Tr BOC, m(<O) =90°
Teorema lui Pitagora =>
[tex] {bc}^{2} = {oc}^{2} + {ob}^{2} \\ 625 = 400 + {ob}^{2} \\ {ob}^{2} = 625 - 400 = 225 \\ ob = 15cm \\ bd = 2ob = 30cm \\ [/tex]
[tex]a = \frac{bd \times ac}{2} = \frac{30 \times 40}{2} = 30 \times 20 = 600 \: {cm}^{2} [/tex]
