Please solve this:
A cylindrical tank has a base with a area of 4.(3)^(1/2) meters and an equilateral triangle painted on the interior side of the base. A grain of sand is dropped into the tank, and has an equal probability of landing on any particular point on the base. If the probability of the grain of sand landing on the portion of the base outside the triangle is 3/4, what is the length of a side of the triangle?
a. 4
b. 2
c. 3^1/2
d. 1/2
e. 2/3
manhattan 700+ quistion
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Area of base is a circle = 4√3
Probability of the grain of sand landing on the portion of the base outside the triangle is 3/4, so probability of the grain of sand landing on the portion of the base inside the triangle is 1/4.
So, area of equilateral triangle = (1/4)(Area of circle) = √3
If a is the side of equilateral triangle, area of equilateral triangle = (1/2)(base)(height) = (1/2)(a)(a√3/2) = (a^2)(√3/4)
We earlier found that area of equilateral triangle = √3, so (a^2)(√3/4) = √3, which implies a^2 = 4 or a=2
The correct answer is (B).
Probability of the grain of sand landing on the portion of the base outside the triangle is 3/4, so probability of the grain of sand landing on the portion of the base inside the triangle is 1/4.
So, area of equilateral triangle = (1/4)(Area of circle) = √3
If a is the side of equilateral triangle, area of equilateral triangle = (1/2)(base)(height) = (1/2)(a)(a√3/2) = (a^2)(√3/4)
We earlier found that area of equilateral triangle = √3, so (a^2)(√3/4) = √3, which implies a^2 = 4 or a=2
The correct answer is (B).
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Here's the figure, it shows the base of cylindrical tank. Hope this helps.ankurmit wrote:thanks Rahul..nice explanation..I am not able to draw its figure. Can you please help me
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welcomeankurmit wrote:Nice figure
Thanks again
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