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Question Number 165900 by cherokeesay last updated on 09/Feb/22

Commented by som(math1967) last updated on 10/Feb/22

$$Isthearea$$$$\{\frac{1}{3}\times \pi \times 8-\pi {(4\sqrt{3}-6)}^2-2\sqrt{3}\}{cm}^2?$$

Commented by cherokeesay last updated on 10/Feb/22

$$\backsimeq 2,2$$$$googanswer!$$$$thankyou.$$

Commented by som(math1967) last updated on 10/Feb/22

Commented by som(math1967) last updated on 10/Feb/22

$$cos\mathrm{\angle }CAB=\frac{AB}{AC}=\frac{2}{4}\Rightarrow \mathrm{\angle }CAB=60°$$$$BC=AB\times tan60=2\sqrt{3}cm$$$$arofsectorCAE=\frac{60}{360}\times \pi \times 4^2=\frac{1}{3}\pi \times 8$$$$ar.of△ABC=\frac{1}{2}\times 2\times 2\sqrt{3}=2\sqrt{3}{cm}^2$$$$AO=4-rAL=2+r$$$${(4-r)}^2={(2+r)}^2+r^2$$$$r^2+12r-12=0$$$$r=4\sqrt{3}-6$$$$ar.ofhatchpart$$$$=\frac{1}{3}\times 8\pi -2\sqrt{3}-\pi {(4\sqrt{3}-6)}^2{cm}^2$$

Commented by Tawa11 last updated on 10/Feb/22

$$\mathrm{Weldone}\mathrm{sir}$$

Answered by Eulerian last updated on 10/Feb/22

$$$$$$\mathbf{Solution}:$$$$\mathrm{Given}\mathrm{that}:$$$$\mathrm{a}=2\mathrm{cm}$$$$\mathrm{b}=\sqrt{4^2-2^2}=2\sqrt{3}\mathrm{cm}$$$$\sin \left(\theta \right)=\frac{1}{2}\Rightarrow \theta =30°$$$${(4-\mathrm{r})}^2={\mathrm{r}}^2+{(2+\mathrm{r})}^2$$$${\mathrm{r}}^2+12\mathrm{r}-12=0$$$$\mathrm{By}\mathrm{quadratic}\mathrm{formula}:$$$$\mathrm{r}=\frac{-12+\sqrt{{12}^2+4\left(12\right)}}{2}=4\sqrt{3}-6\mathrm{cm}$$$$$$$$\therefore$$$${\mathrm{Area}}_{\mathrm{hatched}}=\frac{\pi \left(4^2\right)}{4}-[\pi {(4\sqrt{3}-6)}^2+\frac{2\left(4\right)}{2}+\left(\frac{30°}{360°}\right)\cdot \pi {\left(4\right)}^2]=(48\pi \sqrt{3}-\frac{244\pi }{3}-4){\mathrm{cm}}^2$$$$$$

Commented by cherokeesay last updated on 10/Feb/22

$$please,tryagain!$$

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