A rhombus-shaped collapsible ladle has sides of $20 \mathrm{~cm}$, and its longer diagonal is $32 \mathrm{~cm}$. When we slightly compress the structure along the longer diagonal, the other diagonal becomes longer, specifically 1.2 times the length of the compression. What are the lengths of the diagonals?
Solution. Let's denote the vertices of the rhombus as $A, B, C, D$, and the intersection point of the diagonals as $M$. We know that the diagonals of a rhombus bisect each other at right angles. From the right triangle $BCM$, we can calculate half the length of the shorter diagonal using the Pythagorean theorem:

$$
\begin{aligned}
C M^{2} & =20^{2}-16^{2}=144 \\
C M & =12
\end{aligned}
$$

The length of the shorter diagonal is 24.

![](https://cdn.mathpix.com/cropped/2024_05_02_bbaa1dee036301a6fd51g-1.jpg?height=471&width=509&top_left_y=499&top_left_x=797)

After the compression along the longer diagonal, we get another right triangle. Let the decrease in the length of the longer diagonal be $x$, and the increase in the length of the shorter diagonal be $1.2x$.

In the new right triangle, $M B^{\prime}=16-0.5 x, M C^{\prime}=12+0.6 x$, and $B^{\prime} C^{\prime}=20$. Let's apply the Pythagorean theorem again:

$$
(16-0.5 x)^{2}+(12+0.6 x)^{2}=20^{2}
$$

Perform the operations. We get:

$$
0.61 x^{2}-1.6 x=0
$$

Factoring, we get $x(0.61 x-1.6)=0$, from which $x=0$ (not a solution to the problem), or $x=2.62$. The lengths of the diagonals are:

$$
32-x \approx 29.38, \quad \text { and } \quad 24+1.2 x \approx 27.14
$$