# 2010 AMC 8 Problems/Problem 18

## 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.

~ pi_is_3.14

## Video Solution by WhyMath

~savannahsolver

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