Question Number 138758 by EnterUsername last updated on 17/Apr/21
Answered by mr W last updated on 18/Apr/21
Commented by mr W last updated on 18/Apr/21
$${AD}=\mathrm{6}−\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{4}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${AB}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{4}}{\mathrm{2}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${DB}=\sqrt{\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{6}} \\ $$$${EC}=\frac{\mathrm{6}}{\mathrm{2}}−\frac{\mathrm{4}}{\mathrm{2}}=\mathrm{1} \\ $$$${BC}=\frac{\mathrm{6}}{\mathrm{2}}+\frac{\mathrm{4}}{\mathrm{2}}=\mathrm{5} \\ $$$${BE}=\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{6}} \\ $$$${l}={DB}+{BE}=\sqrt{\mathrm{6}}+\mathrm{2}\sqrt{\mathrm{6}}=\mathrm{3}\sqrt{\mathrm{6}} \\ $$