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Question Number 32659 by Joel578 last updated on 30/Mar/18

Commented by Joel578 last updated on 30/Mar/18

Red circle with radius 2(√2) cm  Blue circle with radius 4 cm  C is center of blue circle  AB is diameter of red circle  Find the area of shaded part

$$\mathrm{Red}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{cm} \\ $$$$\mathrm{Blue}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{4}\:\mathrm{cm} \\ $$$${C}\:\mathrm{is}\:\mathrm{center}\:\mathrm{of}\:\mathrm{blue}\:\mathrm{circle} \\ $$$${AB}\:\mathrm{is}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{red}\:\mathrm{circle} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{part} \\ $$

Answered by MJS last updated on 30/Mar/18

AC=BC=4 ∧ AB=4(√2) ⇒  ⇒ AC^2 +BC^2 =AB^2  ⇒  ⇒ black area=(((2(√2))^2 π)/2)−(((4^2 π)/4)−(4^2 /2))=  =4π−(4π−8)=8

$${AC}={BC}=\mathrm{4}\:\wedge\:{AB}=\mathrm{4}\sqrt{\mathrm{2}}\:\Rightarrow \\ $$$$\Rightarrow\:{AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} ={AB}^{\mathrm{2}} \:\Rightarrow \\ $$$$\Rightarrow\:{black}\:{area}=\frac{\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \pi}{\mathrm{2}}−\left(\frac{\mathrm{4}^{\mathrm{2}} \pi}{\mathrm{4}}−\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}\right)= \\ $$$$=\mathrm{4}\pi−\left(\mathrm{4}\pi−\mathrm{8}\right)=\mathrm{8} \\ $$

Commented by Joel578 last updated on 01/Apr/18

thank you very mucb

$${thank}\:{you}\:{very}\:{mucb} \\ $$

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