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In figura 1, VABC este o piramida triunghiulara regulata cu AM = 6 radical din 3 si VO = 2 radical din 6. Sa se afle: a) AB = ... cm ; b) VM = ... cm ; c)
Aria laterala = ... cm3


Răspuns :

a) AB=2*AM=2*6√3=12√3
b) ΔABC-echilateral⇒MC=[tex] \frac{l \sqrt{3} }{2} [/tex]
MC=[tex] \frac{12 \sqrt{3 }* \sqrt{3} }{2} = \frac{6}{3} =18[/tex]
OM=[tex] \frac{1}{3} *MC[/tex]
MO=6
In ΔVOM-aplici Pitagora⇒VM²=VO²+OM²
VM²=36+24=60
VM=√60=2√15
c) Al=[tex] \frac{Pb*ap}{2} [/tex]
Pb=12√3*3=36√3
ap=VM=2√15
Al=[tex] \frac{36 \sqrt{3} *2 \sqrt{15} }{2} =36 \sqrt{45} =36*9 \sqrt{5} =324 \sqrt{5} [/tex]