# Difference between revisions of "2017 AMC 8 Problems/Problem 25"

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Let the centers of the circles containing arcs <math>\overarc{SR}</math> and <math>\overarc{TR}</math> be <math>S'</math> and <math>T'</math>, respectively. Extend <math>\overline{US}</math> and <math>\overline{UT}</math> to <math>S'</math> and <math>T'</math>. The area of the figure is equal to the area of equilateral triangle <math>\triangle US'T'</math> minus the combined area of the <math>2</math> sectors of the circles. The area of <math>\triangle US'T'</math> is <math>\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.</math> The combined area of the <math>2</math> sectors is <math>2 \cdot \frac 16 \cdot \pi \cdot 2^2 = \frac{4\pi}{3}.</math> Our final answer is then <math>\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.</math> | Let the centers of the circles containing arcs <math>\overarc{SR}</math> and <math>\overarc{TR}</math> be <math>S'</math> and <math>T'</math>, respectively. Extend <math>\overline{US}</math> and <math>\overline{UT}</math> to <math>S'</math> and <math>T'</math>. The area of the figure is equal to the area of equilateral triangle <math>\triangle US'T'</math> minus the combined area of the <math>2</math> sectors of the circles. The area of <math>\triangle US'T'</math> is <math>\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.</math> The combined area of the <math>2</math> sectors is <math>2 \cdot \frac 16 \cdot \pi \cdot 2^2 = \frac{4\pi}{3}.</math> Our final answer is then <math>\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.</math> | ||

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## Revision as of 13:51, 22 November 2017

## Problem 25

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?

[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("", (2,3.464), N); label("", (1,1.732), W); label("", (3,1.732), E); label("", (2,0), S);[/asy]

## Solution

Let the centers of the circles containing arcs and be and , respectively. Extend and to and . The area of the figure is equal to the area of equilateral triangle minus the combined area of the sectors of the circles. The area of is The combined area of the sectors is Our final answer is then

~nukelauncher