[tex]148 \sqrt{3} = \frac{Pp \times H}{3} [/tex]
Pole podstawy
[tex]6 \times \frac{ {a}^{2} \sqrt{3} }{4} = 6 \times \frac{ {4}^{2} \sqrt{3} }{4} = 6 \times \frac{16 \sqrt{3} }{4} = 6 \times 4 \sqrt{3} = 24 \sqrt{3} [/tex]
Zatem
[tex]148 \sqrt{3} = \frac{24 \sqrt{3} \times H}{3 } \times 3 \\ 444 \sqrt{3} = 24 \sqrt{3} \times H | \div 24 \sqrt{3} \\ H = 18.5[/tex]
