2010 AMC 8 Problems/Problem 18
Contents
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 1
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:
Solution 2 (more efficient)
We can solve this without the measurement of inches. Let the constant of proportionality between the sides of the rectangle be .
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=657
~ pi_is_3.14
Video by MathTalks
https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
Video Solution by WhyMath
~savannahsolver
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.