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} = \\ = 96 \sqrt{3} {cm}^{2} [/tex]
Wysokość
[tex] {8}^{2} + {x}^{2} = {10}^{2} \\ 64 + {x}^{2} = 100 \\ {x}^{2} = 36 \\ x = \sqrt{36} = 6cm[/tex]
Zatem objętość
[tex]v = \frac{96 \sqrt{3} \times 6}{3} = 96 \sqrt{3} \times 2 = \boxed{ \boxed{192 \sqrt{3} {cm}^{3} }}[/tex]
