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 the figure into an equilateral triangle with side length 4. Using the area formula, this has area (16sqrt3)/4 which is 4sqrt3. We are given TUS as a 60 degree angle so the two things we need to subtract off are 1/6 circles. Their radius is 2, so there area is 1/6*(4pi) which is 2pi/3. Subtract to get 4sqrt3-2pi/3, which is <math>D</math>. | ||
==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 20:29, 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 the figure into an equilateral triangle with side length 4. Using the area formula, this has area (16sqrt3)/4 which is 4sqrt3. We are given TUS as a 60 degree angle so the two things we need to subtract off are 1/6 circles. Their radius is 2, so there area is 1/6*(4pi) which is 2pi/3. Subtract to get 4sqrt3-2pi/3, which is .
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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