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1985 AHSME Problems/Problem 2: Difference between revisions

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==Problem==
==Problem==
In an arcade game, the "monster" is the shaded sector of a [[circle]] of [[radius]] <math> 1 </math> cm, as shown in the figure. The missing piece (the mouth) has central [[angle]] <math> 60^\circ </math>. What is the [[perimeter]] of the monster in cm?
In an arcade game, the "monster" is the shaded sector of a circle of radius <math>1</math> cm, as shown in the figure. The missing piece (the mouth) has central angle <math>\usepackage{gensymb} 60\degree</math>. What is the perimeter of the monster in cm?


<asy>
<asy>
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==Solution==
==Solution==
First of all, the sum of the lengths of the two radii that make up the mouth is <math> 1+1=2 </math>. There are <math> 360^\circ </math> in a circle, so this figure has a circumference <math> \frac{360^\circ-60^\circ}{360^\circ}=\frac{5}{6} </math> of a full circle. A full circle with radius <math> 1 </math> has a circumference of <math> 2(1)(\pi)=2\pi </math>, so this has a circumference of <math> \left(\frac{5}{6}\right)(2\pi)=\frac{5}{3}\pi </math>. Therefore, the total perimeter is <math> \frac{5}{3}\pi+2, \boxed{\text{E}} </math>.
The two straight sides of the "monster" are radii of length <math>1</math> cm, so their total length is <math>1+1 = 2</math> cm. The circular arc forms <math>\frac{360^{\circ}-60^{\circ}}{360^{\circ}} = \frac{5}{6}</math> of the entire circle, and a circle with radius <math>1</math> cm has circumference <math>2\pi(1) = 2\pi</math> cm, so the length of the arc is <math>\left(\frac{5}{6}\right)(2\pi) = \frac{5}{3}\pi</math> cm. Hence the total perimeter in cm is <math>\boxed{\text{(E)} \ \frac{5}{3}\pi+2}</math>.


==See Also==
==See Also==
{{AHSME box|year=1985|num-b=1|num-a=3}}
{{AHSME box|year=1985|num-b=1|num-a=3}}
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 17:35, 19 March 2024

Problem

In an arcade game, the "monster" is the shaded sector of a circle of radius $1$ cm, as shown in the figure. The missing piece (the mouth) has central angle $\usepackage{gensymb} 60\degree$. What is the perimeter of the monster in cm?

[asy] size(100); defaultpen(linewidth(0.7)); filldraw(Arc(origin,1,30,330)--dir(330)--origin--dir(30)--cycle, yellow, black); label("1", (sqrt(3)/4, 1/4), NW); label("$60^\circ$", (1,0));[/asy]

$\mathrm{(A)\ } \pi+2 \qquad \mathrm{(B) \ }2\pi \qquad \mathrm{(C) \ } \frac{5}{3}\pi \qquad \mathrm{(D) \ } \frac{5}{6}\pi+2 \qquad \mathrm{(E) \ }\frac{5}{3}\pi+2$

Solution

The two straight sides of the "monster" are radii of length $1$ cm, so their total length is $1+1 = 2$ cm. The circular arc forms $\frac{360^{\circ}-60^{\circ}}{360^{\circ}} = \frac{5}{6}$ of the entire circle, and a circle with radius $1$ cm has circumference $2\pi(1) = 2\pi$ cm, so the length of the arc is $\left(\frac{5}{6}\right)(2\pi) = \frac{5}{3}\pi$ cm. Hence the total perimeter in cm is $\boxed{\text{(E)} \ \frac{5}{3}\pi+2}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AHSME Problems and Solutions

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