Difference between revisions of "2018 AMC 10B Problems/Problem 7"

(Solution 1 (Work using Answer Choices))
(Solution 2 (More Algebraic Approach))
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Now that we have found the area of both A and B, we can find the ratio. When we divide the area of the small semicircles combined by the area of the larger semicircles, we get <math>\frac{1}{n}</math>. This is equal to <math>\frac{1}{18}</math>. By setting them equal, we find that <math>n = 18</math>. This is our answer, which corresponds to choice <math>\boxed{D}</math>.
 
Now that we have found the area of both A and B, we can find the ratio. When we divide the area of the small semicircles combined by the area of the larger semicircles, we get <math>\frac{1}{n}</math>. This is equal to <math>\frac{1}{18}</math>. By setting them equal, we find that <math>n = 18</math>. This is our answer, which corresponds to choice <math>\boxed{D}</math>.
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Solution by: Archimedes15
  
 
==See Also==
 
==See Also==

Revision as of 17:17, 16 February 2018

In the figure below, $N$ congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let $A$ be the combined area of the small semicircles and $B$ be the area of the region inside the large semicircle but outside the semicircles. The ratio $A:B$ is $1:18$. What is $N$?


[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]


$\textbf{(A) } 16 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 36$

Solution 1 (Work using Answer Choices)

Use the answer choices and calculate them. The one that works is D

Solution 2 (More Algebraic Approach)

Let the number of semicircles be $n$ and let the radius of each semicircle to be $r$. To find the total area of all of the small semicircles, we have $n \cdot \frac{\pi \cdot r^2}{2}$.

Next, we have to find the area of the larger semicircles. The radius of the large semicircle can be deduced to be $n \cdot r$. So, the area of the larger semicircle is $\frac{\pi \cdot n^2 \cdot r^2}{2}$.

Now that we have found the area of both A and B, we can find the ratio. When we divide the area of the small semicircles combined by the area of the larger semicircles, we get $\frac{1}{n}$. This is equal to $\frac{1}{18}$. By setting them equal, we find that $n = 18$. This is our answer, which corresponds to choice $\boxed{D}$.

Solution by: Archimedes15

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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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