Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

Commented by Joel578 last updated on 30/Mar/18

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

Answered by MJS last updated on 30/Mar/18

$$AC=BC=4\wedge AB=4\sqrt{2}\Rightarrow$$$$\Rightarrow {AC}^2+{BC}^2={AB}^2\Rightarrow$$$$\Rightarrow blackarea=\frac{{\left(2\sqrt{2}\right)}^2\pi }{2}-(\frac{4^2\pi }{4}-\frac{4^2}{2})=$$$$=4\pi -(4\pi -8)=8$$

Commented by Joel578 last updated on 01/Apr/18

$$thankyouverymucb$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com