Math, asked by prasantabarik380, 2 months ago

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

Answers

Answered by IIJustAWeebII
3

\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!!

Answered by XxArmyGirlxX
1

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

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