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Question Number 100378 by bemath last updated on 26/Jun/20

Commented by bemath last updated on 26/Jun/20

$$\mathrm{find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{OACB}.\mathrm{where}\mathrm{radius}=4$$

Answered by john santu last updated on 26/Jun/20

Commented by john santu last updated on 26/Jun/20

$$\mathrm{OM}=\sqrt{1^2+2^2}=\sqrt{5}$$$$\mathrm{OA}=\sqrt{16-5}=\sqrt{11}$$$$\frac{\mathrm{OC}}{4}=\frac{4}{\sqrt{5}}\Rightarrow \mathrm{OC}=\frac{16}{\sqrt{5}}$$$$\mathrm{Area}\mathrm{OACB}=2\times \mathrm{area}\mathrm{OAC}$$$$=2\times \frac{1}{2}\times \frac{16}{\sqrt{5}}\times \sqrt{11}=\frac{16\sqrt{11}}{\sqrt{5}}$$$$=\frac{16\sqrt{55}}{5}$$

Commented by bemath last updated on 26/Jun/20

$$\mathrm{thank}\mathrm{you}$$

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