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

(Solution)
(almost everything)
Line 7: Line 7:
 
<math>\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}</math>
 
<math>\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}</math>
 
==Solution==
 
==Solution==
Extend into an equilateral triangle with area 4sqrt3 by the area formula. We must subtract off the two 1/3-circles, which have radius 2, so there area is 4pi/3. So it's 4sqrt-4pi/3, or B.
+
We can extend <math>\overline{US}</math> and <math>\overline{UT}</math> into an equilateral triangle of side length <math>4</math> with area <math>\dfrac{\sqrt{3}}{4}\cdot4^2=4\sqrt{3}</math>. We then must subtract the two 1/6-circles, which each have radius <math>2</math>, so their total area is <math>2\cdot\dfrac{\pi\cdot2^2}{6}=\dfrac{4\pi}{3}</math>. Our answer is therefore <math>\boxed{\textbf{(B)}\ 4\sqrt{3}-\dfrac{4\pi}{3}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 00:46, 11 November 2019

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

We can extend $\overline{US}$ and $\overline{UT}$ into an equilateral triangle of side length $4$ with area $\dfrac{\sqrt{3}}{4}\cdot4^2=4\sqrt{3}$. We then must subtract the two 1/6-circles, which each have radius $2$, so their total area is $2\cdot\dfrac{\pi\cdot2^2}{6}=\dfrac{4\pi}{3}$. Our answer is therefore $\boxed{\textbf{(B)}\ 4\sqrt{3}-\dfrac{4\pi}{3}}$.

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS