2023 AMC 8 Problems/Problem 12: Difference between revisions
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Start by finding the area of each of the smaller circles. Using the formula /pi~\text{r}^2<math>, we have that each of the smaller shaded circles have area 0.25/pi~\text{units}^2</math>, for a total area of 0.75/pi~\text{units}^2<math>. The bigger shaded circle has area 4/pi~\text{units}^2</math>, and the two medium white circles each have area /pi~\text{units}^2<math>, for a total of 2/pi~\text{units}^2</math>. Adding the shaded values (0.75/pi + 4/pi = 4.75/pi), and subtracting the white value (4.75/pi - 2/pi = 2.75/pi), we get that there is a total of 2.75/pi~\text{units}^2 shaded area. | Start by finding the area of each of the smaller circles. Using the formula /pi~\text{r}^2<math>, we have that each of the smaller shaded circles have area 0.25/pi~\text{units}^2</math>, for a total area of 0.75/pi~\text{units}^2<math>. The bigger shaded circle has area 4/pi~\text{units}^2</math>, and the two medium white circles each have area /pi~\text{units}^2<math>, for a total of 2/pi~\text{units}^2</math>. Adding the shaded values (0.75/pi + 4/pi = 4.75/pi), and subtracting the white value (4.75/pi - 2/pi = 2.75/pi), we get that there is a total of 2.75/pi~\text{units}^2 shaded area. | ||
Using the same formula, we have that the big white circle has area 9/pi~\text{units}^2 | Using the same formula, we have that the big white circle has area 9/pi~\text{units}^2<math>. Dividing /frac{2.75/pi}{9/pi}, we simplify to </math>\boxed{\textbf{(B)}\ \dfrac{11}{36}}$. | ||
==Video Solution by Math-X (How to do this question under 30 seconds)== | ==Video Solution by Math-X (How to do this question under 30 seconds)== | ||
Revision as of 13:05, 21 January 2024
Problem
The figure below shows a large white circle with a number of smaller white and shaded circles in its interior. What fraction of the interior of the large white circle is shaded?
Solution 1
First, the total area of the radius
circle is simply just
when using our area of a circle formula.
Now from here, we have to find our shaded area. This can be done by adding the areas of the
-radius circles and add; then, take the area of the
radius circle and subtract that from the area of the
radius 1 circles to get our resulting complex area shape. Adding these up, we will get
.
So, our answer is
.
~apex304
Solution 2
Pretend each circle is a square. The second largest circle is a square with area
and there are two squares in that square that each has areas of
which add up to
. Subtracting the medium-sized squares' areas from the second-largest square's area, we have
. The largest circle becomes a square that has area
, and the three smallest circles become three squares with area
and add up to
. Adding the areas of the shaded regions, we get
, so our answer is
.
-claregu LaTeX (edits -apex304)
Solution 3 (Similar to 2)
Start by finding the area of each of the smaller circles. Using the formula /pi~\text{r}^2
, for a total area of 0.75/pi~\text{units}^2
, and the two medium white circles each have area /pi~\text{units}^2
. Adding the shaded values (0.75/pi + 4/pi = 4.75/pi), and subtracting the white value (4.75/pi - 2/pi = 2.75/pi), we get that there is a total of 2.75/pi~\text{units}^2 shaded area.
Using the same formula, we have that the big white circle has area 9/pi~\text{units}^2
\boxed{\textbf{(B)}\ \dfrac{11}{36}}$.
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 (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
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing