The perimeter and the length of one of the diagonals of a rhombus are $34\text{cm}$ and $8\text{cm}$ respectively. Find the length of its other diagonal (in $\text{cm}$).

Let the side of the rhombus be $a$.
$$\text{Perimeter}=4a=34⟹a=\frac{34}{4}=8.5\text{cm}$$
We know that the diagonals of a rhombus bisect each other at right angles. Let the diagonals be $d_1=8\text{cm}$ and $d_2$.By Pythagoras theorem in one of the central triangles formed by the intersecting diagonals:
$${\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2=a^2$$
$${\left(\frac{8}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2=(8.5{)}^2$$
$$4^2+{\left(\frac{d_2}{2}\right)}^2=72.25$$
$$16+{\left(\frac{d_2}{2}\right)}^2=72.25$$
$${\left(\frac{d_2}{2}\right)}^2=72.25-16=56.25$$
$$\frac{d_2}{2}=\sqrt{56.25}=7.5$$
$$d_2=7.5\times 2=15\text{cm}$$