Difference between revisions of "1992 AHSME Problems/Problem 26"
m (→See also) |
(Added a solution with explanation) |
||
Line 23: | Line 23: | ||
== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | <math>\fbox{B}</math> The area of the entire outer shape is the area of sector <math>ABE</math>, plus the area of sector <math>ABF</math>, minus the area of triangle <math>ABD</math> (since it is part of both sectors), plus the area of sector <math>DEF</math>. We know <math>AC = CD = 1</math>, so the sector angles for <math>ABE</math> and <math>ABF</math> are <math>45</math> degrees, and the radius of both of them is <math>2</math>. The radius of <math>DEF</math> is <math>DE = BE - BD = 2 - BD</math>, and <math>BD</math> can be found using Pythagoras in triangle <math>BCD</math>, giving <math>BD = \sqrt{2}</math> and <math>DE = 2 - \sqrt{2}</math>, so after doing all the calculations, the area of the entire outer shape is <math>\pi(\frac{5}{2} - \sqrt{2}) - 1</math>. To get the area of the smile, we need to subtract the area of semicircle <math>ABD</math>, which is <math>\frac{1}{2} \pi 1^2 = \frac{\pi}{2}</math>, so the answer is <math>\pi(\frac{5}{2} - \frac{1}{2} - \sqrt{2}) - 1</math> = <math>2\pi - \pi \sqrt{2} - 1</math>. |
== See also == | == See also == |
Latest revision as of 13:29, 20 February 2018
Problem
Semicircle has center and radius . Point is on and . Extend and to and , respectively, so that circular arcs and have and as their respective centers. Circular arc has center . The area of the shaded "smile" , is
Solution
The area of the entire outer shape is the area of sector , plus the area of sector , minus the area of triangle (since it is part of both sectors), plus the area of sector . We know , so the sector angles for and are degrees, and the radius of both of them is . The radius of is , and can be found using Pythagoras in triangle , giving and , so after doing all the calculations, the area of the entire outer shape is . To get the area of the smile, we need to subtract the area of semicircle , which is , so the answer is = .
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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.