A cylindrical tank has a base with a circumference of

A cylindrical tank has a base with a circumference of \(4\sqrt{\pi{\sqrt{3}}}\) 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. \(\sqrt{2{\sqrt{6}}}\)

B. \(\frac{\sqrt{6{\sqrt{6}}}}{2}\)

C. \(\sqrt{2{\sqrt{3}}}\)

D. \(\sqrt{3}\)

E. \(2\)

Given: \(circumference=4\sqrt{\pi{\sqrt{3}}}\) and \(P(out)=\frac{3}{4}\)

Now, as the probability of the grain of sand landing on the portion of the base outside the triangle is 3/4 then the the portion of the base (circle) outside the triangle must be 3/4 of the are of the base and the triangle itself 1/4 of the are of the base.

Next: \(circumference=4\sqrt{\pi{\sqrt{3}}}=2\pi{r}\) --> square both sides --> \(16\pi{\sqrt{3}}=4{\pi}^2{r}^2\) --> \(4{\sqrt{3}}={\pi}{r}^2\) --> \(area_{base}=\pi{r^2}=4{\sqrt{3}}\);

The area of the equilateral triangle is 1/4 of the base: \(area_{equilateral}=\frac{1}{4}*4{\sqrt{3}}=\sqrt{3}\) --> also the ares of the equilateral triangle is \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}\), where \(a\) is the length of a side --> \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=\sqrt{3}\) --> \(a=2\).

Answer: E.