Question Number 166640 by cherokeesay last updated on 23/Feb/22
Answered by som(math1967) last updated on 24/Feb/22
$${Area}\:{with}\:{circle} \\ $$$$=\mathrm{2}×\left\{\frac{\pi}{\mathrm{6}}×\mathrm{4}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{4}^{\mathrm{2}} \right\}{cm}^{\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{16}\pi}{\mathrm{3}}\:−\mathrm{4}\sqrt{\mathrm{3}}\right){cm}^{\mathrm{2}} \: \\ $$$${let}\:{radius}\:{of}\:{circle}\:={r} \\ $$$$\therefore\left(\mathrm{4}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \\ $$$$\:\mathrm{16}−\mathrm{8}{r}+{r}^{\mathrm{2}} ={r}^{\mathrm{2}} +\mathrm{4} \\ $$$$\Rightarrow{r}=\frac{\mathrm{3}}{\mathrm{2}}{cm} \\ $$$$\therefore\:{area}\:{of}\:{hached}\:{region} \\ $$$$=\left(\frac{\mathrm{16}\pi}{\mathrm{3}}\:−\mathrm{4}\sqrt{\mathrm{3}}\right)\:−\pi×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{37}\pi}{\mathrm{12}}\:−\mathrm{4}\sqrt{\mathrm{3}}\right){cm}^{\mathrm{2}} \\ $$
Commented by cherokeesay last updated on 24/Feb/22
$${very}\:{nice} \\ $$$${thank}\:{you}\:{sir}. \\ $$
Commented by Tawa11 last updated on 26/Feb/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$