Let ABCD be a rhombus where each side is 5 cm and one of the diagonals, BD, is 8 cm. Since the diagonals of a rhombus bisect each other at right angles, \[ \therefore OD = \frac{1}{2} \text{ of } BD = 4 \text{ cm} \] Now, in right-angled triangle AOD: \[ AO^2 + OD^2 = AD^2 \quad \text{[By Pythagoras' Theorem]} \Rightarrow AO^2 = AD^2 - OD^2 = 5^2 - 4^2 = 25 - 16 = 9 \Rightarrow AO = 3 \text{ cm} \] \[ \therefore AC = 2 \times AO = 2 \times 3 = 6 \text{ cm} \] \[ \therefore \text{The length of the other diagonal of the rhombus is } 6 \text{ cm}. \]