Question Number 199821 by a.lgnaoui last updated on 09/Nov/23
$$\mathrm{ABFE}\:\:\mathrm{Care} \\ $$$$\mathrm{determiner}\:\boldsymbol{\mathrm{x}}\:\mathrm{en}\:\mathrm{fonction}\:\mathrm{de}\:\mathrm{a}\:\mathrm{etb} \\ $$$$\mathrm{BC}=\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\mathrm{DE}=\:\boldsymbol{\mathrm{b}} \\ $$
Commented by a.lgnaoui last updated on 09/Nov/23
Commented by mr W last updated on 10/Nov/23
$${it}\:{is}\:{solved}. \\ $$$${x}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$ \\ $$$${see}\:{Q}\mathrm{199766} \\ $$