Question

area of arectangle is 51.2dm² and the sides are in the ratio 5:4 find its perimeter​

Answer 1

$\huge{\text{\purple{Solution}}}$

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

$\sf{Area = 51.2 dm {}^{2} }$

$\sf{ = > Length \times Width = 51.2} \: \: \mathcal{\blue{\boxed{Area = Length \times Width}}}$

$\sf{=> 5x×4x = 51.2}$

$\sf{=> 20x {}^{2} = 51.2}$

$\sf{=> {x}^{2} = \frac{5.12}{2} }$

$\sf{=> x {}^{2} = 2.56}$

$\sf{=> x = 1.6 dm}$

$\sf{\green{Length = 5 \times 1.6 = 8 dm}}$

$\sf{\green{Width = 4 × \times 1.6 = 6.4 dm}}$

$\sf{Perimeter = 2( 8+6.4)}\:\:\mathcal{\blue{\boxed{Perimeter = 2(Length + Width}}}$

$\sf{= 2 \times 14.4}$

$\sf{\purple{\boxed{= 28.8 dm}}}$

$\sf{\orange{Hence,\:the\:perimeter\:is\:28.8dm}}$

HOPE THIS HELPS!!

Answer 2

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

Area = 51.2 dm²

=> Length xWidth = 51.2

=> 5a x4a =51.2

=> 20a² = 51.2

=> a² = 5.12/2

=> a² = 2.56

=> a = 1.6 dm

Length = 5 x 1.6 = 8 dm

Width = 4 x 1.6 = 6.4 dm

Perimeter = 2( 8+6.4)

= 2x 14.4

= 28.8 dm