Question

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

Answer 1

Let length of the rectangle = xm
then width = (x - 5)m
Area = x(x - 5)sq.m
In second case,
Length of the second rectangle = x - 9
and width = 2(x - 5)m
Area 
= (x - 9)2(x - 5)
= 2(x - 9)(x - 5)sq.m
According to the condition,
2(x - 9)(x - 5) = x(-5) + 140
⇒ 2(x² - 14x + 45) = x² - 5x + 140
⇒ 2x² - 28x + 90 = x² - 5x + 140
⇒ 2x² - 28x + 90 - x² + 5x - 140 = 0
⇒ x² - 23x - 50 = 0
⇒ x² - 25x + 2x - 50 = 0
⇒ x(x - 25) +2(x - 25) = 0
⇒ (x - 25)(x + 2) = 0
Either x - 25 = 0,
then x = 25
or
x + 2 = 0,
then x = -2,
but it is not possible as it is negative.
∴ Length of the rectangle = 25m
and width = 25 - 5 = 20m.