Question Number 177634 by aurpeyz last updated on 07/Oct/22
Answered by HeferH last updated on 07/Oct/22
$$\: \\ $$$$\:\frac{{CD}}{{AD}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\: \\ $$$$\:\frac{{A}_{\mathrm{1}} }{{A}_{{t}} }\:\:=\:\frac{\mathrm{1}}{\mathrm{9}}\Rightarrow\:\frac{\mathrm{42}}{{A}_{{t}} }\:=\:\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\:\mathrm{42}\centerdot\mathrm{9}\:=\:{A}_{{t}} \:=\:\mathrm{378}\: \\ $$
Commented by aurpeyz last updated on 07/Oct/22
$${sorry}.\:{how}\:{did}\:{you}\:{get}\:\frac{\mathrm{1}}{\mathrm{9}}? \\ $$
Commented by HeferH last updated on 07/Oct/22
$${When}\:{two}\:{triangles}\:{are}\:{similar},{the}\:{ratio} \\ $$$$\:{of}\:{their}\:{areas}\:{is}\:{the}\:{ratio}\:{of}\:{their}\:{sides} \\ $$$$\:{squared}.\:{In}\:{this}\:{case}\:\:\:\:\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{9}} \\ $$
Commented by aurpeyz last updated on 09/Oct/22
$${thanks} \\ $$