10.421 The height of the triangle, equal to 2 cm, divides the angle of the triangle in the ratio $2: 1$, and the base of the triangle into parts, the smaller of which is 1 cm. Find the area of this triangle.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-254.jpg?height=271&width=424&top_left_y=418&top_left_x=102)
$A D=1.2 \cdot \angle A B D=\angle D B C$. Draw the bisector of angle $D B C$. Then $\angle A B D=\angle D B K=\angle K B C \Rightarrow$ $D K=A D=1$. Let $x=K C$. By Theorem 4, $\frac{D K}{K C}=\frac{B D}{B C}$ or $\frac{1}{x}=\frac{2}{\sqrt{B D^{2}+D C^{2}}} \Rightarrow$
$\Rightarrow 2 x=\sqrt{4+(x+1)^{2}}, x=\frac{5}{3}, S_{A B C}=\frac{11}{3}$.

Answer: $\frac{11}{3} \text{ cm}^2$.