Difference between revisions of "2008 AMC 8 Problems/Problem 25"
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</asy> | </asy> | ||
<math> \textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 48\qquad</math> | <math> \textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 48\qquad</math> | ||
| + | |||
| + | ==Solution== | ||
| + | <cmath>\begin{tabular}{c|cc} | ||
| + | \text{circle #} & radius & area \\ \hline | ||
| + | 1 & 2 & 4\pi \\ | ||
| + | 2 & 4 & 16\pi \\ | ||
| + | 3 & 6 & 36\pi \\ | ||
| + | 4 & 8 & 64\pi \\ | ||
| + | 5 & 10 & 100\pi \\ | ||
| + | 6 & 12 & 144\pi | ||
| + | \end{tabular}</cmath> | ||
| + | |||
| + | The entire circle's area is <math>144\pi</math>. The area of the black regions is <math>(100-64)\pi + (36-16)\pi + 4\pi = 60\pi</math>. The percentage of the design that is black is <math>\frac{60\pi}{144\pi} = \frac{5}{12} \approx \boxed{\textbf{(A)}\ 42}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=24|after=Last Problem}} | {{AMC8 box|year=2008|num-b=24|after=Last Problem}} | ||
Revision as of 03:20, 25 December 2012
Problem
Margie's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Approximately what percent of the design is black?
Solution
\[\begin{tabular}{c|cc}
\text{circle #} & radius & area \\ \hline
1 & 2 & 4\pi \\
2 & 4 & 16\pi \\
3 & 6 & 36\pi \\
4 & 8 & 64\pi \\
5 & 10 & 100\pi \\
6 & 12 & 144\pi
\end{tabular}\] (Error compiling LaTeX. Unknown error_msg)
The entire circle's area is
. The area of the black regions is
. The percentage of the design that is black is
.
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 | ||