5. Find the area of a triangle inscribed in a parallelogram, given that the remaining part of the parallelogram consists of three triangles of unit area.
Solution. Let $ABCD$ be a parallelogram. To inscribe a triangle in it so that the remaining part of the parallelogram represents three triangles, there is only one way: by placing one of the vertices of the triangle at a vertex of the parallelogram (for example, at $C$), and the other two vertices on the sides adjacent to vertex $A$ (see figure). Let these vertices be denoted by $F$ and $E$ ($F$ lies on side $AD$). Suppose also that $AD=BC=a$, $AB=CD=b$, and $\angle ABC=\alpha$. Then $S_{CFD}=\frac{1}{2} \cdot DF \cdot b \cdot \sin \alpha$, from which $DF=\frac{2}{b \sin \alpha}$ and $AF=a-\frac{2}{b \sin \alpha}$. Similarly, $AE=b-\frac{2}{a \sin \alpha}$. Finally,

$$
S_{AEF}=\frac{1}{2} \cdot AF \cdot AE \cdot \sin \alpha=\frac{(a b \sin \alpha-2)^{2}}{2 a b \sin ^{2} \alpha} \cdot \sin \alpha=\frac{(S-2)^{2}}{2 S}=1
$$

where $S=ab \sin \alpha$ is the area of the parallelogram $ABCD$. We obtain the quadratic equation $2S=(S-2)^{2}$, solving which, we find that $S=3 \pm \sqrt{5}$. But $S>3$, so $S_{CEF}=\sqrt{5}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c56c4346fae2c9c16492g-2.jpg?height=320&width=623&top_left_y=1685&top_left_x=291)

Answer: $\sqrt{5}$.