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

Answer 2

Answer:

(I) Diagonal of rhombus = 12cm and

16cm half of diagonal = 6cm and 8cm

by using Pythagoras theorem

at the point O where diagnols meet at that point there are

angle formed 90°

so,a (side) of rhombus

= √p² + b²

a = √6² + 8²

= √36+64

=√ 100

= 10cm

now , as we know that Perimeter of rhombus = 4a = 4×10 = 40cm

(iI) Let ABCD be a rhombus with AC and BD as its diagonals.

We know that the diagonals of a rhombus bisect each other at right angles.

Let O be the intersecting point of both the diagonals.

Let AC=30cm and BD=40cm

OA=AC/2

OA= 30/2=15cm

OB=BD/2

OB=40/2=20cm

In rt.ΔAOB by Pythagoras theorem we have

AB²=OA²+OB²

=(15)²+(20)²

=225+400

=625

AB=25cm

Hence, each side of the rhombus is of length 25cm