Difference between revisions of "2010 AMC 8 Problems/Problem 18"
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<math> \textbf{(A)}\ 2:3\qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi\qquad\textbf{(D)}\ 9:\pi\qquad\textbf{(E)}\ 30 :\pi </math> | <math> \textbf{(A)}\ 2:3\qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi\qquad\textbf{(D)}\ 9:\pi\qquad\textbf{(E)}\ 30 :\pi </math> | ||
| − | ==Solution== | + | ==Solution 1== |
We can set a proportion: | We can set a proportion: | ||
| Line 24: | Line 24: | ||
<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> | ||
| − | + | ==Solution 2 (more efficient)== | |
| + | We can solve this without the measurement of <math>30</math> inches. Let the constant of proportionality between the sides of the rectangle be <math>k</math>. | ||
| + | <math>\frac{A_{rect}}{A_{circle}}=\frac{3k*2k}{\pi(\frac{2k}{2})^2}=\frac{6k^2}{\pi k^2}=\frac{6}{\pi}\Rightarrow\boxed{\textbf{(C)}\ 6:\pi}</math> | ||
==Video Solution by OmegaLearn== | ==Video Solution by OmegaLearn== | ||
Latest revision as of 20:35, 23 October 2024
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?
![[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)); dot((0,0),ds); label("$A$",(-0.26,-0.23),NE*lsf); dot((2.5,0),ds); label("$B$",(2.61,-0.26),NE*lsf); dot((0,4),ds); label("$D$",(-0.26,4.02),NE*lsf); dot((2.5,4),ds); label("$C$",(2.64,3.98),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](http://latex.artofproblemsolving.com/1/b/d/1bd7bb101d1f617dcc2d568edfde1a17ef3345b4.png)
Solution 1
We can set a proportion:
![\[\dfrac{AD}{AB}=\dfrac{3}{2}\]](http://latex.artofproblemsolving.com/9/d/c/9dc8ec1f9de87f93331eca33e68b9a9bfb4fe1be.png)
We substitute with 30 and solve for .
![\[\dfrac{AD}{30}=\dfrac{3}{2}\]](http://latex.artofproblemsolving.com/b/d/c/bdc8fcd7c68f8cda8f0d49383feb93a5f8415894.png)
![\[AD=45\]](http://latex.artofproblemsolving.com/f/1/e/f1e60700abae8840538f90cd4b41361fd259e31e.png)
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:
![\[\dfrac{1350}{225\pi}=\dfrac{6}{\pi}\Rightarrow\boxed{\textbf{(C)}\ 6:\pi}\]](http://latex.artofproblemsolving.com/e/2/4/e2419c749922191e5f080078e22383b0ce4f17ae.png)
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. 
