Odpowiedź :
Pole podstawy
[tex]6 \times \frac{ {a}^{2} \sqrt{3} }{4} [/tex]
[tex]6 \times \frac{ {4}^{2} \sqrt{3} }{4} = 6 \times \frac{16 \sqrt{3} }{4} = 6 \times 4 \sqrt{3} = \boxed{ 24 \sqrt{3} }[/tex]
Objętość
[tex]V = \frac{Pp \times H}{3} [/tex]
[tex]V = \frac{24 \sqrt{3} \times 7 }{3} = \frac{168 \sqrt{3} }{3} = \boxed{56 \sqrt{3} }[/tex]
[tex]Pp=6*\frac{a^{2}\sqrt{3} }{4} \\Pp=6*\frac{4^{2}\sqrt{3} }{4} \\Pp=6*\frac{16\sqrt{3} }{4} \\Pp=6*4\sqrt{3} \\Pp=24\sqrt{3}[/tex]
[tex]V=\frac{1}{3}Pp*H\\ V=\frac{1}{3}*24\sqrt{3}*7\\\\V=8\sqrt{3}*7\\ V=56\sqrt{3}[/tex]