Difference between revisions of "2023 AMC 8 Problems/Problem 12"
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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. | 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 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>. | + | 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 * \pi}{4}</math>. |
So, our answer is <math>\frac {\frac{11}{4} \pi}{9 \pi} = \boxed{\textbf{(B)}\ \frac{11}{36}}</math>. | So, our answer is <math>\frac {\frac{11}{4} \pi}{9 \pi} = \boxed{\textbf{(B)}\ \frac{11}{36}}</math>. | ||
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==Solution 2== | ==Solution 2== | ||
− | Pretend each circle is a square. The | + | Pretend each circle is a square. The large shaded circle is a square with area <math>16~\text{units}^2</math>, and the two white circles inside it each have areas of <math>4~\text{units}^2</math>, which adds up to <math>8~\text{units}^2</math>. The three small shaded circles become three squares with area <math>1~\text{units}^2</math>, and add up to <math>3~\text{units}^2</math>. Adding the areas of the shaded circles (19) and subtracting the areas of the white circles (8), we get <math>11~\text{units}^2</math>. Since the largest white circle in which all these other circles are becomes a square that has area <math>36~\text{units}^2</math>, our answer is <math>\boxed{\textbf{(B)}\ \dfrac{11}{36}}</math>. |
-claregu | -claregu | ||
− | LaTeX (edits -apex304 | + | LaTeX (edits -apex304, CoOlPoTaToEs) |
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==Video Solution by Math-X (How to do this question under 30 seconds)== | ==Video Solution by Math-X (How to do this question under 30 seconds)== | ||
https://youtu.be/Ku_c1YHnLt0?si=stUHQ9nHZZE_x-CC&t=1852 ~Math-X | https://youtu.be/Ku_c1YHnLt0?si=stUHQ9nHZZE_x-CC&t=1852 ~Math-X | ||
+ | ==Video Solution (Solve under 60 seconds!!!)== | ||
+ | https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=539 | ||
+ | ~hsnacademy | ||
==Video Solution (HOW TO THINK CREATIVELY!!!) == | ==Video Solution (HOW TO THINK CREATIVELY!!!) == | ||
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~harungurcan | ~harungurcan | ||
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+ | ==Video Solution by Dr. David== | ||
+ | https://youtu.be/2Ih7F0XHmls | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=11|num-a=13}} | {{AMC8 box|year=2023|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:09, 27 October 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution by Math-X (How to do this question under 30 seconds)
- 5 Video Solution (Solve under 60 seconds!!!)
- 6 Video Solution (HOW TO THINK CREATIVELY!!!)
- 7 Video Solution (Animated)
- 8 Video Solution by Magic Square
- 9 Video Solution by SpreadTheMathLove
- 10 Video Solution by Interstigation
- 11 Video Solution by harungurcan
- 12 Video Solution by Dr. David
- 13 See Also
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 large shaded circle is a square with area , and the two white circles inside it each have areas of , which adds up to . The three small shaded circles become three squares with area , and add up to . Adding the areas of the shaded circles (19) and subtracting the areas of the white circles (8), we get . Since the largest white circle in which all these other circles are becomes a square that has area , our answer is .
-claregu LaTeX (edits -apex304, CoOlPoTaToEs)
Video Solution by Math-X (How to do this question under 30 seconds)
https://youtu.be/Ku_c1YHnLt0?si=stUHQ9nHZZE_x-CC&t=1852 ~Math-X
Video Solution (Solve under 60 seconds!!!)
https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=539
~hsnacademy
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education the Study of everything
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
Video Solution by Interstigation
https://youtu.be/DBqko2xATxs&t=1137
Video Solution by harungurcan
https://www.youtube.com/watch?v=oIGy79w1H8o&t=1154s
~harungurcan
Video Solution by Dr. David
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.