Try this beautiful problem on area of circle from SMO, Singapore Mathematics Olympiad, 2010.
Let ABCD be a rectangle with AB = 10 . Draw circles \(C_1 \) and \(C_2\) with diameters AB and CD respectively. Let P,Q be the intersection points of \(C_1\) and \(C_2\) . If the circle with diameter PQ is tangent to AB and CD , then what is the area of the shaded region ?
Area of Circle
2D - Geometry
Area of Rectangle
Answer: 25
Singapore Mathematics Olympiad
Challenges and thrills - Pre - college Mathematics
If you are really got stuck with this sum then we can start from here:
The diagram will be like this . So let 'N' be the midpoint of CD .
so \(\angle {PNQ} = 90^\circ\)
so PQ = 5 \(\sqrt {2}\)
Now let us try to find the area of the shaded region
A = \(2 [ \frac {1}{2}\pi (\frac {PQ)}{2})^2 + \frac {1}{2}(PN)^2 - \frac {1}{4} \pi (PN)^2]\)
= \(2 [\frac {1}{2}\pi (\frac {5\sqrt{2}}{2})^2+\frac {1}{2}.5^2 - \frac {1}{4}\pi . 5^2]\)
= 25
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