Odpowiedź :
[tex]Zadanie\\\\a)\\\\P=a^2\ \ \mid\sqrt{}\\\\\sqrt{P}=a \\\\a=\sqrt{P}\\\\b)\\\\P=\frac{1}{2}\cdot a\cdot h\ \ \mid\cdot2 \\\\2P=a\cdot h\ \ \mid:a\\\\\frac{2P}{a}=h\\\\h=\frac{2P}{a}[/tex]
[tex]a)\\\\P = a^{2}\\\\a^{2} = P\\\\a = \sqrt{P}[/tex]
[tex]b)\\\\P = \frac{1}{2}ah \ \ /\cdot2\\\\ah = 2P \ \ /:a\\\\h = \frac{2P}{a}[/tex]