The lengths of the diagonals of a rhombus are(i) 16 cm and 12 cm(ii) 3

1.	The lengths of the diagonals of a rhombus are(i) 16 cm and 12 cm(ii) 30 cm and 40 cmFind the perimeter of the rhombus.
Answer» ✬ (i) Perimeter = 40 cm ✬ ✬ (ii) Perimeter = 100 cm ✬ Step-by-step explanation: 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

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

Answer»

✬ (i) Perimeter = 40 cm ✬

✬ (ii) Perimeter = 100 cm ✬

Step-by-step explanation:

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