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Junior | Next Rank: 30 Posts
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by atarafder » Sat May 12, 2007 1:28 pm
A equilateral traingle inscribed in circle divides it into 3 equal arcs. So total circumference = 32.
2pir = 32
2r = 10.19 appx to 11 = diiameter.

So the answer is 11. Is my reasoning correct?
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Re: GMAT prep software question

by jayhawk2001 » Sat May 12, 2007 6:43 pm
atarafder wrote:A equilateral traingle inscribed in circle divides it into 3 equal arcs. So total circumference = 32.
2pir = 32
2r = 10.19 appx to 11 = diiameter.

So the answer is 11. Is my reasoning correct?
How did you get circumference to be 32 ?

Arc ABC covers 2/3 of the circle.

2/3 * 2*pi*r = 24

Solve for 2r and you'll get 11
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by atarafder » Sat May 12, 2007 6:55 pm
my bad, I meant 36 and not 32. In a hurry, I calculated 24/2 as 8 and not 12.
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