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.

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