Difference between revisions of "2008 AMC 10B Problems/Problem 19"
m (→Problem) |
|||
| Line 3: | Line 3: | ||
==Solution== | ==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 <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>. | ||
| + | |||
| + | Thus the outer angle <math>CAB</math> has size <math>360^\circ - 2\cdot 60^\circ = 240^\circ</math>. Hence the non-shaded part consists of <math>\frac{240^\circ}{360^\circ} = \frac 23</math> of the circle, plus 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>AC=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>. | ||
| + | |||
| + | 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>. | ||
| + | |||
==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}} | ||
Revision as of 08:42, 29 January 2009
Problem
A cylindrical tank with radius 
feet and height 
feet is lying on its side. The tank is filled with water to a depth of 
feet. What is the volume of water, in cubic feet?
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]](http://latex.artofproblemsolving.com/6/7/a/67a702ddd98af1b8d46b21db963afe5d1ae0d985.png)
The volume of water can be computed as the height of the tank times the area of the shaded part.
Let 
be the size of the smaller angle 
. We then have 
, hence 
.
Thus the outer angle 
has size 
. Hence the non-shaded part consists of 
of the circle, plus the area of the triangle 
.
Using the Pythagorean theorem we can compute that 
. Thus 
, and the area of the triangle is 
.
The area of the shaded part is then 
, and the volume of water is 
.
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 | ||
