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

(Video Solution(Based on Solution 1))
 
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==Problem==
 
==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?
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A cylindrical tank with radius <math>4</math> feet and height <math>9</math> feet is lying on its side. The tank is filled with water to a depth of <math>2</math> feet. What is the volume of water, in cubic feet?
  
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<math>
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\mathrm{(A)}\ 24\pi - 36 \sqrt {2}
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\qquad
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\mathrm{(B)}\ 24\pi - 24 \sqrt {3}
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\qquad
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\mathrm{(C)}\ 36\pi - 36 \sqrt {3}
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\qquad
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\mathrm{(D)}\ 36\pi - 24 \sqrt {2}
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\qquad
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\mathrm{(E)}\ 48\pi - 36 \sqrt {3}
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</math>
  
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==Video Solution(Based on Solution 1)==
 +
https://www.youtube.com/watch?v=I9RRrumPRK4
  
 
==Solution==
 
==Solution==
{{solution}}
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Any vertical cross-section of the tank parallel with its base looks as follows:
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<asy>
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unitsize(0.8cm);
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defaultpen(0.8);
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pair s=(0,0), bottom=(0,-4), mid=(0,-2);
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pair x[]=intersectionpoints( (-10,-2)--(10,-2), circle(s,4) );
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fill( arc(s,x[0],x[1]) -- cycle, lightgray );
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draw( circle(s,4) );
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dot(s);
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draw( s -- bottom );
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label( "$2$", (mid+bottom)/2, E );
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draw ( s -- x[0] -- x[1] -- s );
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label( "$4$", (s+x[0])/2, NW );
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label( "$4$", (s+x[1])/2, NE );
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label( "$A$", s, N );
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label( "$B$", x[0], W );
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label( "$C$", x[1], E );
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label( "$D$", mid, NW );
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label( "$E$", bottom, S );
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</asy>
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The volume of water can be computed as the height of the tank times the area of the shaded part.
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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>.
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Thus the angle <math>CAB</math> has size <math>360^\circ - 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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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>.
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The area of the shaded part is then <math>\frac{4^2\pi}3 - 4\sqrt 3</math>, and the volume of water is <math>9\cdot\left(\frac{4^2\pi}3 - 4\sqrt 3\right) = \boxed{48\pi - 36\sqrt 3}</math>. The answer is <math>\text{E}</math>.
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Alternatively, we can solve for DC and see it is congruent to a 30 60 90 triangle by SSS ~Williamgolly
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=18|num-a=20}}
 
{{AMC10 box|year=2008|ab=B|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 12:20, 28 December 2020

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 $\theta=60^\circ$.

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

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 10 Problems and Solutions

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