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Question-107521




Question Number 107521 by otchereabdullai@gmail.com last updated on 11/Aug/20
Answered by bemath last updated on 11/Aug/20
∠XZY = (1/2)×45°=22.5°   area of shaded segment = (1/8)πr^2 −(1/2)r^2 sin 45°  = (1/8).((22)/7).49−(1/2).49.(1/2)(√2)  = 49(((11)/(28)) − ((√2)/4)) = 49(((11−7(√2))/(28)))  = ((77−49(√2))/4) cm^2
$$\angle{XZY}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{45}°=\mathrm{22}.\mathrm{5}°\: \\ $$$${area}\:{of}\:{shaded}\:{segment}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\pi{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \mathrm{sin}\:\mathrm{45}° \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{8}}.\frac{\mathrm{22}}{\mathrm{7}}.\mathrm{49}−\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{49}.\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}} \\ $$$$=\:\mathrm{49}\left(\frac{\mathrm{11}}{\mathrm{28}}\:−\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\right)\:=\:\mathrm{49}\left(\frac{\mathrm{11}−\mathrm{7}\sqrt{\mathrm{2}}}{\mathrm{28}}\right) \\ $$$$=\:\frac{\mathrm{77}−\mathrm{49}\sqrt{\mathrm{2}}}{\mathrm{4}}\:{cm}^{\mathrm{2}} \\ $$

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