Difference between revisions of "2025 AMC 8 Problems/Problem 18"
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| + | The area of the shaded region in the circle on the left is the area of the circle minus the area of the square, or <math>\big(\pi-2)</math>. The shaded area in the circle on the right is <math>\dfrac{1}{4}</math> of the area of the circle minus the area of the square, or <math>\dfrac{\pi R^2-2R^2}{4}</math>, which can be factored as <math>\dfrac{R^2(\pi-2)}{4}</math>. Since the shaded areas are equal to each other, we have <math>\pi-2=\dfrac{R^2(\pi-2)}{4}</math>, which simplifies to <math>R^2=4</math>. Taking the square root, we have <math>R=\boxed{\text{(B)\ 2}}</math> | ||
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| + | ~mrtnvlknv | ||
Revision as of 21:15, 29 January 2025
Solution
The area of the shaded region in the circle on the left is the area of the circle minus the area of the square, or 
. The shaded area in the circle on the right is 
of the area of the circle minus the area of the square, or 
, which can be factored as 
. Since the shaded areas are equal to each other, we have 
, which simplifies to 
. Taking the square root, we have 
~mrtnvlknv
