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GMAT PREP PS question

by alex.gellatly » Tue Apr 17, 2012 12:09 am
In the figure above, equilateral triangle ABC is inscribed in the circle (sorry can't upload the pic, but it's just a standard inscribed triangle with A at the bottom left, B at the top, and C at bottom right). 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

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by aneesh.kg » Tue Apr 17, 2012 12:16 am
If O is the centre of the circle, Since it is an equilateral triangle, both the sectors OAB and OBC comprise (1/3)rd of the Circle's area. So the larger sector OAC covers (2/3)rd of circle's area.

That's why

length of arc(ABC) = 2/3 of circumference = 2/3 * 2(pi)r = 24
so, r = 18/(pi)

or, diameter = 2 * 18 / (pi) = 36 / (pi) = 11 (approximately)

(C) is the answer
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by Anurag@Gurome » Tue Apr 17, 2012 12:24 am
alex.gellatly wrote:In the figure above, equilateral triangle ABC is inscribed in the circle (sorry can't upload the pic, but it's just a standard inscribed triangle with A at the bottom left, B at the top, and C at bottom right). 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

Thanks
Image

As triangle ABC is an equilateral triangle, so minor arcs AB, BC and CA are equal.
Let their length be x.
Sum of these 3 arcs = x + x + x = 3x = circumference of the circle
Arc AB + Arc BC = x + x = 2x = Arc ABC
2x = Arc ABC = (2/3) * 3x = (2/3) * circumference = (2/3) * 2(pi) * radius = 24
(2/3) * 2(pi) * radius = 24 implies radius = 36/(pi) = 11.46 or 11 approx.

The correct answer is C.
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