Math, asked by ravikantuttasara95, 8 months ago

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.​

Answers

Answered by pandaXop
11

(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 : = +

\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

Attachments:
Answered by rohit50003
6

✬ (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 ★

⟹ AB² = AO² + BO²

⟹ AB² = 8² + 6²

⟹ AB² = 64 + 36

⟹ 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:

⟹ AB² = AO² + BO²

⟹ AB² = 15² + 20²

⟹ AB² = 225 + 400

⟹ AB = √625

⟹ AB = 25 cm

Hence,

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

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