2013 UNCO Math Contest II Problems/Problem 1

Problem

$[asy] filldraw((-4,-3)--(-4,3)--(4,3)--(4,-3)--cycle,grey); filldraw(circle((-4,0),3),white); filldraw(circle((4,0),3),white); draw((-7,3)--(7,3),black); draw((-7,-3)--(7,-3),black); [/asy]$

In the diagram, the two circles are tangent to the two parallel lines. The distance between the centers of the circles is $8$ , and both circles have radius $3$ . What is the area of the shaded region between the circles?

Solution

Notice that the shaded area is equivalent to the area of a rectangle subtracted by the area of 2 semicircles. Since we know that the radius of the circles are $3$ , we can find the width of the rectangle to be $6$ . Solving for the shaded area, we get $\begin{align*} (8)(6) - 2\left(\dfrac12 \cdot 3^2\pi\right) &= 48 - 2\left(\dfrac12 \cdot 9\pi\right) \\ &=\boxed{48-9\pi}. \end{align*}$

2013 UNCO Math Contest II (Problems • Answer Key • Resources)
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2013 UNCO Math Contest II Problems/Problem 1

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