pole podstawy
[tex] Pp=a^2=2^2=4 j^2\\
h_{b} = \sqrt{ {b}^{2} - ( \frac{1}{2} a {)}^{2} } = \sqrt{( \sqrt{17 })^{2} - 1^{2} } = \sqrt{16} = 4 \: \: j\\ \\ Pb = 4 \times \frac{1}{2} a \times h_{b} = 4 \times 1 \times 4 = 16\: {j}^{2} \\ Pc = Pp + Pb = 4 + 16 = 20 \: {j}^{2} [/tex]
