Question

Perimeter of a rhombus is40cm,the distance between two opposite side is 9.6cm.find the area and the sum of diagonals of the rhombus

Answer 1

Answer: Area = 96 cm^2

The sum of diagonals of the rhombus = 28 cm

Step-by-step explanation:

Perimeter of a rhombus is 40cm

Then, side = 40/4= 10

Distance (height) = 9.6

Then, Area = base × height

= 10×9.6

= 96 cm^2

Area = 1/2× d1 × d2

96 = 1/2× d1 × d2

96×2= d1× d2

192= d1 × d2........ (1)

4× (Side) ^2 = (d1)^2+(d2)^2

4× 10^2 = (d1)^2+(d2)^2

400 = (d1)^2+(d2)^2 ............. (2)

But, (a+b)^2 = a^2+b^2+2ab

(d1+d2)^2= (d1)^2+(d2)^2 + 2 × d1 × d2

Using (1) & (2), we get

(d1+d2)^2= 400 + 2 × 192 = 400+ 384= 784

((d1+d2)^2= 784 = 28×28= 28^2

d1+d2 = √(784) = 28 cm