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Source: — Problem Solving |
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by mike22629 » Wed May 13, 2009 11:52 am
I could be wrong, but I think that this problem is around the 600 level.

First thing that you need to do is find the circumference of the circle. An equilateral triange divides the circumference into 3 equal parts.
ABC (which is 2/3 of circumference) = 24

So C*(2/3) = 24
C = 36

Now use the formula C = Pie(d)

36 = Pie*D
D = 36/Pie

D = approximately 11

Note: you do not have to find the exact answer, since pie = 3.14..., divide 36 by 3 = 12, then realize that since pie is a little more than 3, the answer will be approximately 11.

Other things good to know:
square root of 2 = approx. 1.4
square root of 3 = approx. 1.7
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Master | Next Rank: 500 Posts
Posts: 113
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Location: New Jersey
GMAT Score:650

by KICKGMATASS123 » Thu May 14, 2009 10:41 am
mike22629 wrote:I could be wrong, but I think that this problem is around the 600 level.

First thing that you need to do is find the circumference of the circle. An equilateral triange divides the circumference into 3 equal parts.
ABC (which is 2/3 of circumference) = 24

So C*(2/3) = 24
C = 36

Now use the formula C = Pie(d)

36 = Pie*D
D = 36/Pie

D = approximately 11

Note: you do not have to find the exact answer, since pie = 3.14..., divide 36 by 3 = 12, then realize that since pie is a little more than 3, the answer will be approximately 11.

Other things good to know:
square root of 2 = approx. 1.4
square root of 3 = approx. 1.7

I solved this problem a little differently.

AB + BC + AC = 2pi(r)

If you look at arc AB= the central angle is 120 degrees
therefore (2 pi r)- 120/360 (2 pi r) -120/360 (2 pi r)
will give us

2 pi r - (4/3 pi r)

however we know that 4/3 pi r = 24
therefore we can find r is approx 6.

the closest answer is 11
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