Question Number 9298 by tawakalitu last updated on 29/Nov/16
Answered by mrW last updated on 29/Nov/16
$$\mathrm{radius}\:\mathrm{r}=\left(\mathrm{7}+\mathrm{3}\right)/\mathrm{2}=\mathrm{5}\:\mathrm{cm} \\ $$$$\mathrm{height}\:\mathrm{of}\:\mathrm{segment}\:\mathrm{h}=\mathrm{3}\:\mathrm{cm} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{segment}\:=\:\mathrm{r}^{\mathrm{2}} \mathrm{cos}\:^{−\mathrm{1}} \left(\frac{\mathrm{r}−\mathrm{h}}{\mathrm{r}}\right)−\left(\mathrm{r}−\mathrm{h}\right)\sqrt{\mathrm{2rh}−\mathrm{h}^{\mathrm{2}} } \\ $$$$=\mathrm{5}^{\mathrm{2}} \mathrm{cos}\:^{−\mathrm{1}} \left(\frac{\mathrm{5}−\mathrm{3}}{\mathrm{5}}\right)−\left(\mathrm{5}−\mathrm{3}\right)\sqrt{\mathrm{2}×\mathrm{5}×\mathrm{3}−\mathrm{3}^{\mathrm{2}} } \\ $$$$=\mathrm{25cos}\:^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{5}}\right)−\mathrm{2}\sqrt{\mathrm{21}} \\ $$$$\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}\:\mathrm{is}\:\mathrm{half}\:\mathrm{of}\:\mathrm{the}\:\mathrm{segment} \\ $$$$=\frac{\mathrm{25}}{\mathrm{2}}\mathrm{cos}\:^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{5}}\right)−\sqrt{\mathrm{21}}\approx\mathrm{9}.\mathrm{91}\:\mathrm{cm}^{\mathrm{2}} \\ $$
Commented by tawakalitu last updated on 29/Nov/16
$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$
Commented by Nayon last updated on 29/Mar/17
$${How}\:{did}\:{you}\:{realise}\:{that}\:{shaded}\: \\ $$$${area}\:{is}\:{half}\:{of}\:{the}\:{segment}? \\ $$
Commented by mrW1 last updated on 29/Mar/17
$${the}\:{picture}\:{shows}\:{only}\:{the}\:{half}\:{of}\:{the} \\ $$$${circle}. \\ $$