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, 

AD2 = AM2 + MD2

⇒AD2 = 152 + 82

⇒AD2 = 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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 Let diagonal AC (16cm) perpendicular to diagonal BD (30cm).

Therefore, AB (l) = Underroot { (square of half AC) + (square of half BD) }

                             = Underroot { (square of 8) + (square of 15) }  =  Underroot (64 + 225)  =  Underroot 289

Now, perimeter = 4 X l  = 4 X (Underroot 289)

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thanks

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