Question

Find the perimeter of a rhombus whose diagonals are 16 cm and 12 cm

Answer 1

Refers to the attachments:

Given:

• Diagonal,d1 = 16 cm

• Diagonal,d2 = 12 cm

To find out:

Find the perimeter of a rhombus ?

Solution:

We know, Diagonal of rhombus bisect each other at right angles.

AC = 16 cm, OC = 1/2 AC ⇒ 1/2 × 16 = 8 cm

BD = 12 cm, OD = 1/2 BD ⇒ 1/2 × 12 = 6 cm

By using pythagoras therome

DC² = OC² + OD²

DC² = 8² + 6²

DC² = 64 + 36

DC² = 100

DC = 10 cm

Therefore,

DC = BC = AD = AB = 10 cm [ All sides are equal ]

Now,

Perimeter of rhombus = 4 ( Side )

= 4 × 10

= 40 cm

Answer 2

Answer

Let ABCD be Rhombus

AC and BD are diagonals of 16 and 12 cm

Half of diagonals are 8 and 6 cm

Now we can get side of rhombus by using pythagoras theorem with these half of diagonals.

Let the side of rhombus be ' s '

⇒ s² = 8² + 6²

⇒ s² = 64 + 36

⇒ s² = 100

⇒ s = ± 10

But lengths can't be negative.

So , side of rhombus , s = 10 cm

Now perimeter of rhombus = 4 * side

⇒ 4 * 10

⇒ 40 cm

So the perimeter of rhombus is 40 cm