re: Two circles with equal radius pass through centre of each other as sho [#permalink]

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New post 21 Sep 2015, 02:00
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Two circles with equal radius pass through centre of each other as sho [#permalink]

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New post 23 Jul 2016, 10:25
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PareshGmat wrote:
Two circles with equal radius pass through centre of each other as shown in figure. If radius = r, the area of the overlapping region would be:
Attachment:
Circle.jpg


A: \(2\pi r^2 - \frac{\sqrt{3}}{2}\)

B: \(2\pi r^2 - \frac{\pi}{8}\)

C: \(r^2(\frac{4\pi}{3} - \frac{\sqrt{3}}{4})\)

D: \(r^2(\frac{\pi}{3} - \sqrt{2})\)

E: \(r^2(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})\)


The area of the overlapping segment would be less than the area of any of the circles. (lets suppose 3.14r*r-p which will be less than 3.14 if we assume r=1 for solving purpose)
If we feed r=1 in all the options, we get only option E which satisfies the condition, area less than 3.14.
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Re: Two circles with equal radius pass through centre of each other as sho [#permalink]

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New post 25 Aug 2017, 09:03
Hello from the GMAT Club BumpBot!

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Re: Two circles with equal radius pass through centre of each other as sho   [#permalink] 25 Aug 2017, 09:03
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