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Question-114379




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

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