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

Let ABCD be the given rhombus.

We know that the Diagonals of a rhombus are perpendicular bisector of each other. Let the point of intersection of diagonals AC and BD be M.

Given AC = 30 cm and BD = 16 cm

Now, AM = AC/2 = 30/2 = 15 cm

and DM = BD/2 = 16/2 = 8 cm

Now, in right triangle AMD, by pythagoras theorem,

AD^{2} = AM^{2} + MD^{2}

⇒AD^{2} = 15^{2} + 8^{2}

⇒AD^{2} = 225 + 64 = 289

⇒ AD = √289 = 17 cm

Again, all the sides of a rhombus are equal.

Therefore, AB = BC = CD = AD = 17 cm

Now the perimeter of a rhombus = sum of all sides = AB + BC + CD + AD = 17 + 17 + 17 + 17 = 68 cm

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