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Question Number 169745 by cherokeesay last updated on 07/May/22

Answered by mr W last updated on 07/May/22

Commented by mr W last updated on 08/May/22

$$AB=\sqrt{3}$$$$OA=OB=OD=R=\frac{AB}{2}=\frac{\sqrt{3}}{2}$$$$\left[ABC\right]=\frac{1\times \sqrt{3}}{2}=\frac{\sqrt{3}}{2}$$$$\left[AOD\right]=\frac{1}{2}{\left(\frac{\sqrt{3}}{2}\right)}^2\sin 120°=\frac{3\sqrt{3}}{16}$$$$\left[O\stackrel{⌢}{BD}\right]=\frac{\pi }{6}{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{\pi }{8}$$$$hatchedarea=\frac{\sqrt{3}}{2}-\frac{3\sqrt{3}}{16}-\frac{\pi }{8}=\frac{5\sqrt{3}-2\pi }{16}$$

Commented by cherokeesay last updated on 07/May/22

$$verynice!$$$$thankyousir!$$

Commented by Tawa11 last updated on 08/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by som(math1967) last updated on 07/May/22

$$AB=BCtan60=\sqrt{3}cm$$$$OB=\frac{\sqrt{3}}{2}cm$$$$\measuredangle BOL=2\times 30=60$$$$OB=OL\Rightarrow △OBLequilateral$$$$\mathrm{\angle }LBC=90-60=30$$$$ar.of△OBL=\frac{\sqrt{3}}{4}\times {\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3\sqrt{3}}{16}{cm}^2$$$$ar.of△BLC=\frac{1}{2}\times 1\times \frac{\sqrt{3}}{2}sin\mathrm{\angle }LBC$$$$=\frac{\sqrt{3}}{8}{cm}^2$$$$ar.ofsectorBOL=\frac{1}{6}\pi \times {\left(\frac{\sqrt{3}}{2}\right)}^2{cm}^2$$$$\frac{\pi }{8}{cm}^2$$$$hatchedarea=(\frac{3\sqrt{3}}{16}+\frac{\sqrt{3}}{8})-\frac{\pi }{8}$$$$=(\frac{5\sqrt{3}}{16}-\frac{\pi }{8}){cm}^2$$

Commented by som(math1967) last updated on 07/May/22

Commented by cherokeesay last updated on 07/May/22

$$sobeautiful$$$$thankyousomuch!$$

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