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

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

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New post 10 Feb 2012, 07:09
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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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Re: area of the shaded region  [#permalink]

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New post 10 Feb 2012, 07:35
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nafishasan60 wrote:
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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) \(\frac{\pi}{6}\)


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 \(6-\pi{\sqrt{3}}\).

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

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New post 10 Feb 2012, 08:00
option D 6 - \sqrt{3}\pi
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Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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New post 06 Mar 2014, 02:34
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Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

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

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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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