Pole rombu
[tex] \frac{e \times f}{2} = \frac{10 \times 24}{2} = \frac{240}{2} = 120 {cm}^{2} [/tex]
Bok rombu
[tex] {5}^{2} + {12}^{2} = {x}^{2} \\ 25 + 144 = {x}^{2} \\ {x}^{2} = 169 \\ x = \sqrt{169} \\ x = 13m[/tex]
Wysokość rombu
[tex]120 = 13 \times h | \div 13 \\ h = \frac{120}{13} \\ h = 9 \frac{3}{13} cm[/tex]
Obwód rombu
[tex]4 \times 13 = 52cm[/tex]
