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

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#### Solution

Let ABCD be a rhombus (all sides are of equal length) and its diagonals, AC and BD, are intersecting each other at point O. Diagonals in a rhombus bisect each other at 90°. It can be observed that

`AO = (AC)/2 = 16/2 = 8` cm

`BO = (BD)/2 = 30/2 = 15` cm`

By applying Pythagoras theorem in ΔAOB,

OA^{2} + OB^{2} = AB^{2}

8^{2} + 15^{2} = AB^{2}

64 + 225 = AB^{2}

289 = AB2

AB = 17

Therefore, the length of the side of rhombus is 17 cm.

Perimeter of rhombus = 4 × Side of the rhombus = 4 × 17 = 68 cm

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