Difference between revisions of "2013 UNCO Math Contest II Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | <math>48-9\pi</ | + | 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 <math>3</math>, we can find the width of the rectangle to be <math>6</math>. Solving for the shaded area, we get |
| + | <cmath> | ||
| + | \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*} | ||
| + | </cmath> | ||
== See Also == | == See Also == | ||
Revision as of 13:12, 24 February 2022
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]](http://latex.artofproblemsolving.com/c/e/5/ce5a7d2dddce5c0d8880196cb2f63aa28e94203e.png)
In the diagram, the two circles are tangent to the two parallel lines. The
distance between the centers of the circles is 
, and both circles have
radius 
. 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 , we can find the width of the rectangle to be 
. Solving for the shaded area, we get
See Also
| 2013 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
