Question Number 114379 by A8;15: last updated on 18/Sep/20
Commented by A8;15: last updated on 18/Sep/20
please help
Answered by mr W last updated on 19/Sep/20
Commented by mr W last updated on 19/Sep/20
$${AC}={x} \\ $$$${AE}={HF}=\sqrt{\left(\mathrm{4}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{4} \\ $$$${CE}={x}−\mathrm{4}={CG} \\ $$$${BG}={BC}−{CG}=\sqrt{{x}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} }−{x}+\mathrm{4} \\ $$$${BF}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\mathrm{65} \\ $$$${BF}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +{BG}^{\mathrm{2}} =\mathrm{1}+\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} }−{x}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}+\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} }−{x}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{65} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} }−{x}+\mathrm{4}=\mathrm{8} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} }=\mathrm{4}+{x} \\ $$$$\Rightarrow{x}=\mathrm{6} \\ $$
Commented by A8;15: last updated on 19/Sep/20
thank you sir!