Odpowiedź:
R-promień okręgu opisanego na Δ
R=2/3 h
[tex]\displaystyle R=\frac{2}{3} h=\frac{2}{3} \frac{a\sqrt{3} }{2} =\frac{a\sqrt{3} }{3} =\frac{8\sqrt{3} }{3} \\P_{\circ}=\pi R^2=\pi \cdot\left(\frac{8\sqrt{3} }{3} \right)^2=\pi \frac{64\cdot3}{9} =\frac{64\pi }{3} \\P_{\bigtriangleup}=\frac{a^{2} \sqrt{3} }{4} =\frac{8^2\sqrt{3} }{4} =16\sqrt{3} \\P_F=\frac{P_{\circ}-P_{\bigtriangleup}}{3} =\frac{\frac{64\pi }{3} -16\sqrt{3}/\cdot3 }{3/\cdot3} =\frac{64\pi -48\sqrt{3} }{9}[/tex]
A
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Odpowiedź:
R-promień okręgu opisanego na Δ
R=2/3 h
[tex]\displaystyle R=\frac{2}{3} h=\frac{2}{3} \frac{a\sqrt{3} }{2} =\frac{a\sqrt{3} }{3} =\frac{8\sqrt{3} }{3} \\P_{\circ}=\pi R^2=\pi \cdot\left(\frac{8\sqrt{3} }{3} \right)^2=\pi \frac{64\cdot3}{9} =\frac{64\pi }{3} \\P_{\bigtriangleup}=\frac{a^{2} \sqrt{3} }{4} =\frac{8^2\sqrt{3} }{4} =16\sqrt{3} \\P_F=\frac{P_{\circ}-P_{\bigtriangleup}}{3} =\frac{\frac{64\pi }{3} -16\sqrt{3}/\cdot3 }{3/\cdot3} =\frac{64\pi -48\sqrt{3} }{9}[/tex]
A