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In the figure above, A and B are the centers of the two circles. If ea

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In the figure above, A and B are the centers of the two circles. If ea  [#permalink]

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New post 16 May 2018, 05:02
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In the figure above, A and B are the centers of the two circles. If each circle has radius X, what is the area of the shaded region?


A. \(\frac{(2\pi - \sqrt{3}) x^2}{6}\)

B. \(\frac{(4\pi - 3 \sqrt{3}) x^2}{12}\)

C. \(\frac{(4\pi - 3 \sqrt{3}) x^2}{6}\)

D. \(\frac{(4\pi - \sqrt{3}) x^2}{6}\)

E. \(\frac{(6\pi - 1) x^2}{6}\)


Attachment:
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Re: In the figure above, A and B are the centers of the two circles. If ea  [#permalink]

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New post 16 May 2018, 05:51
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[quote="Bunuel"]Image
In the figure above, A and B are the centers of the two circles. If each circle has radius X, what is the area of the shaded region?


A. \(\frac{(2\pi - \sqrt{3}) x^2}{6}\)

B. \(\frac{(4\pi - 3 \sqrt{3}) x^2}{12}\)

C. \(\frac{(4\pi - 3 \sqrt{3}) x^2}{6}\)

D. \(\frac{(4\pi - \sqrt{3}) x^2}{6}\)

E. \(\frac{(6\pi - 1) x^2}{6}\)

First we need to understand that half of the shaded area (Blue colored) can be calculated by subtracting the area of the triangle (yellow coloured) from the sector that is making an angle of 120º at the centre of the circle with radius \(x\)

So the total intersection = 2* [Area of sector - Area of Yellow Triangle]

i.e. Area of intersection = \(2*[(120/360)*πx^2 - (√3/4)*x^2]\) = \(\frac{(4\pi - 3 \sqrt{3}) x^2}{6}\)

Answer: Option C
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Re: In the figure above, A and B are the centers of the two circles. If ea  [#permalink]

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New post 24 Jul 2018, 06:10
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I could not not understand the problem can anyone please explain it a bit more in detail?
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Re: In the figure above, A and B are the centers of the two circles. If ea  [#permalink]

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New post 26 Jul 2018, 04:57
GMATinsight

How do you know tha the Central angle is 120 degree?

Also if we know the know the central angle can we just find the area of the sector and multiply it to 2 to find the shaded region area?
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Re: In the figure above, A and B are the centers of the two circles. If ea  [#permalink]

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New post 08 Aug 2018, 12:40
Can someone provide another solution? I did not really understand the solution above. How can we determine the angle of the diamond shape inside the shaded area?
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Re: In the figure above, A and B are the centers of the two circles. If ea  [#permalink]

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New post 08 Aug 2018, 20:31
Bunuel wrote:
Image
In the figure above, A and B are the centers of the two circles. If each circle has radius X, what is the area of the shaded region?


A. \(\frac{(2\pi - \sqrt{3}) x^2}{6}\)

B. \(\frac{(4\pi - 3 \sqrt{3}) x^2}{12}\)

C. \(\frac{(4\pi - 3 \sqrt{3}) x^2}{6}\)

D. \(\frac{(4\pi - \sqrt{3}) x^2}{6}\)

E. \(\frac{(6\pi - 1) x^2}{6}\)


Attachment:
circles.jpg



I could not understand why the angle is taken 120. Please help me with its solution.
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In the figure above, A and B are the centers of the two circles. If ea  [#permalink]

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New post 09 Aug 2018, 00:20
Bunuel wrote:
In the figure above, A and B are the centers of the two circles. If each circle has radius X, what is the area of the shaded region?

A. \(\frac{(2\pi - \sqrt{3}) x^2}{6}\)

B. \(\frac{(4\pi - 3 \sqrt{3}) x^2}{12}\)

C. \(\frac{(4\pi - 3 \sqrt{3}) x^2}{6}\)

D. \(\frac{(4\pi - \sqrt{3}) x^2}{6}\)

E. \(\frac{(6\pi - 1) x^2}{6}\)


Please refer the enclosed diagram,

It is given that C1C2=AC1=AC2=C1D=C2D=radius=x,

So, AC1C2 is an equilateral triangle. You know, In an equilateral triangle, each of the three angles are equal. So, \(\angle{AC1C2}=60^{\circ}\)
Similarly, DC1C2 is an equilateral triangle. So, \(\angle{ADC1C2}=60^{\circ}\)
Hence, So, \(\angle{AC1D}=\angle{AC1C2}+\angle{ADC1C2}=120^{\circ}\), which is the central angle of circle with center C1.
Similarly, \(\angle{AC2D}=120^{\circ}\), which is the central angle of circle with center C2.

Since both the circles subtend an equal angle at their centers, hence the sector formed are of equal area.

So area of any of the sectors under discussion= \(\pi\)\(x^2*(\frac{120}{360})\)=\(\pi\)\(\frac{x^2}{3}\)

Now Area of shaded region=Area of sector of 1st circle+Area of sector of 2nd circle-4*Area of \(\triangle\)AC1B-----(1)
(We are subtracting the area of triangles since total area of both the sectors contains additional area of these 4 triangles whose areas is equal)
From the figure, each of the triangles is a 30-60-90 triangle with sides in the the ratio \(1:\sqrt{3}:2\)
In the \(\triangle\)AC1B, \(C1B=\frac{x}{2}\), hence \(AB=\sqrt{3}C1B=\sqrt{3}\frac{x}{2}\)

So, Area of \(\triangle\)AC1B=\(\frac{1}{2}*\frac{x}{2}*\sqrt{3}\frac{x}{2}\)=\(\sqrt{3}\frac{x^2}{8}\)
From(1), we have AREA OF SHADED PORTION=\(2*\pi\)\(\frac{x^2}{3}\) - \(4*\sqrt{3}\frac{x^2}{8}\)=\(\frac{(4\pi - 3 \sqrt{3}) x^2}{6}\)

Ans. (C)

P.S:- I have changed the nomenclature from 'A' 'B' of original figure to 'C1' 'C2' for easy understanding.
kanthaliya, Sandy56, rraman, Srija221 Hope it helps.
Open to further queries(if any).
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In the figure above, A and B are the centers of the two circles. If ea   [#permalink] 09 Aug 2018, 00:20
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