Difference between revisions of "2010 AMC 8 Problems/Problem 18"
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<cmath>\dfrac{AD}{AB}=\dfrac{3}{2}</cmath> | <cmath>\dfrac{AD}{AB}=\dfrac{3}{2}</cmath> | ||
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<cmath>\dfrac{AD}{30}=\dfrac{3}{2}</cmath> | <cmath>\dfrac{AD}{30}=\dfrac{3}{2}</cmath> |
Revision as of 18:43, 5 September 2021
Problem
A decorative window is made up of a rectangle with semicircles at either end. The ratio of to
is
. And
is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircles?
Solution
We can set a proportion:
We substitute with 30 and solve for
.
We calculate the combined area of semicircle by putting together semicircle and
to get a circle with radius
. Thus, the area is
. The area of the rectangle is
. We calculate the ratio:
Note that we could have solved this without the measurement of inches.
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.