Difference between revisions of "2000 AMC 8 Problems/Problem 19"
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+ | == Problem == | ||
Three circular arcs of radius <math>5</math> units bound the region shown. Arcs <math>AB</math> and <math>AD</math> are quarter-circles, and arc <math>BCD</math> is a semicircle. What is the area, in square units, of the region? | Three circular arcs of radius <math>5</math> units bound the region shown. Arcs <math>AB</math> and <math>AD</math> are quarter-circles, and arc <math>BCD</math> is a semicircle. What is the area, in square units, of the region? | ||
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<math> \text{(A)}\ 25\qquad\text{(B)}\ 10+5\pi\qquad\text{(C)}\ 50\qquad\text{(D)}\ 50+5\pi\qquad\text{(E)}\ 25\pi </math> | <math> \text{(A)}\ 25\qquad\text{(B)}\ 10+5\pi\qquad\text{(C)}\ 50\qquad\text{(D)}\ 50+5\pi\qquad\text{(E)}\ 25\pi </math> | ||
− | ==Solution 1== | + | == Solutions == |
− | + | === Solution 1 === | |
Draw two squares: one that has opposing corners at <math>A</math> and <math>B</math>, and one that has opposing corners at <math>A</math> and <math>D</math>. These squares share side <math>\overline{AO}</math>, where <math>O</math> is the center of the large semicircle. | Draw two squares: one that has opposing corners at <math>A</math> and <math>B</math>, and one that has opposing corners at <math>A</math> and <math>D</math>. These squares share side <math>\overline{AO}</math>, where <math>O</math> is the center of the large semicircle. | ||
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Adding the two areas gives a total area of <math>50</math>, for an answer of <math>\boxed{C}</math> | Adding the two areas gives a total area of <math>50</math>, for an answer of <math>\boxed{C}</math> | ||
− | ==Solution 2== | + | === Solution 2 === |
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Draw line <math>\overline{BD}</math>. Then draw <math>\overline {CO}</math>, where <math>O</math> is the center of the semicircle. You have two quarter circles on top, and two quarter circle-sized "bites" on the bottom. Move the pieces from the top to fit in the bottom like a jigsaw puzzle. You now have a rectangle with length <math>\overline {BD}</math> and height <math>\overline {AO}</math>, which are equal to <math>10</math> and <math>5</math>, respectively. Thus, the total area is <math>50</math>, and the answer is <math>\boxed{C}</math>. | Draw line <math>\overline{BD}</math>. Then draw <math>\overline {CO}</math>, where <math>O</math> is the center of the semicircle. You have two quarter circles on top, and two quarter circle-sized "bites" on the bottom. Move the pieces from the top to fit in the bottom like a jigsaw puzzle. You now have a rectangle with length <math>\overline {BD}</math> and height <math>\overline {AO}</math>, which are equal to <math>10</math> and <math>5</math>, respectively. Thus, the total area is <math>50</math>, and the answer is <math>\boxed{C}</math>. | ||
− | ==See Also== | + | == See Also == |
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{{AMC8 box|year=2000|num-b=18|num-a=20}} | {{AMC8 box|year=2000|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:31, 19 October 2020
Problem
Three circular arcs of radius units bound the region shown. Arcs and are quarter-circles, and arc is a semicircle. What is the area, in square units, of the region?
Solutions
Solution 1
Draw two squares: one that has opposing corners at and , and one that has opposing corners at and . These squares share side , where is the center of the large semicircle.
These two squares have a total area of , but have two quarter circle "bites" of radius that must be removed. Thus, the bottom part of the figure has area
This is the area of the part of the figure underneath . The part of the figure over is just a semicircle with radius , which has area of
Adding the two areas gives a total area of , for an answer of
Solution 2
Draw line . Then draw , where is the center of the semicircle. You have two quarter circles on top, and two quarter circle-sized "bites" on the bottom. Move the pieces from the top to fit in the bottom like a jigsaw puzzle. You now have a rectangle with length and height , which are equal to and , respectively. Thus, the total area is , and the answer is .
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.