[tex]V = Pp \times H[/tex]
Zatem pole podstawy
[tex]6 \times \frac{ {a}^{2} \sqrt{3} }{4} \\ \\ 6 \times \frac{ {6}^{2} \sqrt{3} }{4} = 6 \times \frac{36 \sqrt{3} }{4} = 6 \times 9 \sqrt{3} = \\ = \boxed{ 54 \sqrt{3} {cm}^{2} }[/tex]
Objętość
[tex]V = 54 \sqrt{3} {cm}^{2} \times 5cm = 270 \sqrt{3} {cm}^{3} [/tex]
Pole powierzchni całkowitej
[tex]2 \times Pp + 6 \times Psb[/tex]
[tex]2 \times 54 \sqrt{3} {cm}^{2} + 6(6cm \times 5cm) = 108 \sqrt{3} {cm}^{2} + 6 \times 30 {cm}^{2} = \boxed{ 108 \sqrt{3} {cm}^{2} + 180 {cm}^{2} }[/tex]
