7.
b)
[tex]|AB| = a, \ \ |BC| = b, \ \ |AC| = c\\\\Z \ tw. Pitagorasa:\\\\7^{2}+1^{2} = a^{2}\\\\49+1 = a^{2}\\\\a^{2} = 50\\\\a = \sqrt{50} = \sqrt{25\cdot2}\\\\\underline{a = 5\sqrt{2}}\\\\\\3^{2}+5^{2} = b^{2}\\\\9+25 = a^{2}\\\\a^{2} = 34\\\\\underline{b = \sqrt{34}}[/tex]
[tex]6^{2}+4^{2} = c^{2}\\\\36+16 = c^{2}\\\\c^{2} = 52\\\\c = \sqrt{4\cdot13}\\\\\underline{c = 2\sqrt{13}}\\\\Obw = a+b+c\\\\\underline{Obw = (5\sqrt{2}+\sqrt{34}+2\sqrt{13}) \ [j]}[/tex]
Pole trójkąta ABC liczę jako różnicę pola prostokąta o bokach 6 i 7 oraz trzech mniejszych trójkątów prostokątnych:
[tex]P = 6\cdot7 - (\frac{6\cdot4}{2}+\frac{7\cdot1}{2}+\frac{3\cdot5}{2}) = 42 - \frac{24+7+15}{2} = 42-\frac{46}{2} = 42-23\\\\\underline{P = 19 \ [j^{2}]}[/tex]
