$\left|\angle BSC\right|=\left|\angle ACB\right|$    (bo trójkąt SCB jest równoboczny)

$0,5\left|AC\right|=\left|SC\right|$

$\left|SC\right|=\left|BC\right|$

$\left|BC\right|=0,5\left|AC\right|=0,5\cdot 12=6$

 

Dla trójkąta prostokątnego ABC:

${\left|AB\right|}^2+{\left|BC\right|}^2={\left|AC\right|}^2$

${\left|AB\right|}^2+6^2={12}^2$

${\left|AB\right|}^2=144-36$

${\left|AB\right|}^2=108$

$\left|AB\right|=\sqrt{108}=\sqrt{36}\cdot \sqrt{3}=6\sqrt{3}$

$P=\left|BC\right|\cdot \left|AB\right|=6\cdot 6\sqrt{3}=36\sqrt{3}$