Math, asked by sania0512, 1 year ago

A rectangle has the same area as another, whose length is 6 m more and breadth
IS 4 m less. It has also the same area as the third, whose length is 8 m more and
breadth 5 m less. Find the length and breadth of the original rectangle.​

Answers

Answered by AneesKakar
6

The length of the original rectangle is 24 m and breath is 20 m

LET THE AREA OF THE THREE RECTANGLE = A

ACCORDING TO THE GIVEN CONDITION

AREA OF ORIGINAL RECTANGLE =A= XY

AREA OF ORIGINAL RECTANGLE=A=(X+6)*(Y-4)

AREA OF ORIGINAL RECTANGLE=A=(X+8)*(Y-5)

(X+6)*(Y-4)=XY

XY + -4X+ 6Y-24 =XY

4X-6Y= -24 --eq1

(X+8)*(Y-5)= XY

XY -5X+8Y -40 =XY

5X -8Y=-40 eq2

solving eq1 and eq2 simultaneously

[4X-6Y= 24]*5   20X -30Y=120 ; 20X =-120 + 30Y

[5X -8Y=40 ]*4 20X - 32Y = -160 ; 20X =- 160 + 32Y

THEREFORE,

-120 + 30Y = - 160 + 32Y

2Y= 40

Y= 20 m

By eq2 we 5x=8y -40

5X= 160-40

X=120/5= 24m

therefore length of the original rectangle is X =24m

and breath of the original rectangle is Y =20m

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