[tex]Zadanie\\\\a=13\\c=\sqrt{394} \\b=?\\\\a^{2}+b^{2}=c^{2}\\\\13^{2}+b^{2}=(\sqrt{394})^{2} \\\\169+b^{2}=394\ \ \ \mid-169\\\\b^{2}=225\\\\b=\sqrt{225}\\\\ b=15\ (j)[/tex]
[tex]P=\frac{1}{2}\cdot a\cdot b\\\\P= \frac{1}{2}\cdot 13\cdot 15=\frac{195}{2} =97,5\ (j^{2})[/tex]
