Odpowiedź:
Pp = [tex]6*\frac{a^{2}\sqrt{3} }{4} = 6 *\frac{8^{2} \sqrt{3} }{4} = 6*\frac{64\sqrt{3} }{4} = 6*16\sqrt{3} =96\sqrt{3} cm^{2}[/tex]
Pb = [tex]4 * \frac{ah}{2} = \frac{8*\sqrt{33} }{2} = 4\sqrt{33} cm^{2}[/tex]
z tw. Pitagorasa
[tex]h^{2} + 4^{2} = 7^{2} \\h^{2} + 16 = 49 \\h^{2} = 49 - 16\\h^{2} = 33\\h = \sqrt{33}[/tex]
Pc = Pp + Pb = [tex]96\sqrt{3}cm^{2} +4\sqrt{33} cm^{2}[/tex]
Odp. Pole powierzchni wynosi [tex]96\sqrt{3}cm^{2} +4\sqrt{33} cm^{2}[/tex].
