Difference between revisions of "2008 AMC 8 Problems/Problem 21"
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The slice is cutting the cylinder into two equal wedges with equal area. The cylinder's volume is <math>\pi r^2 h = \pi (4^2)(6) = 96\pi</math>. The area of the wedge is half this which is <math>48\pi \approx \boxed{\textbf{(C)}\ 151}</math>. | The slice is cutting the cylinder into two equal wedges with equal area. The cylinder's volume is <math>\pi r^2 h = \pi (4^2)(6) = 96\pi</math>. The area of the wedge is half this which is <math>48\pi \approx \boxed{\textbf{(C)}\ 151}</math>. | ||
| − | == Video Solution == | + | == Video Solution by Omega Learn == |
https://youtu.be/FDgcLW4frg8?t=2556 | https://youtu.be/FDgcLW4frg8?t=2556 | ||
Revision as of 02:05, 9 November 2022
Problem
Jerry cuts a wedge from a 
-cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?
![[asy] defaultpen(linewidth(0.65)); real d=90-63.43494882; draw(ellipse((origin), 2, 4)); fill((0,4)--(0,-4)--(-8,-4)--(-8,4)--cycle, white); draw(ellipse((-4,0), 2, 4)); draw((0,4)--(-4,4)); draw((0,-4)--(-4,-4)); draw(shift(-2,0)*rotate(-d-5)*ellipse(origin, 1.82, 4.56), linetype("10 10")); draw((-4,4)--(-8,4), dashed); draw((-4,-4)--(-8,-4), dashed); draw((-4,4.3)--(-4,5)); draw((0,4.3)--(0,5)); draw((-7,4)--(-7,-4), Arrows(5)); draw((-4,4.7)--(0,4.7), Arrows(5)); label("$8$ cm", (-7,0), W); label("$6$ cm", (-2,4.7), N);[/asy]](http://latex.artofproblemsolving.com/0/d/b/0db552e969b4c23a74bc8bb14087a79249bd36a1.png)
Solution
The slice is cutting the cylinder into two equal wedges with equal area. The cylinder's volume is 
. The area of the wedge is half this which is 
.
Video Solution by Omega Learn
https://youtu.be/FDgcLW4frg8?t=2556
~ pi_is_3.14
https://www.youtube.com/watch?v=q7OCpOn1abY
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
