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October 24, 2014

October 24, 2014

Posted by **Hannah** on Sunday, May 5, 2013 at 7:57pm.

- math -
**Reiny**, Sunday, May 5, 2013 at 8:17pmIf the area of the unshaded region in your diagram is 3 m^2, then the whole circle must be 4 m^2 (simple ratio)

and the area of the sector, which is your shaded region, must be 1 m^2

so πr^2 = 4

r^2 = 4/π

r = 2/√π

area of your triangle within your shaded region

= (1/2)r^2

= (1/2)4/π = 2/π

area of segment = 1 - 2/π or (π - 2)/π

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