Daca AC perpendicular pe BC⇒in ΔACB <C=90³ iar <A = 30³ a) sin B=AC/AB=√3/2⇒AC/24=√3/2⇒AC=24√3/2=12√3 cm Conform teoremei <30³⇒BC=AB/2=24/2=12 cm b)Ridicam inaltimile trapezului AP si AN In ΔCNB <C=30³ si conform teoremei <de 30³⇒NB=BC/2⇒NB=12/2⇒NB=6 cm NB=AP= 6 cm⇒CD=AB-(AP+NB)=24-12=12 cm P=AB+BC+CD+AD=12+12+12+24=60 cm c)Nu stiu cum ....Sper ca te-am ajutat cat de cat!