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2014-06-15T14:07:27+03:00


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 :*


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