Question

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

Answer 1

Let the length and width be 5a and 4a respectively

Area = 51.2 dm^2

=> Length ×Width = 51.2

=> 5a ×4a = 51.2

=> 20a^2 = 51.2

=> a^2 = 5.12 / 2

=> a^2 = 2.56

=> a = 1.6 dm

Length = 5 × 1.6 = 8 dm

Width = 4 × 1.6 = 6.4 dm

Perimeter = 2( 8+6.4)

= 2× 14.4

= 28.8 dm

Answer 2

hii!!

here's ur answer...

let the length of the rectangle be 5x and breadth of the rectangle be 4x.

given area of the rectangle is = 51.2dm²

therefore l × b = 51.2dm²

==> 5x × 4x = 51.2dm²

==> 20x² = 51.2dm²

==> x² = 51.2/20

==> x² = 2.56

==> x = √2.56

==> x = 1.6dm

hence, length of the rectangle = 5x

= 5 × 1.6

= 8dm

breadth of the rectangle = 4x

= 4 × 1.6

= 6.4dm

VERIFICATION:-

area of the rectangle = l × b

= 8 × 6.4

= 51.2dm²

hence verified

now,

the perimeter of the rectangle = 2 ( l + b )

= 2 ( 8 + 6.4 )

= 2 × 14.4

= 28.8m

hope this helps u..!