Question

If the area of a rhombus is 68 cm^2 and one of its diagonal is 8 cm , then find perimeter of the rhombus.

Answer 1

It is given that,

• Area of rhombus = 68 cm²

• Length of one diagonal = 8 cm

We know that,

Area of rhombus = 1/2 x Diagonal 1 x Diagonal 2

Substitute the given values. we get,

68 =  1/2 x 8 x Diagonal 2

⇒ 68 = 4 x Diagonal 2

⇒ Diagonal 2 = 68/4

∴ Diagonal 2 = 17 cm

Let ' a ' be the side of rhombus.

∵ Diagonals of rhombus are perpendicular bisector of each other.

So, we can find the side by using Pythagoras theorem.

a² = ( first diagonal/2)² + ( second diagonal/2)²

⇒ a² = ( 8/2 )² + ( 17/2 )²

⇒ a² = 64/4 + 289/4

⇒ a² = 353/4

⇒ a² = 88.5

⇒ a = √88.5

∴ a = 9.4 cm

Also, perimeter of rhombus = 4 × side

⇒ 4 × 9.4

⇒ 37.6 cm

Therefore, perimeter of the rhombus is 37.6 cm.

$\rule{200}2$