Difference between revisions of "2008 AMC 10B Problems/Problem 19"
Person1133 (talk | contribs) |
Kat22vic27 (talk | contribs) (→Video Solution(Based on Solution 1)) |
||
(10 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
==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? | + | 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? |
+ | <math> | ||
+ | \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} | ||
+ | </math> | ||
+ | ==Video Solution(Based on Solution 1)== | ||
+ | https://www.youtube.com/watch?v=I9RRrumPRK4 | ||
==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 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:20, 28 December 2020
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?
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:
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 angle has size . Hence the non-shaded part consists of of the circle, minus 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 . The answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.