In a rhombus, one diagonal is $40\text{cm}$ and the other diagonal is $22\text{cm}$ longer than half the first diagonal. Find the length of each side of the rhombus.

Let the first diagonal be $d_1=40\text{cm}$.
According to the problem, the second diagonal $d_2$ is:

$$d_2=\frac{d_1}{2}+22=\frac{40}{2}+22=20+22=42\text{cm}$$

The diagonals of a rhombus bisect each other at right angles ($90^{\circ }$).
Therefore, the side $a$ of the rhombus forms the hypotenuse of a right-angled triangle with legs equal to half of the diagonals:

$${\text{Leg}}_1=\frac{d_1}{2}=\frac{40}{2}=20\text{cm}$$

$${\text{Leg}}_2=\frac{d_2}{2}=\frac{42}{2}=21\text{cm}$$

Using Pythagoras theorem:

$$a^2=({\text{Leg}}_1{)}^2+({\text{Leg}}_2{)}^2$$

$$a^2=20^2+21^2$$

$$a^2=400+441=841$$

$$a=\sqrt{841}=29\text{cm}$$

Thus, the length of each side of the rhombus is $29\text{cm}$.