5. Zadatak

$$\frac{|AC|}{|MC|}=\frac{40}{16}=\frac52,\quad \frac{|BC|}{|NC|}=\frac{30}{12}=\frac52,$$

Dakle, $\frac{|AC|}{|MC|}=\frac{|BC|}{|NC|}$ i zajednički kut pri vrhu C pa prema poučku S-K-S o sličnosti slijedi $$\triangle MNC\sim\triangle ABC.$$

$$\frac{P_{\triangle ABC}}{P_{\triangle MNC}}=\frac52\cdot\frac52$$ $$\frac{P_{\triangle MNC}+252}{P_{\triangle MNC}}=\frac{25}{4}$$ $$4\cdot P_{\triangle MNC}+1008=25\cdot P_{\triangle MNC}$$ $$21\cdot P_{\triangle MNC}=1008$$ $$P_{\triangle MNC}=1008:21=48\ \mathrm{cm}^2.$$

Površina trokuta MNC je 48 cm².