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> | ||
− | + | ==Solution== | |
+ | Extend into an equilateral triangle with area 4sqrt3 by the area formula. We must subtract off the two 1/3-circles, which have radius 2, so there area is 4pi/3. So it's 4sqrt-4pi/3, or B. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2017|num-b=24|after=Last Problem}} | {{AMC8 box|year=2017|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:31, 10 November 2019
Problem 25
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
Extend into an equilateral triangle with area 4sqrt3 by the area formula. We must subtract off the two 1/3-circles, which have radius 2, so there area is 4pi/3. So it's 4sqrt-4pi/3, or B.
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
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