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 cmÂ
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
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