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Question Number 112467 by Aina Samuel Temidayo last updated on 08/Sep/20

Answered by 1549442205PVT last updated on 08/Sep/20

$$\mathrm{Side}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{equal}\mathrm{to}\frac{\mathrm{a}}{\sqrt{2}}$$$$\mathrm{Width}\mathrm{of}\mathrm{rectangle}\mathrm{equal}\mathrm{to}\frac{2\mathrm{a}}{\sqrt{2}}=\mathrm{a}\sqrt{2}$$$$\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{rectangle}\mathrm{is}{\mathrm{S}}_1=2\mathrm{a}.\mathrm{a}\sqrt{2}={2\mathrm{a}}^2\sqrt{2}\left(1\right)$$$$\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{semicircle}\mathrm{equal}\mathrm{to}\frac{1}{2}\pi {\left(\frac{\mathrm{a}}{2\sqrt{2}}\right)}^2$$$$=\frac{\pi {\mathrm{a}}^2}{16}\left(2\right).\mathrm{Side}\mathrm{of}\mathrm{the}\mathrm{right}\mathrm{triangle}\mathrm{equal}\mathrm{to}$$$$\frac{\mathrm{a}\sqrt{2}}{\sqrt{2}}=\mathrm{a}-\mathrm{this}\mathrm{is}\mathrm{also}\mathrm{side}\mathrm{of}\mathrm{two}\mathrm{squares}$$$$\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{two}\mathrm{squares}\mathrm{and}$$$$\mathrm{triangle}\mathrm{equal}\mathrm{to}{2\mathrm{a}}^2+\frac{1}{2}{\mathrm{a}}^2$$$$=\frac{5}{2}.{\mathrm{a}}^2=\left(3\right)$$$$\mathrm{From}\left(1\right)\left(2\right)\left(3\right)\mathrm{we}\mathrm{infer}\mathrm{grey}\mathrm{area}$$$$\mathrm{equal}\mathrm{to}{2\mathrm{a}}^2\sqrt{2}+\frac{\pi {\mathrm{a}}^2}{16}+\frac{5}{2}{\mathrm{a}}^2-\frac{{\mathrm{a}}^2}{2}$$$$=\frac{{\mathbf{a}}^2(\mathit{\pi }+32+32\sqrt{2})}{16}$$

Commented by Aina Samuel Temidayo last updated on 08/Sep/20

$$\mathrm{Ok}.\mathrm{Thanks}.$$

Commented by 1549442205PVT last updated on 13/Sep/20

$$\mathrm{You}\mathrm{are}\mathrm{welcome}$$

Commented by Aina Samuel Temidayo last updated on 13/Sep/20

$$\mathrm{Is}\mathrm{your}\mathrm{User}\mathrm{ID}{}^{\prime }..{.}^{\prime }?$$

Commented by Rasheed.Sindhi last updated on 21/Sep/20

$$widthoftherectangle$$$$=diagonalofthesquare=a\ne \mathrm{a}\sqrt{2}$$

Commented by 1549442205PVT last updated on 21/Sep/20

$$\mathrm{In}\mathrm{question}\mathrm{given}\mathrm{side}\mathrm{of}\mathrm{square}\mathrm{equal}$$$$\mathrm{to}\mathrm{a}\Rightarrow \mathrm{diagonal}\mathrm{of}\mathrm{sq}.=\mathrm{a}\sqrt{2}$$

Commented by Rasheed.Sindhi last updated on 21/Sep/20

$$Thenthediagramisincorrect.$$

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