[tex]a) d = a\sqrt{2} \\a = \frac{d}{\sqrt2} \\d = 2\sqrt2cm\\a = \frac{2\sqrt2}{\sqrt2} = 2cm\\P = a^2\\P = 2cm^2 = 4cm^2\\b) d = 8cm\\a = \frac{8}{\sqrt2} = \frac{8\sqrt2}{2} = 4\sqrt2[/tex]
[tex]4\sqrt2 ^2 = 4 * 2 = 8cm^2\\[/tex]
Zadanie z komentarza:
oblicz pole trójkąta równobocznego o boku a) a=1,2cm b)= 3 pierwiastek z 3cm
[tex]P = \frac{a^2\sqrt3}{4} \\P = \frac{1,2^2\sqrt3}{4} P = \frac{1,44\sqrt3}{4} = 0,36\sqrt{3}cm\\\\P = \frac{3\sqrt3 ^2 \sqrt3}{4} \\P = \frac{3 * 3 * \sqrt{3} }{4} = \frac{9 \sqrt3}{4} = 2,25\sqrt3[/tex]cm
