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Let's call the 3 areas from left to right as A1,A2,A3 and the overall area as A. Such that A1+A2+A3 = A. You need to solve for A3. A1 is simply a quarter of a circle and A2 is a right angled triangle. I suppose you know the formulae's for both A1 and A2.
if I draw AD line, then ADO İS equilateral triangle. m(DOA)=60. Area of half circle is 2pi. Colored area= 2pi.120/360=2pi/3 but answer is 4pi/3 tell me where i'm wrong.
If area of full circle is 4pi then area of 120 degree sector would be 4pi*120/360
so we need to consider the who circle again thank you :)
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Btw, brilliant job in finding that ADO is equilateral !
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