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

(Created page with "The ratio of AD to AB is 3:2 and AB=30 inches. <math>If 30=2, then x=3. x=45. </math> The area of the rectangle is <math>45*30= 1350</math>. The semicircles, when combined, have...")
 
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The ratio of AD to AB is 3:2 and AB=30 inches.
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==Problem==
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<math>If 30=2, then x=3. x=45.
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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.
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</math>
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<asy>
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The area of the rectangle is <math>45*30= 1350</math>.
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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));
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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);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy>
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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>
  
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The semicircles, when combined, have the area of a regular circle.
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==Solution==
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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>
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We can set a proportion:
  
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<math>1350/225 \pi</math> is equivalent to <math>6/ \pi </math>
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<cmath>\dfrac{AD}{AB}=\dfrac{3}{2}</cmath>
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[http://i.imgur.com/5qEb2.png]
 
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The image given in the test is given above for reference.
 
  
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Answer is <math> \box{C: 6:\pi} </math>
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We substitute <math>AB</math> with 30 and solve for AD.
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<cmath>\dfrac{AD}{30}=\dfrac{3}{2}</cmath>
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<cmath>AD=45</cmath>
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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;
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<cmath>\dfrac{1350}{225\pi}=\dfrac{6}{\pi}\Rightarrow\boxed{\textbf{(C)}\ 6:\pi}</cmath>
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==See Also==
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{{AMC8 box|year=2010|num-b=17|num-a=19}}

Revision as of 16:25, 5 November 2012

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 semicircle. [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;

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

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
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