Odpowiedź:
[tex]\boxed{P_{c} = 12(\sqrt{3}+9)} \ \ [j^{2}]\\\\\boxed{V = 27\sqrt{3}} \ \ [j^{3}][/tex]
Szczegółowe wyjaśnienie:
[tex]a = 2\\H = 9\\\\P_{c} = 2P_{p}+P_{b}\\\\P_{p} = 6\cdot\frac{a^{2}\sqrt{3}}{4} = \frac{3a^{2}\sqrt{3}}{2}\\\\P_{s} = aH\\\\P_{b}= 6P_{s} = 6aH\\\\P_{c} = 2\cdot\frac{3a^{2}\sqrt{3}}{4}+6aH = 3a^{2}\sqrt{3}+6aH\\\\P_{c} = 3\cdot2^{2}\sqrt{3}+6\cdot2\cdot9 = 3\cdot4\sqrt{3}+108\\\\P_{c} = 12\sqrt{3}+108\\\\\boxed{P_{c} = 12(\sqrt{3}+9)}[/tex]
[tex]V = P_{p}\cdot H\\\\V = \frac{3a^{2}\sqrt{3}}{4}\cdot H\\\\V = \frac{3\cdot2^{2}\sqrt{3}}{4}\cdot9=\frac{3\cdot4\sqrt{3}}{4}\cdot9\\\\\boxed{V = 27\sqrt{3}}[/tex]
