Difference between revisions of "2000 AMC 8 Problems/Problem 19"
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This is the area of the part of the figure underneath <math>\overline{BD}</math>. The part of the figure over <math>\overline{BD}</math> is just a semicircle with radius <math>5</math>, which has area of <math>\frac{1}{2}\pi\cdot 5^2 = \frac{25\pi}{2}</math> | This is the area of the part of the figure underneath <math>\overline{BD}</math>. The part of the figure over <math>\overline{BD}</math> is just a semicircle with radius <math>5</math>, which has area of <math>\frac{1}{2}\pi\cdot 5^2 = \frac{25\pi}{2}</math> | ||
− | 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>\boxed{50}</math>, for an answer of <math>\boxed{C}</math> |
=== Solution 2 === | === Solution 2 === |
Revision as of 16:57, 12 January 2024
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
.
Video Solution
https://youtu.be/kLH0ql186UE Soo, DRMS, NM
https://www.youtube.com/watch?v=-Sysux6_wXo ~David
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.