Question

The length of the diagonals of a rhombus are 24 cm and 32 cm. The perimeter
of the rohombus is
​

Answer 1

rohombus side is 4 side

adding but they are 4 side and 2 , 2 sides are equal

= 24+24+32+32

= 112

Answer 2

SOLUTION:

Let ABCD be a rhombus where AC is (diagonal 1) = 32 cm

And BD is (diagonal 2) = 24 cm

Now,

We have to find the perimeter of the rhombus.

Some properties of the rhombus :

• All sides of the rhombus are equal.
• The diagonals bisect each other at 90°.

So,

As the diagonals bisect each other at 90°.

So,

AO = OC = 1/2 of AC

= $\frac{32}{2}$ = 16 cm

So,

AO = OC = 16 cm

Similarly,

BO = OD = 1/2 of BD

= $\frac{24}{2}$ = 12 cm

So,

BO = OD = 12 cm

Now,

We get four right-angled triangles,

ΔAOB, ΔBOC, ΔCOD, ΔDOA

Now,

Taking any triangle,

ΔAOB, here,

AO = 16

BO = 12

And angle AOB = 90°

By using Pythagoras theorem, we get,

AB² = AO² +  BO²

⇒ AB² = (16)² + (12)²

⇒ AB² = 256 + 144

⇒ AB² = 400

⇒ AB = √400

⇒ AB = 20 cm

So,

We got the value of one side of the rhombus,

Now from the properties o the rhombus we get that, 'All the sides fo the rhombus are equal.'

Thus,

AB = BC = CD = CA = 20 cm each

So,

The perimeter of the rhombus = Sum of all sides units

OR

= 4*side units

Therefore,

= 20 + 20 + 20 + 20

= 20 (1+1+1+1)

[Taking 20 as common]

= 20*(4)

= 80 cm

Hence,

The perimeter of the rhombus is 80 cm.