Difference between revisions of "1985 AHSME Problems/Problem 2"
Sevenoptimus (talk | contribs) m (Improved formatting) |
|||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | In an arcade game, the "monster" is the shaded sector of a | + | 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> | ||
Line 12: | Line 12: | ||
==Solution== | ==Solution== | ||
− | + | 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 cm, as shown in the figure. The missing piece (the mouth) has central angle . What is the perimeter of the monster in cm?
Solution
The two straight sides of the "monster" are radii of length cm, so their total length is cm. The circular arc forms of the entire circle, and a circle with radius cm has circumference cm, so the length of the arc is cm. Hence the total perimeter in cm is .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.