Question Number 107521 by otchereabdullai@gmail.com last updated on 11/Aug/20
Answered by bemath last updated on 11/Aug/20
$$\angle{XZY}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{45}°=\mathrm{22}.\mathrm{5}°\: \\ $$$${area}\:{of}\:{shaded}\:{segment}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\pi{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \mathrm{sin}\:\mathrm{45}° \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{8}}.\frac{\mathrm{22}}{\mathrm{7}}.\mathrm{49}−\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{49}.\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}} \\ $$$$=\:\mathrm{49}\left(\frac{\mathrm{11}}{\mathrm{28}}\:−\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\right)\:=\:\mathrm{49}\left(\frac{\mathrm{11}−\mathrm{7}\sqrt{\mathrm{2}}}{\mathrm{28}}\right) \\ $$$$=\:\frac{\mathrm{77}−\mathrm{49}\sqrt{\mathrm{2}}}{\mathrm{4}}\:{cm}^{\mathrm{2}} \\ $$