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The area of a rectangle gets reduced by 80 sq. units if its length is reduced by 5 units and the breadth is increased by 2 units. If we increase the length by 10 units and decrease the by 5 units, the area is increased by 50 sq. units. Find the length and breadth of the rectangle.

Last updated date: 13th Jun 2024
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Hint: To solve this problem, we will first assume some variables for the length and the breadth of the rectangle. Then we are given information about the changes in the dimension of the length and the breadth and the effect these changes have on the area of the rectangle. We know that the area of a rectangle is given by the relation A = lb, where l is the length and b is the breadth. We will use the above information to form two equations in length and breadth. Once we have two equations in two variables, we will solve the equations for the values of length and breadth.

Let us assume that the length of the rectangle is l and the breadth of the rectangle is b.
Therefore, the area of the original rectangle will be given as A = lb.
The figure will be as follows:

It is given to us that the area of a rectangle gets reduced by 80 sq. units if its length is reduced by 5 units and the breadth is increased by 2 units.
Thus, the modified length is (l – 5) and the modified breadth is (b + 2). The modified area is (lb – 80).
$\Rightarrow$ (l – 5)(b + 2) = lb – 80
$\Rightarrow$ lb – 5b + 2l – 10 = lb – 80
$\Rightarrow$ 2l – 5b = – 70……(1)
Now the second condition given to us is that if we increase the length by 10 units and decrease the by 5 units, the area is increased by 50 sq. units.
Thus, the modified length is (l + 10) and the modified breadth is (b – 5). The modified area is (lb + 50).
$\Rightarrow$ (l + 10)(b – 5) = lb + 50
$\Rightarrow$ lb + 10b – 5l – 50 = lb + 50
$\Rightarrow$ – l + 2b = 20……(2)
(1) and (2) are two equations in l and b. To find the values of l and b, we will solve (1) and (2).
We will solve the equation by substitution method.
From equation (2)
$\Rightarrow$ l = 2b – 20……(3)
We will substitute (3) in (1).
$\Rightarrow$ 2(2b – 20) – 5b = – 70
$\Rightarrow$ 4b – 40 – 5b = – 70
$\Rightarrow$ – b = – 30
$\Rightarrow$ b = 30 units
We will substitute b = 30 in equation (3).
$\Rightarrow$ l = 2(30) – 20
$\Rightarrow$ l = 40 units

Hence, the length of the rectangle is 40 units and breadth of the rectangle is 30 units.

Note: There is no compulsion as to how we solve equation (1) and (2), which are equations in two variables. Students can choose any method they are comfortable with, but it is good practice to try and be able to solve the equations with all the methods.