Question

Area of a rectangle is 51.2 dm2 and the sides are in the ratio 5: 4. Find its perimeter. ​

Answer 1

Answer:

given,

A =51.2dm^2,(A=area of rectangle)

let, 'l' and 'b' be the length and breadth of the rectangle.

according to question,

l*b=51.2

and

l/b=5/4

l^2={5*(51.2)}/4

l^2=5*(12.8)

l^2=64

l=8

b=51.2/8=6.4

perimeter=2*(l+b)=2*(8+6.4)=2*14.4=28.8 dm

now

Answer 2

Let the main ratio be x

Therefore,

First side = 5x = Length (l)

First side = 5x = Length (l) Second side = 4x = Breadth (b)

According to the Question,

Given that the area of rectangle is 51.2 dm²

Area of Rectangle = l × b

51.2 = 5x × 4x

20x² = 51.2

x² = 51.2/20

x²= 2.56

x = √2.56

x = 1.6

Hence the length of rectangle = 5x = 8dm

The breadth of rectangle = 4x = 6.4dm

Now, the perimeter of Rectangle =

2 (l + b) = 2 (8 + 6.4)

= 2 × 14.4

= 28.8dm

Hence the perimeter of the rectangle is 28.8dm

In cms, *if required*

1dm = 10cm

28.8dm = 28.8×10

= 288cm is the perimeter.

$...$

Hope this helps you

~Thank you~