2018 AMC 10B Problems/Problem 7
In the figure below, 
congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let 
be the combined area of the small semicircles and 
be the area of the region inside the large semicircle but outside the semicircles. The ratio 
is 
. What is ?
![[asy] draw((0,0)--(18,0)); draw(arc((9,0),9,0,180)); filldraw(arc((1,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((3,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((5,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((7,0),1,0,180)--cycle,gray(0.8)); label("...",(9,0.5)); filldraw(arc((11,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((13,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((15,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((17,0),1,0,180)--cycle,gray(0.8)); [/asy]](http://latex.artofproblemsolving.com/9/0/1/901531f4b527616e8cc1a1884115f9781f713e0f.png)
Solution 1
Use the answer choices and calculate them. The one that works is D
Solution 2
Let the radius of the smaller semicircles be 
. This means that the radius of the larger semicircle is 
. The sum of the areas of the smaller semicircles is $\frac{\pinr^2}{2}$ (Error compiling LaTeX. Unknown error_msg), while the area of the larger semicircle subtracted by the sum of the areas of the smaller semicircle is $\frac{\pin^2r^2}{2}-\frac{\pinr^2}{2}=\frac{\pir^2(n^2-n)}{2}$ (Error compiling LaTeX. Unknown error_msg). Taking the ratio of the areas of the smaller semicircles to this value, we get 
. Solving, we get 
, or 
.
Solution by ElectroVortex
See Also
| 2018 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 AMC 10 Problems and Solutions | ||
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