Odpowiedź:
[tex] v = \frac{1}{3} pp \times h[/tex]
V -objetosc
pp-pole podstawy
h-wysokosc ostrosłupa
a-bok trojkata rownobocznego
H=3cm
a=8cm
[tex]pp = \frac{ {a}^{2} \sqrt{3} }{4}[/tex]
[tex]pp = \frac{ {8}^{2} \sqrt{3 } }{4} = \frac{64 \sqrt{3} }{4} = 16 \sqrt{3} [/tex]
[tex]v = \frac{1}{3} \times 16 \sqrt{13} \times 3 = 16 \sqrt{3} {cm}^{2} [/tex]
