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?

$3\sqrt{3}-\pi$

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

$2\sqrt{3}$

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

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

Answer: B

Concepts:equilateral triangle sector area decomposition

Difficulty rating: 1750

Video solution:

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Written solution:

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.

2017 AMC 8 Problem 25

Answer: B

Difficulty rating: 1750

Video solution:

Written solution:

Problem 25 in Other Years