Difference between revisions of "2010 AMC 8 Problems/Problem 18"

Latest revision as of 09:04, 10 March 2023

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

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:

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

Note that we could have solved this without the measurement of $30$ 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

@@ Line 1: / Line 1: @@
-The ratio of AD to AB is 3:2 and AB=30 inches.
+==Problem==
-<math>If 30=2, then x=3. x=45.
+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?
-</math>
-The area of the rectangle is <math>45*30= 1350</math>.
-The semicircles, when combined, have the area of a regular circle.
+<asy>
-Radius is 15 (<math>30/2=15</math>, for the record). Area of a circle is <math>\pi r^2</math> , so it's equal to 225<math>\pi</math>
+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>
-<math>1350/225 \pi</math> is equivalent to <math>6/ \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>
-[http://i.imgur.com/5qEb2.png]
-The image given in the test is given above for reference.
-Answer is <math> \box{C: 6:\pi} </math>
+==Solution==
+We can set a proportion:
+<cmath>\dfrac{AD}{AB}=\dfrac{3}{2}</cmath>
+We substitute <math>AB</math> with 30 and solve for <math>AD</math>.
+<cmath>\dfrac{AD}{30}=\dfrac{3}{2}</cmath>
+<cmath>AD=45</cmath>
+We calculate the combined area of semicircle by putting together semicircle <math>AB</math> and <math>CD</math> to get a circle with radius <math>15</math>. Thus, the area is <math>225\pi</math>. The area of the rectangle is <math>30\cdot 45=1350</math>. We calculate the ratio:
+<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==
+{{AMC8 box|year=2010|num-b=17|num-a=19}}
+{{MAA Notice}}

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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