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Difference between revisions of "2023 AMC 8 Problems/Problem 12"
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Pretend each circle is a square. The second largest circle is a square with area <math>16\text{units}^2</math> and there are two squares in that square that each have area <math>4\text{units}^2</math> which add up to 8. Subtracting the medium-sized squares' areas from the second-largest square's area, we have <math>8\text{units}^2</math>. The largest circle becomes a square that has area <math>36\text{units}^2</math>, and the three smallest circles become three squares with area <math>8\text{units}^2</math> and add up to <math>3
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text{units}^2</math>. Adding the areas of the shaded regions we get <math>11\text{units}^2</math>, so our answer is <math>\boxed{\text{(B)}\dfrac{11}{36}}</math>.
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-claregu
  
  

Revision as of 21:57, 24 January 2023

First the total area of the $3$ radius circle is simply just $9* \pi$. 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 $3$ $\frac{1}{2}$ radius circles and add then take the area of the $2$ radius circle and subtracting that from the area of the $2$, 1 radius circles to get our resulting complex area shape. Adding these up we will get $3 * \frac{1}{4} \pi + 4 \pi -\pi - \pi = \frac{3}{4} \pi + 2 \pi = \frac{11}{4}$

Our answer is $\frac {\frac{11}{4} \pi}{9 \pi} = \boxed{\text{(B)}\frac{11}{36}}$

~apex304

Pretend each circle is a square. The second largest circle is a square with area $16\text{units}^2$ and there are two squares in that square that each have area $4\text{units}^2$ which add up to 8. Subtracting the medium-sized squares' areas from the second-largest square's area, we have $8\text{units}^2$. The largest circle becomes a square that has area $36\text{units}^2$, and the three smallest circles become three squares with area $8\text{units}^2$ and add up to $3 text{units}^2$. Adding the areas of the shaded regions we get $11\text{units}^2$, so our answer is $\boxed{\text{(B)}\dfrac{11}{36}}$.

-claregu


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