The length of a rectangle exceeds its breadth by 5m. If the breadth were doubled and the length reduced by 9m, the area of the rectangle would have increased by 140 m². Find its dimensions.

Let breadth of rectangle be x meters

Since , length of rectangle exceeds breadth by 5 meters so, length = (x + 5) meters

Given, if breadth were doubled and the length reduced by 9m, the area of the rectangle would have increased by 140 m

∴ Area of new rectangle = Length $\times$ Breadth = (x + 5 - 9)(2x)

$\Rightarrow (x + 5 - 9)(2x) - x(x + 5) = 140 \\[0.5em] \Rightarrow (x - 4)(2x) - (x^2 + 5x) = 140 \\[0.5em] \Rightarrow 2x^2 - 8x - x^2 - 5x = 140 \\[0.5em] \Rightarrow x^2 - 13x - 140 = 0 \\[0.5em] \Rightarrow x^2 - 20x + 7x - 140 = 0 \\[0.5em] \Rightarrow x(x - 20) + 7(x - 20) = 0 \\[0.5em] \Rightarrow (x + 7)(x - 20) = 0 \\[0.5em] \Rightarrow x + 7 = 0 \text{ or } x - 20 = 0 \\[0.5em] x = -7 \text{ or } x = 20$

Since, breadth cannot be negative hence x ≠ -7

∴ x = 20 and x + 5 = 25

Length of rectangle = 25 m , Breadth of rectangle = 20 m.