[tex]obliczamy\ pole\ trojkata\ ABC \ z wysokoscia\ opadajaca\ na\ bok\ AB\\\\P=\frac{1}{\not{2}^1}*\not{10}^5*6=30 \\\\ obliczamy\ dlugosc\ boku: |AC|,\ wysokosc\ opadajaca\ na\ ten\ bok\ wynosi\ 4 \\\\P=\frac{1}{2}*|AC|*4\\\\ \frac{1}{\not{2}^1}*|AC| *\not{4}^2=30\\\\ 2|AC|=30\ \ |:2\\\\|AC|=15\\\\odp.\ C.\ \ 15[/tex]
