Question

8. The diagonals of a rhombus are 48 cm and 20 cm long. Find the perimeter of the rhombus.

Answer 1

✬ Perimeter = 104 cm ✬

Step-by-step explanation:

Given:

• Length of the Diagonals of the rhombus are 48 cm and 20 cm respectively.

To Find:

• What is the perimeter of the rhombus?

Solution: Let ABCD be a rhombus in which

• AC = diagonal = 48 cm

∴ OA = OC = 1/2 x 48 = 24 cm

• BD = diagonal = 20 cm

∴ OB = OD = 1/2 x 20 = 10 cm

• We know that diagonals of a rhombus bisect each other at 90° •

Therefore, Applying Pythagoras Theorem in ∆AOB

$\implies{\rm }$ Base² + Perpendicular ² = Hypotenuse ²

$\implies{\rm }$ OB² + OA² = AB²

$\implies{\rm }$ 10² + 24² = AB²

$\implies{\rm }$ 100 + 576 = AB²

$\implies{\rm }$ 676 = AB²

$\implies{\rm }$ √676 = AB

$\implies{\rm }$ 26 cm = AB

Since, All the four sides of a rhombus are equal to each other, so AB = BC = CD = DA = 26 cm

★ Perimeter of rhombus = (4 $\times$ side) ★

$\implies{\rm }$ Perimeter = 4 $\times$ 26 cm

$\implies{\rm }$ 104 cm

Hence, The perimeter of the rhombus is 104 cm.

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→ Area of rhombus = (1/2 x Product of diagonals)

→ All sides of a rhombus are equal.

→ Diagonals of a rhombus bisect each other perpendicularly.

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