Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

2010 AMC 8 Problems/Problem 18: Difference between revisions

← Older edit
Mathway (talk | contribs)
254 edits
Social climber (talk | contribs)
editor
33 edits
 
(10 intermediate revisions by 10 users not shown)
Line 1: Line 1:
==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 semicircle.
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));
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);
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>
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy>
<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:


<cmath>\dfrac{AD}{AB}=\dfrac{3}{2}</cmath>
<cmath>\dfrac{AD}{AB}=\dfrac{3}{2}</cmath>


We substitute <math>AB</math> with 30 and solve for AD.
We substitute <math>AB</math> with 30 and solve for <math>AD</math>.


<cmath>\dfrac{AD}{30}=\dfrac{3}{2}</cmath>
<cmath>\dfrac{AD}{30}=\dfrac{3}{2}</cmath>
Line 21: Line 23:


<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==
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==
https://youtu.be/zoWJrxTCYPM
~savannahsolver


==See Also==
==See Also==
{{AMC8 box|year=2010|num-b=17|num-a=19}}
{{AMC8 box|year=2010|num-b=17|num-a=19}}
{{MAA Notice}}

Latest revision as of 20:35, 23 October 2024

Problem

A decorative window is made up of a rectangle with semicircles at either end. The ratio of $AD$ to $AB$ is $3:2$. And $AB$ 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]

$\textbf{(A)}\ 2:3\qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi\qquad\textbf{(D)}\ 9:\pi\qquad\textbf{(E)}\ 30 :\pi$

Solution 1

We can set a proportion:

\[\dfrac{AD}{AB}=\dfrac{3}{2}\]

We substitute $AB$ with 30 and solve for $AD$.

\[\dfrac{AD}{30}=\dfrac{3}{2}\]

\[AD=45\]

We calculate the combined area of semicircle by putting together semicircle $AB$ and $CD$ to get a circle with radius $15$. Thus, the area is $225\pi$. The area of the rectangle is $30\cdot 45=1350$. We calculate the ratio:

\[\dfrac{1350}{225\pi}=\dfrac{6}{\pi}\Rightarrow\boxed{\textbf{(C)}\ 6:\pi}\]

Solution 2 (more efficient)

We can solve this without the measurement of $30$ inches. Let the constant of proportionality between the sides of the rectangle be $k$. $\frac{A_{rect}}{A_{circle}}=\frac{3k*2k}{\pi(\frac{2k}{2})^2}=\frac{6k^2}{\pi k^2}=\frac{6}{\pi}\Rightarrow\boxed{\textbf{(C)}\ 6:\pi}$

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

https://youtu.be/zoWJrxTCYPM

~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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_18&oldid=230464"