obwod-trojkata-abc-przedstawionego-na-rysunku-jest-rowny

$\displaystyle \left(3+ \frac{\sqrt{3}}{2} \right)a$\displaystyle \left(3+ \frac{\sqrt{3}}{2} \right)a

$\displaystyle \left(2+ \frac{\sqrt{2}}{2} \right)a$\displaystyle \left(2+ \frac{\sqrt{2}}{2} \right)a

$\left(3+\sqrt{3}\right)a$\left(3+\sqrt{3}\right)a

$\left(2+\sqrt{2}\right)a$\left(2+\sqrt{2}\right)a