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)); | ||
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> | ||
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<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> | ||
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<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> | ||
+ | |||
+ | Note that we could have solved this without the measurement of <math>30</math> inches. | ||
+ | |||
+ | ==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}} | {{MAA Notice}} |
Latest revision as of 09:04, 10 March 2023
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?
Solution
We can set a proportion:
We substitute with 30 and solve for
.
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:
Note that we could have solved this without the measurement of inches.
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.