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Question-198184




Question Number 198184 by Blackpanther last updated on 13/Oct/23
Answered by Rasheed.Sindhi last updated on 13/Oct/23
Sector DFG=quarter of the circle of radius 3  =π(3)^2 /4=((9π)/4)  Shaded area in one circle=Sector DFG−▲DFG         =((9π)/4)−((3×3)/2)=((9π−18)/4)  Shaded area in all four circles=4×((9π−18)/4)=9π−18 ✓
$$\mathrm{Sector}\:\mathrm{DFG}=\mathrm{quarter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{3} \\ $$$$=\pi\left(\mathrm{3}\right)^{\mathrm{2}} /\mathrm{4}=\frac{\mathrm{9}\pi}{\mathrm{4}} \\ $$$${Shaded}\:{area}\:{in}\:{one}\:{circle}={Sector}\:\mathrm{DFG}−\blacktriangle\mathrm{DFG} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{9}\pi}{\mathrm{4}}−\frac{\mathrm{3}×\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{9}\pi−\mathrm{18}}{\mathrm{4}} \\ $$$${Shaded}\:{area}\:{in}\:{all}\:{four}\:{circles}=\mathrm{4}×\frac{\mathrm{9}\pi−\mathrm{18}}{\mathrm{4}}=\mathrm{9}\pi−\mathrm{18}\:\checkmark \\ $$

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