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July 21st, 2012, 12:12 AM   #1
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find area

[attachment=0:1p8raamh]semicicle.jpg[/attachment:1p8raamh]
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 July 21st, 2012, 12:51 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,898 Thanks: 1093 Math Focus: Elementary mathematics and beyond Re: find area $A\,=\,2\pi\,-\,$$\frac{\theta r^2}{2}\,-\,4\sqrt{3}$$,\,\theta\,=\,\frac{\pi}{3},\,r\,=\,4 .$
July 21st, 2012, 01:17 AM   #3
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Re: find area

[attachment=0:3qj7s5n3]semicicle.jpg[/attachment:3qj7s5n3]

The triangle is equilateral, hence the area of the circular sector, minus that of the triangle is:

$\frac{1}{2}\cdot4^2\cdot\frac{\pi}{3}-\frac{1}{2}4^2\sin$$\frac{\pi}{3}$$=\frac{8\pi}{3}-4\sqrt{3}$

Now, subtract this from the area of the smaller semi-circle:

$\frac{1}{2}\pi(2)^2-$$\frac{8\pi}{3}-4\sqrt{3}$$=4\sqrt{3}-\frac{2\pi}{3}$
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