Option b

Step-by-step explanation:

GIVEN:-

In a rhombus, the length of the two diagonals are 3 meters and 4 meters respectively.

To find:-

Find its PERIMETER?

Solution:-

The lengths of the two diagonals are 3 m and 4 m

Consider a ABCD rhombus

AC = 3m and BD = 4 m

We know that

The diagonals bisect to each other at 90°

AO = OC

AO = AC/2 = 3/2 CM = 1.5 m

BO=OD

BD = BO/2 = 4/2 = 2 m

∆AOB is a right angled triangle

By Pythagoras THEOREM:

The square of the hypotenuse is equal to the sum of the squares of the other two sides

=>AB^2 = AO^2+OB^2

=>AB^2 = (1.5)^2+2^2

=>AB^2 = 2.25 +4

=>AB^2 = 6.25

=>AB=√6.25

=>AB=2.5 m

The length of the side=2.5 m

We know that

All sides are equal in a rhombus

=>AB=BC=CD=DA=2.5m

Perimeter of a rhombus = Sum of all sides

=>Perimeter=4×length of its side

=>P=4×2.5m

=>P=10 m

Answer:-

Perimeter of the given rhombus is 10 m

Used formulae:-

• The diagonals bisect to each other at 90°
• All sides are equal in a rhombus
• Pythagoras theorem:

The square of the hypotenuse is equal to the sum of the squares of the other two sides

• Perimeter of a rhombus=4×length of its side