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

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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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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 = 20Ï€
Hence, radius of the base = (20Ï€)/(2Ï€) = 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.
Anurag Mairal, Ph.D., MBA
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