Difference between revisions of "2023 AMC 8 Problems/Problem 12"
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Themathguyd (talk | contribs) (→Problem: Replaced pixelated image with asy perfection.) |
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interior. What fraction of the interior of the large white circle is shaded? | interior. What fraction of the interior of the large white circle is shaded? | ||
− | + | <asy> | |
+ | // Diagram by TheMathGuyd | ||
+ | size(6cm); | ||
+ | draw(circle((3,3),3)); | ||
+ | filldraw(circle((2,3),2),lightgrey); | ||
+ | filldraw(circle((3,3),1),white); | ||
+ | filldraw(circle((1,3),1),white); | ||
+ | filldraw(circle((5.5,3),0.5),lightgrey); | ||
+ | filldraw(circle((4.5,4.5),0.5),lightgrey); | ||
+ | filldraw(circle((4.5,1.5),0.5),lightgrey); | ||
+ | int i, j; | ||
+ | for(i=0; i<7; i=i+1) | ||
+ | { | ||
+ | draw((0,i)--(6,i), dashed+grey); | ||
+ | draw((i,0)--(i,6), dashed+grey); | ||
+ | } | ||
+ | </asy> | ||
+ | |||
<math>\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{11}{36} \qquad \textbf{(C)}\ \frac{1}{3} \qquad \textbf{(D)}\ \frac{19}{36} \qquad \textbf{(E)}\ \frac{5}{9}</math> | <math>\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{11}{36} \qquad \textbf{(C)}\ \frac{1}{3} \qquad \textbf{(D)}\ \frac{19}{36} \qquad \textbf{(E)}\ \frac{5}{9}</math> |
Revision as of 15:37, 25 January 2023
Contents
Problem
The figure below shows a large white circle with a number of smaller white and shaded circles in its interior. What fraction of the interior of the large white circle is shaded?
Solution 1
First the total area of the radius circle is simply just
. Using our area of a circle formula.
Now from here we have to find our shaded area. This can be done by adding the areas of the
radius circles and add then take the area of the
radius circle and subtracting that from the area of the
, 1 radius circles to get our resulting complex area shape. Adding these up we will get
Our answer is
~apex304
Solution 2
Pretend each circle is a square. The second largest circle is a square with area and there are two squares in that square that each have area
which add up to
. Subtracting the medium-sized squares' areas from the second-largest square's area, we have
. The largest circle becomes a square that has area
, and the three smallest circles become three squares with area
and add up to
. Adding the areas of the shaded regions we get
, so our answer is
.
-claregu LaTeX edits -apex304
Video Solution (Animated)
~Star League (https://starleague.us)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=4590
See Also
2023 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.