Question

The lengths of the diagonals of a rhombus are
(i) 16 cm and 12 cm
(ii) 30 cm and 40 cm
Find the perimeter of the rhombus.​

Answer 1

✬ (i) Perimeter = 40 cm ✬

✬ (ii) Perimeter = 100 cm ✬

Given:

(1) Length of diagonals of rhombus are 16 & 12 cm respectively.

To Find:

What is the perimeter of rhombus ?

Solution:

Let ABCD be a rhombus where

AB || DC & BC || AD

AC = 16 cm ( Diagonal )

BD = 12 cm ( Diagonal )

[ The diagonals of a rhombus bisect each other perpendicularly i.e at 90° ]

So,

➯ AO = OC = 1/2(AC)

➯ AO = OC = 1/2(16)

➯ AO = OC = 8 cm

and

➯ BO = OD = 1/2(BD)

➯ BO = OD = 1/2(12)

➯ BO = OD = 6 cm

Now, in right angled ∆AOB

AB = Hypotenuse

BO = Base

AO = Perpendicular

∠AOB = 90°

Applying Pythagoras Theorem:

★ Pythagoras Theorem : H² = P² + B² ★

$\implies{\rm }$ AB² = AO² + BO²

$\implies{\rm }$ AB² = 8² + 6²

$\implies{\rm }$ AB² = 64 + 36

$\implies{\rm }$ AB = √100 = 10

Hence,

=> Perimeter of rhombus = 4(Side)

=> Perimeter = 4(10) = 40 cm

_____________________

Now in same rhombus taking diagonals as 30 cm and 40 cm.

➯ AO = OC = 1/2(AC)

➯ AO = OC = 1/2(30)

➯ AO = OC = 15 cm

and

➯ BO = OD = 1/2(BD)

➯ BO = OD = 1/2(40)

➯ BO = OD = 20 cm

In ∆AOB ,

AO = Perpendicular , BO = Base , AB = Hypotenuse

Applying Pythagoras Theorem:

$\implies{\rm }$ AB² = AO² + BO²

$\implies{\rm }$ AB² = 15² + 20²

$\implies{\rm }$ AB² = 225 + 400

$\implies{\rm }$ AB = √625

$\implies{\rm }$ AB = 25 cm

Hence,

=> Perimeter of rhombus = 4(25) = 100 cm