10.145. The height of the rhombus is $12 \mathrm{~cm}$, and one of its diagonals is $15 \mathrm{~cm}$. Find the area of the rhombus.
Solution.

Let $B K=12$ (cm), $B D=15$ (cm) (Fig. 10.130). From $\triangle B K D$ $\sin \alpha=\frac{B K}{B D}=\frac{12}{15}=\frac{4}{5} ;$ since $0<\alpha<90^{\circ}$, then $\cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$.

From $\triangle A O D \cos \alpha=\frac{O D}{A D}=\frac{B D}{2 A D}$, hence $A D=\frac{B D}{2 \cos \alpha}=\frac{15 \cdot 5}{2 \cdot 3}=\frac{75}{6}$ (cm). The area of the rhombus $S=A D \cdot B K=\frac{75}{6} \cdot 12=150\left(\mathrm{~cm}^{2}\right)$.

Answer: $150 \mathrm{~cm}^{2}$.