Difference between revisions of "2017 AMC 8 Problems/Problem 25"
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<math>\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}</math> | <math>\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}</math> | ||
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==Solution 2== | ==Solution 2== | ||
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==Video Solutions== | ==Video Solutions== | ||
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https://youtu.be/sVclz6EmpEU | https://youtu.be/sVclz6EmpEU | ||
~savannahsolver | ~savannahsolver | ||
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Latest revision as of 11:06, 24 July 2024
Problem
In the figure shown, and
are line segments each of length 2, and
. Arcs
and
are each one-sixth of a circle with radius 2. What is the area of the region shown?
Solution 2
In addition to the given diagram, we can draw lines and
The area of rhombus
is half the product of its diagonals, which is
. However, we have to subtract off the circular segments. The area of those can be found by computing the area of the circle with radius 2, multiplying it by
, then finally subtracting the area of an equilateral triangle with a side length 2 from the sector. The sum of the areas of the circular segments is
The area of rhombus
minus the circular segments is
~PEKKA
Video Solutions
~savannahsolver