1.3. The base of an isosceles triangle is $4 \sqrt{2} \mathrm{~cm}$, and the median of the lateral side is $5 \mathrm{~cm}$. Find the lengths of the lateral sides.
1.3. Since $\triangle A B C$ is isosceles, the height $B D$ is also a median (Fig. 1.2); the point of intersection of the medians divides each of them in the ratio $2: 1$, hence $A O=\frac{10}{3} \quad$ (cm). From $\triangle A O D$ we find $O D=\sqrt{A O^{2}-A D^{2}}=$ $=\sqrt{\left(\frac{10}{3}\right)^{2}-(2 \sqrt{2})^{2}}=\frac{\sqrt{28}}{3}$ (cm), i.e., $B D=\sqrt{28}$ (cm). Finally, from $\triangle B D C$ we get $B C=\sqrt{(\sqrt{28})^{2}+(2 \sqrt{2})^{2}}=6(\mathrm{cm})$.

Answer: $6 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-415.jpg?height=463&width=373&top_left_y=1134&top_left_x=46)

Fig. 1.3