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

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label("$S'$", (0,0), W);
 
label("$S'$", (0,0), W);
 
label("$T'$", (4,0), E);</asy>
 
label("$T'$", (4,0), E);</asy>
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We can clearly see that <math>\triangle AS'T'</math> is an equilateral triangle, because two of its angles are <math>60^\circ</math>, which is the degree measure of <math>\frac{1}{6}</math> a circle. 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. Using the area formula for an equilateral triangle, <math>\frac{\sqrt{3}}{4} \cdot a,</math> where <math>a</math> is the side length of the equilateral triangle, 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</math> times one sixth <math>\pi \cdot r^2</math>, which 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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We can clearly see that <math>\triangle US'T'</math> is an equilateral triangle, because two of its angles are <math>60^\circ</math>, which is the degree measure of <math>\frac{1}{6}</math> a circle. 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. Using the area formula for an equilateral triangle, <math>\frac{\sqrt{3}}{4} \cdot a,</math> where <math>a</math> is the side length of the equilateral triangle, 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</math> times one sixth <math>\pi \cdot r^2</math>, which 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>
  
 
==See Also==
 
==See Also==

Revision as of 19:20, 30 January 2018

Problem 25

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]

$\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$

Solution

Let the centers of the circles containing arcs $\overarc{SR}$ and $\overarc{TR}$ be $S'$ and $T'$, respectively. Extend $\overline{US}$ and $\overline{UT}$ to $S'$ and $T'$, and connect point $S'$ with point $T'$. [asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw((1,1.732)--(0,0)--(4,0)--(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); label("$S'$", (0,0), W); label("$T'$", (4,0), E);[/asy] We can clearly see that $\triangle US'T'$ is an equilateral triangle, because two of its angles are $60^\circ$, which is the degree measure of $\frac{1}{6}$ a circle. The area of the figure is equal to the area of equilateral triangle $\triangle US'T'$ minus the combined area of the $2$ sectors of the circles. Using the area formula for an equilateral triangle, $\frac{\sqrt{3}}{4} \cdot a,$ where $a$ is the side length of the equilateral triangle, the area of $\triangle US'T'$ is $\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.$ The combined area of the $2$ sectors is $2$ times one sixth $\pi \cdot r^2$, which is $2 \cdot \frac 16 \cdot \pi \cdot 2^2 = \frac{4\pi}{3}.$ Our final answer is then $\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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