pole-trojkata-prostokatnego-abc-przedstawionego-na-rysunku-jest-rowne

$\displaystyle \frac{32\sqrt{3}}{6}$\displaystyle \frac{32\sqrt{3}}{6}

$\displaystyle \frac{16\sqrt{3}}{6}$\displaystyle \frac{16\sqrt{3}}{6}

$\displaystyle \frac{8\sqrt{3}}{3}$\displaystyle \frac{8\sqrt{3}}{3}

$\displaystyle \frac{4\sqrt{3}}{3}$\displaystyle \frac{4\sqrt{3}}{3}