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Question Number 198184 by Blackpanther last updated on 13/Oct/23

Answered by Rasheed.Sindhi last updated on 13/Oct/23

$$\mathrm{Sector}\mathrm{DFG}=\mathrm{quarter}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{of}\mathrm{radius}3$$$$=\pi {\left(3\right)}^2/4=\frac{9\pi }{4}$$$$Shadedareainonecircle=Sector\mathrm{DFG}-▴\mathrm{DFG}$$$$=\frac{9\pi }{4}-\frac{3\times 3}{2}=\frac{9\pi -18}{4}$$$$Shadedareainallfourcircles=4\times \frac{9\pi -18}{4}=9\pi -18✓$$

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