[tex]H = 4[/tex]
[tex]d = 4 \sqrt{3} [/tex]
[tex]a = \frac{1}{2} d[/tex]
[tex]a = \frac{1}{2} \times 4 \sqrt{3} = 2 \sqrt{3} [/tex]
Pole podstawy
[tex]6 \times \frac{(2 \sqrt{3} {)}^{2} \sqrt{3} }{4} = 6 \times \frac{4 \times 3 \sqrt{3} }{4} = 6 \times \frac{12 \sqrt{3} }{4} = 6 \times 3 \sqrt{3} = 18 \sqrt{3} [/tex]
Objętość
[tex]V = 18 \sqrt{3} \times 4 = \boxed{\underline{72\sqrt{3} }}[/tex]
