Question-9298

Question Number 9298 by tawakalitu last updated on 29/Nov/16

Answered by mrW last updated on 29/Nov/16

radius r=(7+3)/2=5 cm height of segment h=3 cm area of segment = r^2 cos^(−1) (((r−h)/r))−(r−h)(√(2rh−h^2 )) =5^2 cos^(−1) (((5−3)/5))−(5−3)(√(2×5×3−3^2 )) =25cos^(−1) ((2/5))−2(√(21)) the shaded area is half of the segment =((25)/2)cos^(−1) ((2/5))−(√(21))≈9.91 cm^2

$$\mathrm{radius}\:\mathrm{r}=\left(\mathrm{7}+\mathrm{3}\right)/\mathrm{2}=\mathrm{5}\:\mathrm{cm} \\ $$$$\mathrm{height}\:\mathrm{of}\:\mathrm{segment}\:\mathrm{h}=\mathrm{3}\:\mathrm{cm} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{segment}\:=\:\mathrm{r}^{\mathrm{2}} \mathrm{cos}\:^{−\mathrm{1}} \left(\frac{\mathrm{r}−\mathrm{h}}{\mathrm{r}}\right)−\left(\mathrm{r}−\mathrm{h}\right)\sqrt{\mathrm{2rh}−\mathrm{h}^{\mathrm{2}} } \\ $$$$=\mathrm{5}^{\mathrm{2}} \mathrm{cos}\:^{−\mathrm{1}} \left(\frac{\mathrm{5}−\mathrm{3}}{\mathrm{5}}\right)−\left(\mathrm{5}−\mathrm{3}\right)\sqrt{\mathrm{2}×\mathrm{5}×\mathrm{3}−\mathrm{3}^{\mathrm{2}} } \\ $$$$=\mathrm{25cos}\:^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{5}}\right)−\mathrm{2}\sqrt{\mathrm{21}} \\ $$$$\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}\:\mathrm{is}\:\mathrm{half}\:\mathrm{of}\:\mathrm{the}\:\mathrm{segment} \\ $$$$=\frac{\mathrm{25}}{\mathrm{2}}\mathrm{cos}\:^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{5}}\right)−\sqrt{\mathrm{21}}\approx\mathrm{9}.\mathrm{91}\:\mathrm{cm}^{\mathrm{2}} \\ $$

Commented by tawakalitu last updated on 29/Nov/16