Let us assume a rhombus ABCD having diagonals AC and BD

Diagonals of this rhombus ABCD intersects each other at a point O

We know that,

Diagonals of rhombus bisects each other at 90^o

Now,

AO = = = 8 cm

BO = = = 15 cm

By applying Pythagoras theorem, we get:

OA² + OB² = AB²

8² + 15² = AB²

64 + 225 = AB²

AB² = 289

AB =

AB = 17 cm

∴ The length of the side of Rhombus = 17 cm

Perimeter of rhombus = 4 × Side of rhombus

= 4 × 17

= 68 cm