Question Number 210050 by Batmath last updated on 29/Jul/24
Answered by mr W last updated on 29/Jul/24
Commented by mr W last updated on 29/Jul/24
$$\frac{\mathrm{2}}{\mathrm{3}}×\left(\mathrm{6}+\frac{\mathrm{3}{x}}{\mathrm{2}}\right)=\mathrm{3}{x} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$
Answered by A5T last updated on 29/Jul/24
Commented by A5T last updated on 29/Jul/24
$$\frac{{AD}}{{DB}}×\frac{{BC}}{{CG}}×\frac{{GE}}{{EA}}=\mathrm{1}\Rightarrow\frac{{AD}}{{DB}}×\mathrm{2}×\frac{\mathrm{3}{x}}{\mathrm{12}}=\mathrm{1}\Rightarrow\frac{{AD}}{{DB}}=\frac{\mathrm{2}}{{x}} \\ $$$$\frac{{CG}}{{GB}}×\frac{{BA}}{{AD}}×\frac{{DE}}{{EC}}=\mathrm{1}\Rightarrow\mathrm{1}×\frac{\mathrm{2}+{x}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}+{x}=\mathrm{4}\Rightarrow{x}=\mathrm{2} \\ $$
Answered by A5T last updated on 29/Jul/24
Commented by A5T last updated on 29/Jul/24
$${G}\:{and}\:{G}'\:{are}\:{the}\:{centroids}\:{of}\:\bigtriangleup{ABC}\:{and}\:\bigtriangleup{GBC} \\ $$$${respectively}\Rightarrow\frac{{GG}'}{{x}}=\mathrm{2}\:{and}\:\frac{\mathrm{12}}{{GG}'+{x}}=\mathrm{2} \\ $$$$\Rightarrow\frac{\mathrm{12}}{\mathrm{2}{x}+{x}}=\mathrm{2}\Rightarrow{x}=\mathrm{2} \\ $$