10.018. Find the lengths of sides $A B$ and $A C$ of triangle $A B C$, if $B C=8 \mathrm{~cm}$, and the lengths of the heights drawn to $A C$ and $B C$ are 6.4 and 4 cm, respectively.
Solution.

According to the problem, $B C=8 \text{ cm}, A E=4$ cm, $B K=6.4$ cm (Fig. 10.18). From $\triangle B K C \sim \triangle A E C$, it follows that $\frac{6.4}{4}=\frac{8}{A C} ; A C=5$ cm. From $\triangle A E C$, we have $C E=\sqrt{25-16}=3 \text{ cm}, B E=8-3=5 \text{ cm}$; then from $\triangle A E B$, we find $A B=\sqrt{16+25}=\sqrt{41} \text{ cm}$.

Answer: $\sqrt{41}$ cm, 5 cm.