a=26cm - bok
e,f - przekątne
Wyznaczam e
[tex]e-f=14\\\\e=14+f[/tex]
Obliczam f
[tex](\frac{1}{2}e)^2+(\frac{1}{2}f)^2=a^2\\\\\frac{1}{4}e^2+\frac{1}{4}f^2=26^2\\\\\frac{1}{4}e^2+\frac{1}{4}f^2=676\ \ \ |\cdot4\\\\e^2+f^2=2704\\\\(14+f)^2+f^2-2704=0\\\\196+28f+f^2+f^2-2704=0\\\\2f^2+28f-2508=0\ \ \ |:2\\\\f^2+14f-1254=0\\\\\Delta=14^2-2\cdot1\cdot(-1256)=196+2508=2704\\\\\sqrt{\Delta}=\sqrt{2704}=52\\\\f_1=\frac{-14-52}{2}=\frac{-66}{2}=-33 < 0\\\\f_2=\frac{-14+52}{2}=\frac{38}{2}=19cm[/tex]
Oblicza e
[tex]e=14+f\\\\e=14+19\\\\e=33cm[/tex]
Obliczam pole
[tex]P=\frac{ef}{2}\\\\P=\frac{33\cdot19}{2}\\\\P=\frac{627}{2}\\\\P=313,5cm^2[/tex]
