[tex]Pc = Pp + Pb[/tex]
Pole podstawy
[tex]6 \times \frac{ {a}^{2} \sqrt{3} }{4} \\ \\ 6 \times \frac{ {8}^{2} \sqrt{3} }{4} = 6 \times \frac{64 \sqrt{3} }{4} = 6 \times 16 \sqrt{3} = \\ = \boxed{ 96 \sqrt{3} }[/tex]
Pole ściany bocznej
[tex]a \times b \\ \\ 8 \times 3 = \boxed{24}[/tex]
Pole całkowite
[tex]96 \sqrt{3} + 6 \times 24 = \boxed{(96 \sqrt{3} + 144 ){j}^{2} }[/tex]
