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# Understand the problem

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? $[asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("U", (2,3.464), N); label("S", (1,1.732), W); label("T", (3,1.732), E); label("R", (2,0), S);[/asy]$

# 2017 AMC(American Mathematical Contests) 8 Problem 25

Area – Geometry
##### Difficulty Level
Easy

Do you really need a hint? Try it first!

Try to find the virtuals and complete the Figure : $[asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("U", (2,3.464), N); label("S", (1,1.732), W); label("T", (3,1.732), E); label("R", (2,0), S);[/asy]$ $$\\$$To complete the figure emphasize on $$\angle SUT = 60^{\circ}$$ and arcs are $$\frac {1}{6} th$$ of a circle of radius 2.
Completed figure :   $[asy] unitsize(1 cm); pair U,S,T,R,X,Y; U =(2,3.464); S=(1,1.732); T=(3,1.732); R=(2,0); X=(0,0); Y=(4,0); draw(U--S); draw(S--U--T); draw(S--X--Y--T,red); draw(arc(X,R,S),red); draw(arc(Y,T,R),red); label("U",U, N); label("S", S, W); label("T", T, E); label("R", R, S); label("X",X, W); label("Y", Y, E); [/asy]$
$[asy] unitsize(1 cm); pair U,S,T,R,X,Y; U =(2,3.464); S=(1,1.732); T=(3,1.732); R=(2,0); X=(0,0); Y=(4,0); draw(U--S); draw(S--U--T); draw(S--X--Y--T,red); draw(arc(X,R,S),red); draw(arc(Y,T,R),red); label("U",U, N); label("S", S, W); label("T", T, E); label("R", R, S); label("X",X, W); label("Y", Y, E); [/asy]$   We can clearly see that $$\triangle UXT$$ is an equilateral triangle with sidelength $$4$$ as  given $$\angle SUT = 60^{\circ}$$ and $$\angle SXT \ and \ \angle TYR$$ are of $$60^{\circ}$$  .[As  the  $$arc SR \ and \ arc TR \ are \ \frac {1}{6} th \ of \ a \ circle \ of \ radius \ 2$$  ] .
Clearly the area of the figure is equal to  $$area(\triangle UXY )$$ minus the combined area of the 2 sectors of the circles (in red) [i.e. $$SXR \ region \ and \ TRY \ region$$  ]. Using the area formula for an equilateral triangle $$\frac {a^2 \sqrt 3}{4}$$where $$a$$ is the side length of the equilateral triangle, $$area(\triangle UXY )$$  is  $$\frac {4^2 \sqrt 3}{4} = 4 \sqrt 3$$ . The combined area of the $2$ sectors is  $$2 \times \frac {1}{6} \times \pi r^2$$ , which is   $$\frac {1}{3} \pi \times 2^2$$  . Thus, our final answer is  $$4 \sqrt {3} – \frac {4 \pi}{3}$$ .

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