$a)$  

$\text{Z tw. Pitagorasa:}$  

$x^2={12}^2+{\left(\frac{x-3}{2}\right)}^2$  

$x^2=144+\frac{1}{4}{x}^2-\frac{6}{4}x+\frac{9}{4}$  

$\frac{3}{4}{x}^2+\frac{3}{2}x-146\frac{1}{4}=0$   

$3x^2+6x-585=0$  

 

$\Delta =36+7020=7056$   

$\sqrt{\Delta }=84$  

$x_1=\frac{-6-84}{6}=-\frac{90}{6}$  

$x_2=\frac{-6+84}{6}=\frac{78}{6}=13$  

$x>0\Rightarrow x=13$  

$P=\frac{1}{2}\cdot 10\cdot 12=60$  

${\text{Odp. Pole troˊjkąta wynosi 60 cm}}^2.$  

 

 

$b)$  

$P=\frac{1}{2}{h}_2\cdot x=60$  

$\frac{1}{2}\cdot 13\cdot h_2=60$  

$13h_2=120$  

$\underline{h_2=\frac{120}{13}=9\frac{3}{13}}$