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> | ||
+ | ~pineconee | ||
== See Also == | == See Also == |
Latest revision as of 13:13, 24 February 2022
Problem
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 ~pineconee
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 |