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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?

Area – Geometry

Easy

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Try to find the virtuals and complete the Figure : \( \\ \)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 :

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 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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