Question Number 93717 by john santu last updated on 14/May/20
Answered by john santu last updated on 14/May/20
Commented by john santu last updated on 14/May/20
$$\frac{{a}}{\mathrm{2}}\:=\:\frac{{b}+\mathrm{3}}{\mathrm{4}}\:\Rightarrow\:{a}\:=\:\frac{{b}+\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{b}}{\mathrm{3}}\:=\:\frac{{a}+\mathrm{2}}{\mathrm{4}}\:\Rightarrow{a}\:=\:\frac{\mathrm{4}{b}}{\mathrm{3}}−\mathrm{2} \\ $$$$\Rightarrow\frac{{b}+\mathrm{3}}{\mathrm{2}}\:=\:\frac{\mathrm{4}{b}−\mathrm{6}}{\mathrm{3}}\: \\ $$$$\mathrm{8}{b}−\mathrm{12}\:=\:\mathrm{3}{b}+\mathrm{9}\:\Rightarrow\:{b}\:=\:\frac{\mathrm{21}}{\mathrm{5}} \\ $$$$\Rightarrow{a}\:=\:\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{21}}{\mathrm{5}}−\mathrm{2}\:=\:\frac{\mathrm{28}−\mathrm{10}}{\mathrm{5}} \\ $$$${a}\:=\:\frac{\mathrm{18}}{\mathrm{5}}.\:{G}\mathrm{reen}\:\mathrm{area}\:=\:{a}+{b}\:=\:\frac{\mathrm{39}}{\mathrm{5}} \\ $$
Commented by i jagooll last updated on 15/May/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$