Question

please solve this

class 10th

Good question ​

Answer 1

Hope it will help you. . . . . . . . . . .

Answer 2

let the rectangle previous length is x breadth is y and area is xy

ATQ

$(x - 5)(y + 2) = xy - 80 \\ xy + 2x - 5y - 10 = xy - 80 \\ 2x - 5y - 10 + 80 = xy - xy \\ 2x - 5y + 70 = 0......(i)$

Again

$(x + 10)(y - 5) = xy + 50 \\ xy - 5x + 10y - 50 = xy + 50 \\ 10y - 5x = xy - xy + 50 + 50 \\ 10y - 5x = 100 \\ 5(2y - x) = 5 \times 20 \\ 2y - x = 20 \\ 2y - 20 = x$

putting the value of x in equation i

$2x - 5y + 70 = 0 \\ 2(2y - 20) - 5y + 70 = 0 \\ 4y - 40 - 5y + 70 = 0 \\ - y + 30 = 0 \\ - y = - 30 \\ y = 30 \\ x = 2y - 20 \\ x = 2 \times 30 - 20 \\ x = 60 - 20 \\ x = 40$

Hence the length of rectangle is 40

and breadth is 30.