# Q.8 The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Answer :

Let ABCD be a rhombus whose diagonals intersect each other at O such that AC = 16 cm and BD = 30 cm
Since, the diagonals of a rhombus bisect each other at 90°.
$$\therefore$$ OA = OC = 8 cm and OB = OD = 15 cm
In right $$\triangle$$OAB,
$$AB^2 = OA^2 + OB^2$$ (By Pythagoras property)
= $$(8)^2+ (15)^2 = 64 + 225$$ = 289
$$\therefore$$AB =$$\sqrt{289}$$= 17 cm
Since AB = BC = CD = DA (Property of rhombus)
$$\therefore$$Required perimeter of rhombus = 4 × side = 4 × 17 = 68 cm.