2017 AMC 8 Problem 25

Question

In the figure shown,

\overline{US}

and

\overline{UT}

are line segments each of length

2,

and

m\angle TUS = 60^\circ.

Arcs

\overset{\large\frown}{TR}

and

\overset{\large\frown}{SR}

are each one-sixth of a circle with radius

2.

What is the area of the region shown?

3\sqrt{3}-\pi

4\sqrt{3}-\dfrac{4\pi}{3}

2\sqrt{3}

4\sqrt{3}-\dfrac{2\pi}{3}

4+\dfrac{4\pi}{3}

Extend the two straight sides to form an equilateral triangle.

Subtract two

60^\circ

sectors of radius

2

from that equilateral triangle.

Click to load, then click again to play

We can extend

\overline{SU}

and

\overline{TU}

to form the following picture.

The area of this region is the area of an equilateral triangle with side length of

4

minus the area of two-sixths of a circle with radius

2.

The area for an equilateral triangle with side length

s

is

\dfrac{s^2\sqrt{3}}{4}.

This means that the total area is

\dfrac{4^2 \sqrt{3}}{4} - \dfrac{1}{3} \pi \cdot 2^2 = 4 \sqrt{3} - \dfrac{4}{3} \pi.

Thus, B is the correct answer.

Problem 25 in other years