Question

aria latera a unei piramide patrulatere regulate cu muchia bazei 6 si h=3radical din 3 =?

Answer 1

Vo=h=3$\sqrt{3}$
AB=6 -----> OM=Apatra=$\frac{AB}{2}$=3
in Triunghiul VOM masura O=90 grade -------> t pitagora  VM^2=VO^2+OM^2 
Vm^2=27+9
VM=$\sqrt{36}$
Vm=6
P bazei =4AB
            = 24
A lat = $\frac{Pbaza X Apotema piramida }{2}$
       = $\frac{24X3 \sqrt{3} }{2}$
        = 12x3$\sqrt{3}$=36$\sqrt{3}$