The diagonals of a rhombus are $24\text{cm}$ and $10\text{cm}$. The perimeter of the rhombus (in $\text{cm}$) is:

Let the diagonals of the rhombus be $d_1=24\text{cm}$ and $d_2=10\text{cm}$.
The diagonals of a rhombus bisect each other at right angles.
Thus, the side of the rhombus $a$ can be found using the relation:

$$a=\sqrt{{\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2}$$

$$a=\sqrt{{\left(\frac{24}{2}\right)}^2+{\left(\frac{10}{2}\right)}^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{cm}$$

The perimeter of a rhombus is given by $4\times a$:

$$\text{Perimeter}=4\times 13=52\text{cm}$$