2017 AMC 8 Problem 25 Area-Geometry

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]  
Source of the problem

2017 AMC(American Mathematical Contests) 8 Problem 25

Topic
Area – Geometry
Difficulty Level
Easy

Start with hints

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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