Question Number 176092 by a.lgnaoui last updated on 12/Sep/22
$$\angle{AOB}\:\:\:{triangle}\:{equilareral}\:{de}\:{cote}\:\boldsymbol{{a}} \\ $$$$\left({A};\mathrm{B}\right)\::\:\mathrm{c}{e}\mathrm{ntres}\:\mathrm{de}\:\mathrm{cercles}\:\mathrm{de}\:\mathrm{rayon}\:\boldsymbol{\mathrm{r}} \\ $$$$\boldsymbol{\theta}\:=\measuredangle\:\:{GOH}\:\:\:\mathrm{EF}=\boldsymbol{{a}}_{\mathrm{0}} \\ $$$${Determiner}\:{l}\:{aire}\:{de}\:{l}\:{espace}\:{delimite}\:{par} \\ $$$$\:\:\boldsymbol{\mathrm{AFOEBHG}}\:{comme}\:{il}\:{est}\:{marque}\:{sur}\:{l}\:{image}\:{ci}−{joint}\: \\ $$$${en}\:{fonction}\:{de}:\:\boldsymbol{{a}},\boldsymbol{{r}},\boldsymbol{{a}}_{\mathrm{0}} \:{et}\:\boldsymbol{\theta} \\ $$
Commented by a.lgnaoui last updated on 11/Sep/22
Commented by a.lgnaoui last updated on 12/Sep/22
$$\mathrm{A}\:{et}\:\mathrm{B}\:\:\left({centres}\:{de}\:{cercles}\:{de}\:{rayon}\:\left({r}\right)\right) \\ $$
Commented by a.lgnaoui last updated on 12/Sep/22
$$\boldsymbol{{r}}=\boldsymbol{{m}}=\boldsymbol{{n}}\:\:\:\:\:\:\:\:\:\mathrm{GE}=\mathrm{FH} \\ $$