# 2008 AMC 10B Problems/Problem 19

## Problem

A cylindrical tank with radius $4$ feet and height $9$ feet is lying on its side. The tank is filled with water to a depth of $2$ feet. What is the volume of water, in cubic feet? $\mathrm{(A)}\ 24\pi - 36 \sqrt {2} \qquad \mathrm{(B)}\ 24\pi - 24 \sqrt {3} \qquad \mathrm{(C)}\ 36\pi - 36 \sqrt {3} \qquad \mathrm{(D)}\ 36\pi - 24 \sqrt {2} \qquad \mathrm{(E)}\ 48\pi - 36 \sqrt {3}$

## Solution

Any vertical cross-section of the tank parallel with its base looks as follows: $[asy] unitsize(0.8cm); defaultpen(0.8); pair s=(0,0), bottom=(0,-4), mid=(0,-2); pair x[]=intersectionpoints( (-10,-2)--(10,-2), circle(s,4) ); fill( arc(s,x,x) -- cycle, lightgray ); draw( circle(s,4) ); dot(s); draw( s -- bottom ); label( "2", (mid+bottom)/2, E ); draw ( s -- x -- x -- s ); label( "4", (s+x)/2, NW ); label( "4", (s+x)/2, NE ); label( "A", s, N ); label( "B", x, W ); label( "C", x, E ); label( "D", mid, NW ); label( "E", bottom, S ); [/asy]$

The volume of water can be computed as the height of the tank times the area of the shaded part.

Let $\theta$ be the size of the smaller angle $DAC$. We then have $\cos\theta = \frac{AD}{AC}=\frac 12$, hence $\theta=60^\circ$.

Thus the angle $CAB$ has size $360^\circ - 2\cdot 60^\circ = 120^\circ$. Hence the non-shaded part consists of $\frac{120^\circ}{360^\circ} = \frac 13$ of the circle, minus the area of the triangle $ABC$.

Using the Pythagorean theorem we can compute that $CD=\sqrt{AC^2-AD^2}=\sqrt{16-4}=2\sqrt 3$. Thus $BC=4\sqrt 3$, and the area of the triangle $ABC$ is $\frac {2 \cdot 4\sqrt 3} 2 = 4\sqrt 3$.

The area of the shaded part is then $\frac{4^2\pi}3 - 4\sqrt 3$, and the volume of water is $9\cdot\left(\frac{4^2\pi}3 - 4\sqrt 3\right) = \boxed{48\pi - 36\sqrt 3}$. The answer is $\text{E}$.

Alternatively, we can solve for DC and see it is congruent to a 30 60 90 triangle by SSS ~Williamgolly

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 