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# Arcs DE, EF, FD are centered at C, B and A, in equilateral

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Joined: 18 Feb 2011
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Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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10 Feb 2012, 07:09
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74% (02:46) correct 26% (02:53) wrong based on 194 sessions

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Arcs DE, EF, FD are centered at C, B and A, in equilateral triangle ABC, as shown above. If the area of the triangle is 6, what is the area of the shaded region formed by the intersection of the arcs and the triangle?

A) 6 - $$\frac{\sqrt{3}}{3}$$$$\pi$$
B) 6 - $$\frac{\sqrt{3}}{2}$$$$\pi$$
C) $$\frac{\pi}{\sqrt{3}}$$
D) 6 - $$\sqrt{3}\pi$$
E) $$\frac{\pi}{6}$$
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10 Feb 2012, 07:35
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nafishasan60 wrote:

Arcs DE, EF, FD are centered at C, B and A, in equilateral triangle ABC, as shown above. If the area of the triangle is 6, what is the area of the shaded region formed by the intersection of the arcs and the triangle?

A) 6 - $$\frac{\sqrt{3}}{3}$$$$\pi$$
B) 6 - $$\frac{\sqrt{3}}{2}$$$$\pi$$
C) $$\frac{\pi}{\sqrt{3}}$$
D) 6 - $$\sqrt{3}\pi$$
E) $$\frac{\pi}{6}$$

The area of an equilateral triangle is $$area=a^2*\frac{\sqrt{3}}{4}$$, where $$a$$ is the length of a side. Thus $$a^2*\frac{\sqrt{3}}{4}=6$$ --> $$a^2=8*\sqrt{3}$$ --> radius of the circles (which make arcs) is a/2, so $$r^2=\frac{a^2}{4}$$. Now, as the triangle is equilateral each arc is 60 degrees, so the area of white region (non-shaded) is $$\frac{60*3}{360}*\pi{r^2}=\frac{1}{2}*\pi{\frac{8*\sqrt{3}}{4}=\pi{\sqrt{3}}$$ (three 60 degrees arcs).

The area of the shaded region is $$6-\pi{\sqrt{3}}$$.

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Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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10 Feb 2012, 08:00
option D 6 - \sqrt{3}\pi
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+1 if you like my explanation .Thanks

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Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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06 Mar 2014, 02:34
Bumping for review and further discussion.

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Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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27 Jul 2015, 08:50
Bunuel:

Can you please tell me how did u get the radius as a/2.
Is there a formula for calculating radius inside a equilateral triangle or u did just by some calculation??
Can u also post some links having same concept tested.

Thanks
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Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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27 Jul 2015, 09:00
Shree9975 wrote:
Bunuel:

Can you please tell me how did u get the radius as a/2.
Is there a formula for calculating radius inside a equilateral triangle or u did just by some calculation??
Can u also post some links having same concept tested.

Thanks

since all three arcs of circle touch each other externally, all three have same radius as its an equilateral triangle..
if each side is a, the radius of each circle is a/2
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Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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27 Jul 2015, 09:07
Shree9975 wrote:
Bunuel:

Can you please tell me how did u get the radius as a/2.
Is there a formula for calculating radius inside a equilateral triangle or u did just by some calculation??
Can u also post some links having same concept tested.

Thanks

Look at the attached picture. There is no "formula" as such. It is calculated by looking at the figure as shown below.

From the figure, if 'a' is the side of the equilateral triangle and 'r' is the radius of the 2 arcs,

r+r = a ---> r =a/2 (this is true as all the 3 sectors created by the 3 arcs are equal).

Hope this helps.
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2015-07-27_12-01-00.jpg [ 11.07 KiB | Viewed 2315 times ]

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Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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16 Jan 2019, 10:02
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral   [#permalink] 16 Jan 2019, 10:02
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