The area of a rectangle is 247 cm2. If the lengthdecreass by 3 cm and the breadth increases by 3 cm, the rectangle becomes a square. Find the perimeter of the original rectangle

Let's denote the length of the rectangle as L and the breadth as B.

From the problem, we know that:

1. L * B = 247 cm^2 (since the area of a rectangle is length times breadth)

2. If the length decreases by 3 cm and the breadth increases by 3 cm, the rectangle becomes a square. This means that L - 3 = B + 3.

We can simplify the second equation to L - B = 6.

Now we have a system of two equations, and we can solve it step by step.

First, let's express L from the second equation: L = B + 6.

Now we can substitute L in the first equation: (B + 6) * B = 247.

This is a quadratic equation. If we open the brackets and bring all terms to one side, we get: B^2 + 6B - 247 = 0.

The solutions of this equation are the roots of the equation, which can be found using the quadratic formula: B = [-b ± sqrt(b^2 - 4ac)] / (2a), where a = 1, b = 6, and c = -247.

Calculating the discriminant (b^2 - 4ac) gives us 6^2 - 41(-247) = 36 + 988 = 1024.

The square root of 1024 is 32, so the solutions for B are: B1 = [-6 + 32] / 2 = 13 and B2 = [-6 - 32] / 2 = -19.

Since the breadth of a rectangle can't be negative, we discard the second solution. So, B = 13 cm.

Substituting B into the equation L = B + 6, we find that L = 19 cm.

Finally, the perimeter of the original rectangle is 2*(L + B) = 2*(19 cm + 13 cm) = 64 cm.