If the diagonals of a rhombus are 9 cm and 12 cm, then the perimeter of the rhombus(in cm) is:

Let ABCD be rhombus with AC and BD it's diagonals bisecting each other at O ,where

AC=9cm,AO=OC=4.5cm

BD=12cm, BO=OD=6 cm 

In $$\triangle\ $$AOB,

$$\angle\ $$AOB=90$$^{\circ\ }$$(we know diagonals of rhombus bisect each other at 90$$^{\circ\ }$$)

Apply Pythagoras theorem,

$$AO^{ 2}$$+$$OB^{2 }$$=$$AB^{ 2}$$

$$4.5^{ 2}$$+$$6^{ 2}$$=$$AB^{ 2}$$

20.25$$\ +\ $$36=$$AB^{2 }$$

$$AB^{ 2}$$=56.25

AB=7.5

Perimeter of rhombus=4$$\times\ $$7.5=30