A trapezoid with bases equal to $a$ and $b(a>b)$ is cut by a line parallel to the bases into two trapezoids, the areas of which are in the ratio $k: p$. Find the length of the common side of the resulting trapezoids.
Let the length of the segment $M N$ be denoted by $x$ (see figure). Draw through point $M$ the lines $E F \| C D$ and $K L \perp A D$. Denote the length of segment $K M$ by $H_{1}$, and the length of segment $M L$ by $H_{2}$. Then $\frac{(b+x) H_{1}}{(a+x) H_{2}}=\frac{k}{p}$. From the similarity of triangles $E M B$ and $F M A \quad M K: M L=E B: A F$, or $\frac{H_{1}}{H_{2}}=\frac{x-b}{a-x}$. Therefore,

$\frac{x-b}{a-x}=\frac{k(a+x)}{p(b+x)}$, from which $x=\sqrt{\frac{k a^{2}+p b^{2}}{k+p}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-05.jpg?height=554&width=1038&top_left_y=212&top_left_x=516)

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-05.jpg?height=94&width=169&top_left_y=862&top_left_x=13)