[tex]Zadanie\\\\a=4\\b=x\ (druga\ przyprostokatna)\\\\\frac{a}{b}=tg\alpha\\\\\frac{4}{x}=tg30^o\\\\\frac{4}{x}= \frac{\sqrt{3}}{3}\\\\x\sqrt{3}=12\ \ \ \mid\cdot\sqrt{3}\\\\3x =12\sqrt{3}\ \ \ \mid:3\\\\ x=4\sqrt{3}\\\\Szesciokat\ sklada\ sie\ z\ szesciu\ trojkatow\ rownobocznych\ o\ boku:\\\\c=\frac{x}{2}=\frac{4\sqrt{3} }{2}=2\sqrt{3} \\\\P_p=6\cdot\frac{a^{2} \sqrt{3}}{4}\\\\P_p=6\cdot\frac{(2\sqrt{3})^{2}\cdot \sqrt{3}}{4}=6\cdot\frac{12 \sqrt{3} }{4}=6\cdot3\sqrt{3}=18\sqrt{3}\ (j^{2})[/tex]
[tex]Objetosc\ graniastoslupa\\\\H=4\\\\V=P_p\cdot H\\\\V=18\sqrt{3}\cdot4=72\sqrt{3}\ (j^{3})[/tex]
