Odpowiedź:
Pole tego trójkąta P = 48 cm².
Szczegółowe wyjaśnienie:
[tex]P = 200 \ cm^{2}\\i\\P = a^{2}\\\\a^{2} = 200\\\\a = \sqrt{200} = \sqrt{100\cdot2}\\\\a = c = 10\sqrt{2} \ cm[/tex]
Długości boków trójkąta tworzą ciąg arytmetyczny o wyrazach:
[tex]a, \ a - r, \ a - 2r\\\\Z \ twierdzenia \ Pitagorasa:\\\\(a-r)^{2} + (a-2r)^{2} = a^{2}\\\\a^{2}-2ar + r^{2}+a^{2}-4ar + 4r^{2} = a^{2}\\\\5r^{2}-6ar + a^{2} = 0\\\\podstawiam \ a = 10\sqrt{2} \ cm\\\\5r^{2}-6\cdot10\sqrt{2}\cdot r + (10\sqrt{2})^{2} = 0\\\\5r^{2}-60\sqrt{2}r + 200 = 0 \ \ /:5\\\\r^{2}-12\sqrt{2}r + 40 = 0\\\\\Delta = b^{2}-4ac =(-12\sqrt{2})^{2}-4\cdot1\cdot40 = 288-160 = 128\\\\\sqrt{\Delta} = \sqrt{128} = \sqrt{64\cdot2} = 8\sqrt{2}[/tex]
[tex]r_1 = \frac{12\sqrt{2}-8\sqrt{2}}{2} =\frac{4\sqrt{2}}{2} = 2\sqrt{2}\\\\r_2 = \frac{12\sqrt{2}+8\sqrt{2}}{2} = \frac{20\sqrt{2}}{2} = 10\sqrt{2} \ - \ odpada\\\\r = 2\sqrt{2}[/tex]
Boki tego trójkąta to:
[tex]a = 10\sqrt{2}-2\sqrt{2} = 8\sqrt{2} \ cm\\\\b = 10\sqrt{2}-2\cdot2\sqrt{2} = 10\sqrt{2}-4\sqrt{2} = 6\sqrt{2} \ cm\\\\P = \frac{1}{2}ab\\\\P = \frac{1}{2}\cdot8\sqrt{2}\cdot6\sqrt{2} = 48 \ cm^{2}[/tex]
