2017 AMC 8 Problems/Problem 25

Problem 25

In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$ . Arcs $\overarc{TR}$ and $\overarc{SR}$ 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$", (2,0), S);[/asy]$

$\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$

Solution

We can extend $\overline{US}$ and $\overline{UT}$ into an equilateral triangle of side length $4$ with area $\dfrac{\sqrt{3}}{4}\cdot4^2=4\sqrt{3}$ . We then must subtract the two 1/6-circles, which each have radius $2$ , so their total area is $2\cdot\dfrac{\pi\cdot2^2}{6}=\dfrac{4\pi}{3}$ . Our answer is therefore $\boxed{\textbf{(B)}\ 4\sqrt{3}-\dfrac{4\pi}{3}}$ .

2017 AMC 8 (Problems • Answer Key • Resources)
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2017 AMC 8 Problems/Problem 25

Problem 25

Solution

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