1.7. The area of triangle $ABC$ is $30 \, \text{cm}^2$. A point $D$ is taken on side $AC$ such that $AD:DC = 2:3$. The length of the perpendicular $DE$, drawn to side $BC$, is $9 \, \text{cm}$. Find $BC$.
1.7. Draw $B D$ (Fig. 1.6); triangles $A B D$ and $B D C$ have a common height; therefore, their areas are in the ratio of the lengths of their bases, i.e., $S_{\triangle A B D}: S_{\triangle B D C}=2: 3$; then $S_{\triangle B D C}=\frac{3}{5} S_{\triangle A B C}=18 \mathrm{~cm}^{2} . \mathrm{C}$

On the other hand, according to formula (1.1), $S_{\triangle B D C}=\frac{1}{2} B C \cdot D E$, i.e., $18=\frac{1}{2} B C \cdot 9$, from which $B C=4 \mathrm{~cm}$.

Answer: 4 cm.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-417.jpg?height=392&width=479&top_left_y=1467&top_left_x=45)

Fig. 1.7