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Intern  Joined: 18 Feb 2011
Posts: 33
GPA: 3.91
Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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13 00:00

Difficulty:   55% (hard)

Question Stats: 71% (02:46) correct 29% (02:52) wrong based on 215 sessions

### HideShow timer Statistics 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) $$\frac{\pi}{6}$$
Math Expert V
Joined: 02 Sep 2009
Posts: 59726

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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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option D 6 - \sqrt{3}\pi
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+1 if you like my explanation .Thanks Math Expert V
Joined: 02 Sep 2009
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Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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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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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
Math Expert V
Joined: 02 Aug 2009
Posts: 8324
Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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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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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.
Attachments 2015-07-27_12-01-00.jpg [ 11.07 KiB | Viewed 2881 times ]

Non-Human User Joined: 09 Sep 2013
Posts: 13737
Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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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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