Question

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

Answer 1

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...

hope this helps!! cheers!! (:

Answer 2

As the diagonals of rhombus bisects each other by 90°


8²+15²=side²

64+225=side²

289=side²

√289=side

17cm =side


As the sides of rhombus are equal then

Perimeter =4*side

Perimeter =4*17

Perimeter =68cm