The diagonals of a rhombus measure …
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Yogita Ingle 6 years, 1 month ago
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=68cm...
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