$a)$  

$a=32\text{cm}:2=16\text{cm}$

Obliczmy długość wysokości tego trójkąta, korzystając z twierdzenia Pitagorasa.

$h^2+{\left(16\text{cm}\right)}^2={\left(34\text{cm}\right)}^2$  

$h^2+256{\text{cm}}^2=1156{\text{cm}}^2\mid -256{\text{cm}}^2$  

$h^2=900{\text{cm}}^2\mid \sqrt{}$ $h^2=900{\text{cm}}^2\mid \sqrt{}$  

$h=\sqrt{900{\text{cm}}^2}=30\text{cm}$  

$P=\frac{1}{2}\cdot a\cdot h=\frac{1}{{\sout{2}}^1}\cdot {\sout{16}}^8\text{cm}\cdot 30\text{cm}=\underline{240{\text{cm}}^2}$  

$b)$  

${\left(10\text{cm}\right)}^2+x^2={\left(12,5\text{cm}\right)}^2$  

$100{\text{cm}}^2+x^2=156,25{\text{cm}}^2\mid -100{\text{cm}}^2$  

$x^2=56,25{\text{cm}}^2\mid \sqrt{}$  

$x=7,5\text{cm}$  

Długość podstawy jest dwa razy dłuższa i wynosi:

$a=2\cdot 7,5\text{cm}=15\text{cm}$  

$P=\frac{1}{2}\cdot a\cdot h=\frac{1}{{\sout{2}}^1}\cdot 15\text{cm}\cdot {\sout{10}}^5\text{cm}=\underline{75{\text{cm}}^2}$  

 

$c)$ $x=20\text{cm}$  

$a\sqrt{3}=20\text{cm}\mid :\sqrt{3}$  

$a=\frac{20}{\sqrt{3}}\text{cm}=\frac{20}{\sqrt{3}}\text{cm}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{20\sqrt{3}}{3}\text{cm}=6\frac{2}{3}\sqrt{3}\text{cm}$  

$y=6\frac{2}{3}\sqrt{3}\text{cm}$  

$P=\frac{1}{2}\cdot a\cdot h=\frac{1}{{\sout{2}}^1}\cdot \left(20\text{cm}+6\frac{2}{3}\sqrt{3}\text{cm}\right)\cdot {\sout{20}}^{10}\text{cm}=20\text{cm}\cdot 10\text{cm}+6\frac{2}{3}\sqrt{3}\text{cm}\cdot 10\text{cm}=$  

$=200{\text{cm}}^2+\frac{20}{3}\sqrt{3}\text{cm}\cdot 10\text{cm}=200{\text{cm}}^2+\frac{200}{3}\sqrt{3}{\text{cm}}^2=200{\text{cm}}^2+66\frac{2}{3}\sqrt{3}{\text{cm}}^2=\underline{\left(200+66\frac{2}{3}\sqrt{3}\right){\text{cm}}^2}$