Difference between revisions of "2010 AMC 8 Problems/Problem 18"
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<cmath>\dfrac{1350}{225\pi}=\dfrac{6}{\pi}\Rightarrow\boxed{\textbf{(C)}\ 6:\pi}</cmath> | <cmath>\dfrac{1350}{225\pi}=\dfrac{6}{\pi}\Rightarrow\boxed{\textbf{(C)}\ 6:\pi}</cmath> | ||
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+ | Note that we could have solved this without the measurement of <math>30</math> inches. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=17|num-a=19}} | {{AMC8 box|year=2010|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:04, 16 October 2018
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 AD.
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 |
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