[tex]e = 2 \sqrt{10} \\ f = 2 \sqrt{6} [/tex]
Pole
[tex] \frac{2 \sqrt{10} \times 2 \sqrt{6} }{2} = \sqrt{10} \times 2 \sqrt{6} = 2 \sqrt{60} = 4 \sqrt{15} [/tex]
Obliczmy bok
[tex]( \sqrt{10} )^{2} + ( \sqrt{6} ) ^{2} = {x}^{2} \\ 10 + 6 = {x}^{2} \\ 16 = {x}^{2} \\ x = 4[/tex]
Obwód:
[tex]4 \times 4 = 16[/tex]
Myślę że pomogłem ;)
