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 03:05, 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?
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.