Question Number 149865 by Naser last updated on 07/Aug/21
Answered by mr W last updated on 07/Aug/21
Commented by mr W last updated on 07/Aug/21
$${A}_{{total}} ={b}×{b}={b}^{\mathrm{2}} \\ $$$${y}_{{total}} =\frac{{b}}{\mathrm{2}} \\ $$$${A}_{\mathrm{1}} =\frac{{b}}{\mathrm{2}}×\frac{{b}}{\mathrm{2}}=\frac{{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${A}_{\mathrm{3}} =\frac{{b}}{\mathrm{3}}×{b}=\frac{{b}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${A}_{\mathrm{2}} ={A}_{{total}} −{A}_{\mathrm{1}} −{A}_{\mathrm{3}} ={b}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}−\frac{{b}^{\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{5}{b}^{\mathrm{2}} }{\mathrm{12}} \\ $$$${y}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}×\frac{{b}}{\mathrm{2}}=\frac{{b}}{\mathrm{6}} \\ $$$${y}_{\mathrm{2}} =? \\ $$$${y}_{\mathrm{3}} =\frac{\mathrm{3}{b}}{\mathrm{4}} \\ $$$${A}_{\mathrm{1}} {y}_{\mathrm{1}} +{A}_{\mathrm{2}} {y}_{\mathrm{2}} +{A}_{\mathrm{3}} {y}_{\mathrm{3}} ={A}_{{total}} {y}_{{total}} \\ $$$$\frac{{b}^{\mathrm{2}} }{\mathrm{4}}×\frac{{b}}{\mathrm{6}}+\frac{\mathrm{5}{b}^{\mathrm{2}} }{\mathrm{12}}×{y}_{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{\mathrm{3}}×\frac{\mathrm{3}{b}}{\mathrm{4}}={b}^{\mathrm{2}} ×\frac{{b}}{\mathrm{2}} \\ $$$$\Rightarrow{y}_{\mathrm{2}} =\frac{{b}}{\mathrm{2}} \\ $$$${the}\:{y}−{coordinate}\:{of}\:{centroid} \\ $$$${of}\:{shaded}\:{area}\:{is}\:\frac{{b}}{\mathrm{2}}. \\ $$