$h^2+{\left(9\text{cm}\right)}^2={\left(15\text{cm}\right)}^2$  

$h^2+81{\text{cm}}^2=225{\text{cm}}^2\mid -81{\text{cm}}^2$  

$h^2=144{\text{cm}}^2\mid \sqrt{}$  

$h=12\text{cm}$  

 

 

$P=\frac{1}{2}\cdot a\cdot h$  

$P=\frac{1}{{\sout{2}}^1}\cdot {\sout{18}}^9\text{cm}\cdot 12\text{cm}=108{\text{cm}}^2$  

Odp. Pole tego trójkąta wynosi 108 cm².