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 | + | ==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. |
− | </ | + | <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> | ||
+ | <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== | |
− | + | We can set a proportion: | |
− | < | + | <cmath>\dfrac{AD}{AB}=\dfrac{3}{2}</cmath> |
− | |||
− | |||
− | + | We substitute <math>AB</math> with 30 and solve for AD. | |
+ | |||
+ | <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> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2010|num-b=17|num-a=19}} |
Revision as of 17:25, 5 November 2012
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 semicircle.
Solution
We can set a proportion:
We substitute with 30 and solve for AD.
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;
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 |