Question Number 193052 by a.lgnaoui last updated on 02/Jun/23
$$\boldsymbol{\mathrm{determiner}}\:\boldsymbol{\mathrm{la}}\:\boldsymbol{\mathrm{valeur}}\:\boldsymbol{\mathrm{de}}\:\:\boldsymbol{\mathrm{r}} \\ $$
Commented by a.lgnaoui last updated on 02/Jun/23
Answered by deleteduser1 last updated on 02/Jun/23
$${r}\left(\frac{\mathrm{5}+\mathrm{4}+\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}×\mathrm{3}\right)\Rightarrow{r}=\mathrm{1} \\ $$
Commented by a.lgnaoui last updated on 02/Jun/23
$$\mathrm{detail}\:\mathrm{calcul}? \\ $$
Commented by deleteduser1 last updated on 02/Jun/23
$${Area}\:{of}\:\Delta={rs}=\frac{\mathrm{1}}{\mathrm{2}}{bh}\:\:\:\:\:\:\:\left({s}=\frac{{perimeter}}{\mathrm{2}}\right) \\ $$
Commented by a.lgnaoui last updated on 03/Jun/23
$$\mathrm{thanks} \\ $$
Answered by Subhi last updated on 02/Jun/23
$${x}+{y}\:=\:\mathrm{4} \\ $$$${y}+{z}\:=\:\mathrm{3} \\ $$$${x}+{z}\:=\:\mathrm{5} \\ $$$$\mathrm{4}−{x}+{z}\:=\:\mathrm{3}\:\:\:\Rrightarrow\:\:{z}={x}−\mathrm{1} \\ $$$$\mathrm{2}{x}−\mathrm{1}=\mathrm{5}\:\:\:\Rrightarrow\:{x}=\mathrm{3} \\ $$$${y}=\mathrm{1} \\ $$$${z}=\mathrm{2} \\ $$$${radius}\:=\:{y}\:=\:\mathrm{1} \\ $$
Commented by Subhi last updated on 02/Jun/23
Commented by a.lgnaoui last updated on 02/Jun/23
$$\mathrm{exact},\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Subhi last updated on 02/Jun/23
$${You}\:{are}\:{welcome} \\ $$