MitDavidDv wrote:

A cylindrical tank has a base with a circumference of 4sqrt^(pi * sqrt^3) meters and an equilateral triangle painted on the interior side of the base. If a stone is dropped inside the tank and the probability of the stone hitting the portion of the base outside the triangle is 3/4, what is the length of a side of the triangle?

a.) Sqrt^(2*sqrt^6)

b.) Sqrt^(6*sqrt^6)/2

c.) Sqrt^(2*sqrt^3)

d.) Sqrt^3

e.) 2

Since we know the probability of hitting outside of the triangle is 3/4, we then know the area of hitting anywhere inside the triangle must be 1/4. Thus, the Area of triangle = 1/4 the area of the circle. Since we can find the area of the circle from the circumference, we just have to solve for the area of the triangle, and solve for the side from there.

C = \(4 \sqrt{\pi * \sqrt{3}}=2 \pi r\)

\(2 \sqrt{\pi * \sqrt{3}}= \pi r\)

Square both sides:

\(4\pi\sqrt{3}= \pi^2 r^2\)

\(4\sqrt{3} = \pi r^2\)

We don't even need to solve for r; we already have the area here.

\(4\sqrt{3}\) = 4*Area of Triangle, and thus the Area of the Triangle must be \sqrt{3}.

Since we know the formula for the Area of an Equilateral Triangle with side \(s\) is \(\frac{\sqrt{3}}{4} s^2\), we see that \(s^2/4 = 1\), and thus that \(s = 2.\)

Answer: E