Confusing problem....geometry and probablity combined!!

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cypherskull: Senior | Next Rank: 100 Posts; Posts: 94; Joined: Sat Mar 31, 2012 3:39 am; Location: Calcutta; Thanked: 8 times

by cypherskull » Sun Aug 19, 2012 10:34 am

A cylinder has a base with a circumference of 20pi meters and an equilateral triangle inscribed on the interior side of the base. A marker is dropped into the tank with an equal probability of landing on any point on the base. If the probability of the marker landing inside the triangle is (Sqrt 3)/4 , what is the length of a side of the triangle?

A. 3(sqrt 2pi)
B. 3(sqrt 3pi)
C. 10*sqrt (pi)
D. 10(sqrt 3pi)
E. 20 pi

Don't even know where to start. Please help!

Regards,
Sunit

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Source: Beat The GMAT — Problem Solving | Sun Aug 19, 2012 10:34 am

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Sun Aug 19, 2012 11:22 am

cypherskull wrote:A cylinder has a base with a circumference of 20pi meters and an equilateral triangle inscribed on the interior side of the base. A marker is dropped into the tank with an equal probability of landing on any point on the base. If the probability of the marker landing inside the triangle is (Sqrt 3)/4 , what is the length of a side of the triangle?

The probability of the marker landing inside the triangle will be equal to the ratio of the area of the triangle to the area of the base of the cylinder.

Let us assume that the length of a side of the triangle is a.
Hence, are of the triangle = (√3/4)*a²

Circumference of the base = 20π
Hence, radius of the base = (20π)/(2π) = 10
Hence, area of the base = π(10)² = 100π

So, [(√3/4)*a²]/[100π] = (√3/4)
--> a² = 100π
--> a = 10√π

The correct answer is C.

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