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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.

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