Introduction / Context:
Problems that connect rectangles and squares typically ask you to use both shape properties and area relations. Here, a rectangle, after a specific change to length and width, becomes a square whose area equals the original rectangle’s area. We are asked to find the original perimeter, which requires recovering the original dimensions first.
Given Data / Assumptions:
- Original rectangle dimensions: length L, width W (in cm).
- New figure after changes: length decreased by 4 cm and width increased by 3 cm; this new figure is a square.
- The square’s area equals the original rectangle’s area.
Concept / Approach:
Let the square’s side be s. Because the changed figure is a square: s = L - 4 = W + 3. Also, areas are equal: L * W = s^2. Use these equations to solve for s, L, and W, then compute the perimeter 2(L + W).
Step-by-Step Solution:
Let s = L - 4 and s = W + 3 ⇒ L = s + 4 and W = s - 3.Area equality: L * W = s^2 ⇒ (s + 4)(s - 3) = s^2.Expand: s^2 + s - 12 = s^2 ⇒ s - 12 = 0 ⇒ s = 12.Then L = 12 + 4 = 16, W = 12 - 3 = 9.Perimeter = 2(L + W) = 2(16 + 9) = 2 * 25 = 50 cm.
Verification / Alternative check:
Original area: 16 * 9 = 144. Square of side 12 has area 144. Areas match; perimeter computed from valid L, W.
Why Other Options Are Wrong:
20, 30, 40 are inconsistent with the derived dimensions (16, 9), which uniquely satisfy both “square after change” and “equal area”.
Common Pitfalls:
Confusing “becomes a square” with “has equal sides after change but different area”. Both conditions must hold.Setting (L - 4) = (W + 3) but forgetting to also enforce area equality.
Final Answer:
50
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