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Question Number 108626 by I want to learn more last updated on 18/Aug/20

Answered by Aziztisffola last updated on 18/Aug/20

A_((1/2)circle) =(1/2)π((a/2))^2 =((πa^2 )/8) A_(rectangle) =2a^(2 ) A_(triangle) =(A/2) A_(2square) =2A A_(grey) =((πa^2 )/8)+2a^2 +(A/2)+2A−A =(((16+π)/8))a^2 +((3A)/2) with A=(a^2 /2) A_(grey) =(((16+π)/8))a^2 +((3a^2 )/4) =(((22+π)/8))a^2

$$\mathrm{A}_{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{circle}} =\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\mathrm{A}_{\mathrm{rectangle}} =\mathrm{2a}^{\mathrm{2}\:} \\ $$$$\mathrm{A}_{\mathrm{triangle}} =\frac{\mathrm{A}}{\mathrm{2}} \\ $$$$\mathrm{A}_{\mathrm{2square}} =\mathrm{2A} \\ $$$$\mathrm{A}_{\mathrm{grey}} =\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{8}}+\mathrm{2a}^{\mathrm{2}} +\frac{\mathrm{A}}{\mathrm{2}}+\mathrm{2A}−\mathrm{A} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{16}+\pi}{\mathrm{8}}\right)\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{3A}}{\mathrm{2}} \\ $$$$\mathrm{with}\:\mathrm{A}=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{A}_{\mathrm{grey}} =\left(\frac{\mathrm{16}+\pi}{\mathrm{8}}\right)\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{3a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{22}+\pi}{\mathrm{8}}\right)\mathrm{a}^{\mathrm{2}} \\ $$

Commented by I want to learn more last updated on 18/Aug/20

Thanks sir, i really appreciate.

$$\mathrm{Thanks}\:\mathrm{sir},\:\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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