na-trojkacie-prostokatnym-ktorego-przyprostokatne-maja-dlugosci-12-i-9

$\sqrt{108}$\sqrt{108}

$\displaystyle \frac{15}{2}$\displaystyle \frac{15}{2}

$15$15

$\displaystyle \frac{\sqrt{108}}{2}$\displaystyle \frac{\sqrt{108}}{2}