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An equilateral triangle ABC is inscribed in the circle. If

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An equilateral triangle ABC is inscribed in the circle. If [#permalink]

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New post 23 Apr 2011, 15:32
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Question Stats:

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An equilateral triangle ABC is inscribed in the circle. If the length of the arc ABC is 24, what is the approximate diameter of the circle.

A) 5
B) 8
C) 11
D) 15
E) 19

OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/in-the-figure ... 97393.html
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Re: An equilateral triangle ABC is inscribed in the circle. If [#permalink]

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New post 23 Apr 2011, 16:42
Since ABC is an equilateral triangle, arc ABC is 2/3 of circumference, so circumference equals 24*3/2 = 36

\(\pi * Diameter = 36\) , so Diameter ~11
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Re: An equilateral triangle ABC is inscribed in the circle. If [#permalink]

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New post 24 Apr 2011, 21:50
24/pi*D = 240/360 (Angle subtended by arc ABC = 240)

=> D = 36/pi = 36/3.14 ~ 11

Answer - C
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Re: An equilateral triangle ABC is inscribed in the circle. If [#permalink]

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New post 01 Oct 2016, 21:05
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agdimple333 wrote:
An equilateral triangle ABC is inscribed in the circle. If the length of the arc ABC is 24, what is the approximate diameter of the circle.

A) 5
B) 8
C) 11
D) 15
E) 19



Ans C ..plz follow as attached below
Attachments

eq.png
eq.png [ 15.87 KiB | Viewed 962 times ]

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Re: An equilateral triangle ABC is inscribed in the circle. If [#permalink]

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New post 14 Jan 2018, 06:40
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In the figure above, equilateral triangle ABC is inscribed in the circle. If the length of arc ABC is 24, what is the approximate diameter of the circle?

A. 5
B. 8
C. 11
D. 15
E. 19

Arc ABC is \(\frac{2}{3}\) of the circumference (as ABC is equilateral triangle and thus arc AB=arc BC=arc AC, so arc AB+arc BC=arc ABC = 2/3 of circumference) --> \(24=c*\frac{2}{3}\), hence circumference \(c=\frac{24*3}{2}=36=\pi{d}\) --> \(d\approx{11.5}\).

Answer: C.

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Re: An equilateral triangle ABC is inscribed in the circle. If   [#permalink] 14 Jan 2018, 06:40
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