Al=Pb*Ap/2
VABC PIRAMIDA TRIUNGHIULARA REGULATA
O CENTRUL BAZEI ABC
SE DUCE DIN C O PERPENDICULARA PE DREAPTA AB CARE O INTERSECTEAZA IN PUNCTUL M
OC=2/3 *CM
CM=l√3.2
OC=2/3*l√3/2
OC=l√3/9
VC²=VO²+OC²
(2√13)²=4²+(l√3/9)²
52=16+3l²/81
l²/27=52-16
l²/27=36
l²/27=972/27
l²=972
l=18√3
P=3*18√3
p=54√3
VM²=VO²+OM²
OM=1/3*18√3
OM=6√3
VM²=16+108
VM=2√31
Al=54√2*2√31/2
Al=54√62
V=Ab*h/3
Ab=l²√3/4
Ab=(18√3)²√3/4
Ab=972√3/4
Ab=243√3
V=243√3*4/3
V=81√3*4
V=324√3
sper ca te-am ajutat...te pwp :*
