Difference between revisions of "2008 AMC 8 Problems/Problem 25"
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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 | + | The entire circle's area is <math>144\pi</math>. The area of the black region 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)}\ 41.7}</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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:54, 24 October 2020
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?
![[asy] real d=320; pair O=origin; pair P=O+8*dir(d); pair A0 = origin; pair A1 = O+1*dir(d); pair A2 = O+2*dir(d); pair A3 = O+3*dir(d); pair A4 = O+4*dir(d); pair A5 = O+5*dir(d); filldraw(Circle(A0, 6), white, black); filldraw(circle(A1, 5), black, black); filldraw(circle(A2, 4), white, black); filldraw(circle(A3, 3), black, black); filldraw(circle(A4, 2), white, black); filldraw(circle(A5, 1), black, black); [/asy]](http://latex.artofproblemsolving.com/3/4/4/34420ee5e67e87663563a3047fde204ba36cff7f.png)
Solution
![\[\begin{array}{c|cc} \text{circle \#} & \text{radius} & \text{area} \\ \hline 1 & 2 & 4\pi \\ 2 & 4 & 16\pi \\ 3 & 6 & 36\pi \\ 4 & 8 & 64\pi \\ 5 & 10 & 100\pi \\ 6 & 12 & 144\pi \end{array}\]](http://latex.artofproblemsolving.com/8/1/4/8146393d4a39389bc89d9f904b246e939af97714.png)
The entire circle's area is 
. The area of the black region 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
