Zadanie 2
z twierdzenia Pitagorasa:
H^2 =(3a)^2 - (\frac{2}{3}a)^2
H^2 = 9a^2 - \frac{4}{9} * (\frac{a\sqrt3}{2})^2
H^2 = 9a^2 - \frac{4}{9} * \frac{3a^2}{4}
H^2 = \frac{27a^2}{3}-\frac{a^2}{3}
H = a \sqrt{\frac{26}{3}}
V = 1/3 Pp * H
\frac{1}{3} * \frac{a^2\sqrt3}{4} * a{\frac{\sqrt{26}}{\sqrt3}} = 2\sqrt{26}
\frac{a^3 \sqrt{26}}{12} = 2\sqrt{26} |:\sqrt{26}
\frac{a^3}{12}=2
a^3 = 24
a=\sqrt[3]{24}
a = 2\sqrt[3]8 <-- odpowiedź