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Difference between revisions of "2008 AMC 10B Problems/Problem 19"

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Let <math>\theta</math> be the size of the smaller angle <math>DAC</math>. We then have <math>\cos\theta = \frac{AD}{AC}=\frac 12</math>, hence <math>\theta=60^\circ</math>.
 
Let <math>\theta</math> be the size of the smaller angle <math>DAC</math>. We then have <math>\cos\theta = \frac{AD}{AC}=\frac 12</math>, hence <math>\theta=60^\circ</math>.
  
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Thus the angle <math>CAB</math> has size <math>2\cdot 60^\circ = 120^\circ</math>. Hence the non-shaded part consists of <math>\frac{120^\circ}{360^\circ} = \frac 13</math> of the circle, minus the area of the triangle <math>ABC</math>.
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The figure is symmetrical, so the angle <math>CAB</math> has size <math>2\cdot 60^\circ = 120^\circ</math>. Hence the non-shaded part consists of <math>\frac{120^\circ}{360^\circ} = \frac 13</math> of the circle, minus the area of the triangle <math>ABC</math>.
  
 
Using the [[Pythagorean theorem]] we can compute that <math>CD=\sqrt{AC^2-AD^2}=\sqrt{16-4}=2\sqrt 3</math>. Thus <math>BC=4\sqrt 3</math>, and the area of the triangle <math>ABC</math> is <math>\frac {2 \cdot 4\sqrt 3} 2 = 4\sqrt 3</math>.
 
Using the [[Pythagorean theorem]] we can compute that <math>CD=\sqrt{AC^2-AD^2}=\sqrt{16-4}=2\sqrt 3</math>. Thus <math>BC=4\sqrt 3</math>, and the area of the triangle <math>ABC</math> is <math>\frac {2 \cdot 4\sqrt 3} 2 = 4\sqrt 3</math>.

Revision as of 12:20, 7 June 2021

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?

$\mathrm{(A)}\ 24\pi - 36 \sqrt {2} \qquad \mathrm{(B)}\ 24\pi - 24 \sqrt {3} \qquad \mathrm{(C)}\ 36\pi - 36 \sqrt {3} \qquad \mathrm{(D)}\ 36\pi - 24 \sqrt {2} \qquad \mathrm{(E)}\ 48\pi - 36 \sqrt {3}$

Video Solution(Based on Solution 1)

https://www.youtube.com/watch?v=I9RRrumPRK4

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[0],x[1]) -- cycle, lightgray ); draw( circle(s,4) ); dot(s); draw( s -- bottom ); label( "$2$", (mid+bottom)/2, E ); draw ( s -- x[0] -- x[1] -- s ); label( "$4$", (s+x[0])/2, NW ); label( "$4$", (s+x[1])/2, NE ); label( "$A$", s, N ); label( "$B$", x[0], W ); label( "$C$", x[1], 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 $\theta=60^\circ$.

The figure is symmetrical, so the angle $CAB$ has size $2\cdot 60^\circ = 120^\circ$. 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

See also

2008 AMC 10B (Problems • Answer Key • 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
All AMC 10 Problems and Solutions

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