Difference between revisions of "2008 AMC 8 Problems/Problem 21"
Line 18: | Line 18: | ||
label("$8$ cm", (-7,0), W); | label("$8$ cm", (-7,0), W); | ||
label("$6$ cm", (-2,4.7), N);</asy> | label("$6$ cm", (-2,4.7), N);</asy> | ||
+ | |||
+ | <math>\textbf{(A)} \ 48 \qquad | ||
+ | \textbf{(B)} \ 75 \qquad | ||
+ | \textbf{(C)} \ 151\qquad | ||
+ | \textbf{(D)} \ 192 \qquad | ||
+ | \textbf{(E)} \ 603</math> | ||
+ | |||
+ | ==Solution== | ||
+ | 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>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=20|num-a=22}} | {{AMC8 box|year=2008|num-b=20|num-a=22}} | ||
− |
Revision as of 02:56, 25 December 2012
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 .
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 |