A rectangle has length L and width W.\nIf the length is decreased by 4 cm and the width is increased by 3 cm, the new figure becomes a square. Also, this square has the same area as the original rectangle.\nFind the perimeter (in cm) of the original rectangle.

Introduction:
This is an algebraic geometry problem involving a rectangle transformed into a square under changes in dimensions. Two conditions are given: (1) after changing dimensions, the new shape is a square, meaning the new length equals the new width; and (2) the new square has the same area as the original rectangle. These two conditions create a solvable system of equations. The problem tests careful equation setup and handling of area equality correctly. The “same area” condition is crucial; without it, there would be many rectangles that could become a square after adjustments.

Given Data / Assumptions:

• Original length = L cm
• Original width = W cm
• New length = L - 4
• New width = W + 3
• New figure is a square: L - 4 = W + 3
• Area preserved: L * W = (L - 4) * (W + 3)

Concept / Approach:
Use the square condition to express L in terms of W. Then substitute into the area equation. Because L - 4 equals W + 3, the square side can be written as (W + 3), making the area equation simpler. Solve for W, compute L, then perimeter = 2(L + W).

Step-by-Step Solution:
Square condition: L - 4 = W + 3 => L = W + 7Square side = L - 4 = (W + 7) - 4 = W + 3Area equality: L * W = (L - 4) * (W + 3)Substitute L = W + 7: (W + 7)W = (W + 3)(W + 3)W^2 + 7W = W^2 + 6W + 97W - 6W = 9 => W = 9L = W + 7 = 16Perimeter = 2(L + W) = 2(16 + 9) = 50

Verification / Alternative Check:
Original area = 16*9 = 144. New dimensions: L - 4 = 12 and W + 3 = 12, so it becomes a 12 by 12 square. New area = 12*12 = 144, same as original. Both conditions are satisfied, so the perimeter 50 cm is confirmed.

Why Other Options Are Wrong:
20, 30, 40, 60: these perimeters would correspond to different L and W values that cannot satisfy both “becomes a square” and “same area” simultaneously.

Common Pitfalls:
Using only the square condition and forgetting the same-area condition.Making the area equation L*W = (L-4) + (W+3) (wrong; area uses multiplication).Algebra mistakes when expanding (W+3)^2.Computing perimeter as L+W instead of 2(L+W).

Final Answer:
50 cm