Question

the diagonal of rhombus is 48 cm and 10 cm find the perimeter of rhambus​

Answer 1

perimeter of s rhombus = 4xside

side = 1/2 x $\sqrt{48^2+10^2\\}$=24.5

perimeter = 24.5 x 4 = 98 cm

Answer 2

Step-by-step explanation:

Given:-

The diagonal of rhombus is 48 cm and 10 cm .

To find:-

Find the perimeter of rhambus ?

Solution:-

The diagonals of a rhombus = 48 cm and 10 cm .

Let d1 = 10 cm

d2 = 48 cm

Consider a rhombus ABCD,

AC =d2 = 48 cm

BD = d1 = 10 cm

We know that

The Diagonals are bisecting each other in a Rhombus

AO=OC

AO=OC = AC/2=48/2= 24 cm

BO=OD

BO=OD = BD/2 = 10/2=5cm

Now ∆AOB is a right angled triangle since the diagonals perpendicular bisectors to each other

From Pythagoras theorem

AB^2=AO^2+OB^2

=>AB^2=24^2+5^2

=>AB^2=576+25

=>AB^2 = 601

=>AB=√601

=>AB=24.5 cm (approximately)

We know that

all sides are equal in Rhombus

AB=BC=CD=DA = 24.5 cm

Perimeter of a rhombus = 4×Side units

=>P= 4×24.5 cm

=>P = 98 cm

Answer:-

Perimeter of the given rhombus = 98 cm

Used formulae:-

• The Diagonals are bisecting each other in a Rhombus
• the diagonals perpendicular bisectors to each other
• all sides are equal in rhimbus
• Perimeter of a rhombus = 4×Side units
• The square of a hypotenuse is equal to the sum of the squares of the other two sides in a right angled triangle is called Pythagoras theorem