Difference between revisions of "2008 AMC 8 Problems/Problem 25"
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filldraw(circle(A5, 1), black, black); | filldraw(circle(A5, 1), black, black); | ||
</asy> | </asy> | ||
− | <math> \textbf{(A)}\ | + | |
+ | <math> \textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 48\qquad</math> | ||
==Solution== | ==Solution== | ||
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\end{array}</cmath> | \end{array}</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)}\ | + | 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>. |
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/j3QSD5eDpzU?t=3363 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=24|after=Last Problem}} | {{AMC8 box|year=2008|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:38, 2 January 2023
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. Which of the following is closest to the percent of the design that is black?
Solution
Let the smallest circle be 1, the second smallest circle be 2, the third smallest circle be 3, etc.
The entire circle's area is . The area of the black regions is . The percentage of the design that is black is .
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=3363
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.