In a rhombus, its diagonals are 30 cm and 40 cm, its perimeter is :

1. 20 cm

2. 10 cm

3. 60 cm

4. 100 cm

Let ABCD be the rhombus, with diagonals AC and BD intersecting at O.

We know that,

Diagonals of rhombus intersect at right angles.

Let AC = 40 cm and BD = 30 cm.

∴ AO = $\dfrac{AC}{2} = \dfrac{40}{2} = 20 \text{ cm}, BO = \dfrac{BD}{2} = \dfrac{30}{2}$ = 15 cm.

In right angle triangle AOB,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AO² + BO²

⇒ AB² = (20)² + (15)²

⇒ AB² = 400 + 225

⇒ AB² = 625

⇒ AB = $\sqrt{625}$ = 25 cm.

We know that,

All sides of rhombus are equal.

∴ Perimeter of rhombus = 4 × side = 4 × 25 = 100 cm.

Hence, Option 4 is the correct option.