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A circular region centered at C is inscribed in equilateral

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A circular region centered at C is inscribed in equilateral  [#permalink]

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New post 24 Jul 2014, 00:58
1
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A
B
C
D
E

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  65% (hard)

Question Stats:

67% (03:03) correct 33% (03:22) wrong based on 205 sessions

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A circular region centered at C is inscribed in equilateral triangle ABC above. If the area of ΔABC is 12, then what is the area of the shaded region?

A. 6-3π
B. 6-π√3
C. 12-π√3
D. 12-3π
E. 3+π√3
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Re: A circular region centered at C is inscribed in equilateral  [#permalink]

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New post 24 Jul 2014, 05:03
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goodyear2013 wrote:
Image
A circular region centered at C is inscribed in equilateral triangle ABC above. If the area of ΔABC is 12, then what is the area of the shaded region?

A. 6-3π
B. 6-π√3
C. 12-π√3
D. 12-3π
E. 3+π√3


One could also use following approach and solve this question in 30 seconds:

The area of the shaded region = 1/2*(the area of the triangle - the area of the sector) = 1/2*(12 - something with π) = 6 - something with π.

Only A and B fits this format, but A is negative, so we can discard it.

Also:
C: 12 - π√3 = ~7 too big for the area of the tiny shaded region (more than half of the area of the triangle).
D: 12 - 3π = ~3 too big for the area of the tiny shaded region (~1/4 of the area of the triangle).
E: 3 + π√3 = ~8 too big for the area of the tiny shaded region (more than half of the area of the triangle).

Answer: B.
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Re: A circular region centered at C is inscribed in equilateral  [#permalink]

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New post 24 Jul 2014, 02:46
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goodyear2013 wrote:
Attachment:
Triangle.png

A circular region centered at C is inscribed in equilateral triangle ABC above. If the area of ΔABC is 12, then what is the area of the shaded region?
A. 6-3π
B. 6-π√3
C. 12-π√3
D. 12-3π
E. 3+π√3




area of equilateral triangle = √3/4 * a* a = 12.
Area of Sector = 1/2 * Theta * [π/180] * r * r
Here r= Height of equilateral triangle = √3/2 * a

Area of shaded region = 1/2[12 - Area of sector] = 6-π√3
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Re: A circular region centered at C is inscribed in equilateral  [#permalink]

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New post 14 Feb 2016, 15:31
I tried to go beyond logic, and see how to solve the problem...
anyone could help me find out where did I go wrong??
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Re: A circular region centered at C is inscribed in equilateral  [#permalink]

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New post 15 Feb 2016, 08:28
any expert to comment on the steps needed to solve the problem?
VeritasPrepKarishma, EMPOWERgmatRichC, mikemcgarry -would love to hear from you guys.
thanks in advance.
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Re: A circular region centered at C is inscribed in equilateral  [#permalink]

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New post 28 Apr 2016, 21:40
goodyear2013 wrote:
Attachment:
Triangle.png

A circular region centered at C is inscribed in equilateral triangle ABC above. If the area of ΔABC is 12, then what is the area of the shaded region?

A. 6-3π
B. 6-π√3
C. 12-π√3
D. 12-3π
E. 3+π√3


Hi debbiem,
refer your message ..
whenever you see these Qs, look how you can home on to the shaded region.
it is area of triangle - area of that sector of circle, and this should be div by 2 as there are two equal portions left over and we have to pick one of them..

area =\(\sqrt{3}/4*a^2 = 12\)..

Now the radius of circle is equal to altitude AND what is altitude of equilateral triangle = \(\sqrt{3}/2a\)

so area of sector, which is 60 degree = 1/6th of area of circle =\(\frac{1}{6} *pi*(\sqrt{3}/2*a)^2 = \frac{1}{6}*pi*\frac{3}{4}*a^2 = \frac{1}{6}*pi*\sqrt{3}*area-of-circle\)..

=> \(\frac{1}{6}*pi*\sqrt{3}*12 = 2\sqrt{3}*pi.\).

area of shaded portion = 1/2( area of triangle - area of circle) = \(\frac{1}{2} (12-2\sqrt{3}*pi) = 6-\sqrt{3}*pi\)
B
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Re: A circular region centered at C is inscribed in equilateral  [#permalink]

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New post 17 Jan 2019, 07:11
I keep ending up with 6 - πx^2. Could someone help explain how to do this problem? I am almost there but am doing something wrong since I keep ending up with this answer. Any insight would be appreciated. Thank you!

Bunuel wrote:
goodyear2013 wrote:
Image
A circular region centered at C is inscribed in equilateral triangle ABC above. If the area of ΔABC is 12, then what is the area of the shaded region?

A. 6-3π
B. 6-π√3
C. 12-π√3
D. 12-3π
E. 3+π√3


One could also use following approach and solve this question in 30 seconds:

The area of the shaded region = 1/2*(the area of the triangle - the area of the sector) = 1/2*(12 - something with π) = 6 - something with π.

Only A and B fits this format, but A is negative, so we can discard it.

Also:
C: 12 - π√3 = ~7 too big for the area of the tiny shaded region (more than half of the area of the triangle).
D: 12 - 3π = ~3 too big for the area of the tiny shaded region (~1/4 of the area of the triangle).
E: 3 + π√3 = ~8 too big for the area of the tiny shaded region (more than half of the area of the triangle).

Answer: B.
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Re: A circular region centered at C is inscribed in equilateral   [#permalink] 17 Jan 2019, 07:11
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