Question Number 13998 by tawa tawa last updated on 26/May/17
Commented by ajfour last updated on 26/May/17
Answered by ajfour last updated on 26/May/17
$${A}=\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \theta\:\:\: \\ $$$${P}=\mathrm{2}{r}+{r}\theta \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}\pi{R}^{\mathrm{2}} {h}\:\:\:\:\:\:\left({h}=\sqrt{{r}^{\mathrm{2}} −{R}^{\mathrm{2}} }\:\right) \\ $$$$\:\:\:\:\:\:\:\:{where}\:{R}=\frac{{r}\theta}{\mathrm{2}\pi}\:. \\ $$$$\theta=\frac{\mathrm{2}{A}}{{r}^{\mathrm{2}} }\:=\frac{\mathrm{2}×\mathrm{396}}{\mathrm{400}}=\mathrm{1}.\mathrm{98}\:{rad} \\ $$$${r}\theta=\mathrm{20}×\mathrm{1}.\mathrm{98}=\mathrm{39}.\mathrm{6}{cm} \\ $$$${P}=\mathrm{40}+\mathrm{39}.\mathrm{6}=\mathrm{79}.\mathrm{6}{cm} \\ $$$${R}=\frac{\mathrm{20}×\mathrm{9}.\mathrm{8}}{\mathrm{2}\pi}=\mathrm{6}.\mathrm{3}{cm} \\ $$$${h}=\sqrt{\mathrm{400}−\left(\mathrm{6}.\mathrm{3025}\right)^{\mathrm{2}} }=\mathrm{18}.\mathrm{98}{cm} \\ $$$${V}=\frac{\pi\left(\mathrm{6}.\mathrm{3025}\right)^{\mathrm{2}} \left(\mathrm{18}.\mathrm{981}\right)}{\mathrm{3}}=\mathrm{789}.\mathrm{5}{cm}^{\mathrm{3}} . \\ $$$$ \\ $$
Commented by tawa tawa last updated on 26/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$