Difference between revisions of "2017 AMC 8 Problems/Problem 25"
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==Solution== | ==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. | 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. | ||
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==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 20:41, 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?
![[asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("$U$", (2,3.464), N); label("$S$", (1,1.732), W); label("$T$", (3,1.732), E); label("$R$", (2,0), S);[/asy]](http://latex.artofproblemsolving.com/7/9/f/79f3ffcfa12364f64959c7aeadc809ba8d2aa6ac.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
