Difference between revisions of "2010 AMC 8 Problems/Problem 18"
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==Problem== | ==Problem== | ||
− | A decorative window is made up of a rectangle with semicircles at either end. The ratio of <math>AD</math> to <math>AB</math> is <math>3:2</math>. And <math>AB</math> is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircles | + | A decorative window is made up of a rectangle with semicircles at either end. The ratio of <math>AD</math> to <math>AB</math> is <math>3:2</math>. And <math>AB</math> is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircles? |
<asy> | <asy> | ||
import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0)); | import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0)); |
Revision as of 17:31, 20 February 2016
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:
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.