2008 AMC 10B Problems/Problem 19

From AoPSWiki

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?

$\mathrm{(A)}\ 24\pi - 36 \sqrt {2}\qquad\mathrm{(B)}\ 24\pi - 24 \sqrt {3}\qquad\mathrm{(C)}\ 36\pi - 36 \sqrt {3}\qquad\math...$

Solution

Any vertical cross-section of the tank parallel with its base looks as follows:

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 outer angle $CAB$ has size $360^\circ - 2\cdot 60^\circ = 240^\circ$ . Hence the non-shaded part consists of $\frac{240^\circ}{360^\circ} = \frac 23$ of the circle, plus 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 $9\cdot\left(\frac{4^2\pi}3 - 4\sqrt 3\right) = \boxed{48\pi - 36\sqrt 3}$ .

2008 AMC 10B (Problems • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Personal tools

Navigation

Search

Toolbox

2008 AMC 10B Problems/Problem 19

From AoPSWiki

Problem

Solution

See also