Difference between revisions of "2023 AMC 8 Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
− | First the total area of the <math>3</math> | + | First, the total area of the radius <math>3</math> circle is simply just <math>9* \pi</math> when 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 <math>3</math> <math>\frac{1}{2}</math> radius circles and add then take the area of the <math>2</math> radius circle and | + | Now from here, we have to find our shaded area. This can be done by adding the areas of the <math>3</math> <math>\frac{1}{2}</math>-radius circles and add; then, take the area of the <math>2</math> radius circle and subtract that from the area of the <math>2</math> radius 1 circles to get our resulting complex area shape. Adding these up, we will get <math>3 * \frac{1}{4} \pi + 4 \pi -\pi - \pi = \frac{3}{4} \pi + 2 \pi = \frac{11}{4}</math>. |
− | + | So, our answer is <math>\frac {\frac{11}{4} \pi}{9 \pi} = \boxed{\textbf{(B)}\ \frac{11}{36}}</math>. | |
~apex304 | ~apex304 |
Revision as of 02:28, 16 February 2023
Contents
[hide]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 when 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 subtract that from the area of the radius 1 circles to get our resulting complex area shape. Adding these up, we will get .
So, 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
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=UWoUhV5T92Y
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.