A rectangular sheet of paper is folded into two congruent parts along one set of sides, and each of these folded sheets has a perimeter of 34 cm. When the same original sheet is folded into two congruent parts along the other set of sides, the perimeter of each folded part is 38 cm. What is the area of the original sheet of paper in square centimetres?

Introduction:
This question is a clever application of rectangle perimeter and the effect of folding a rectangle in half. The same sheet is folded along two different directions and the perimeters of the resulting congruent parts are given. Using this information, we must determine the original dimensions of the sheet and then compute its area. This tests understanding of perimeter formulas, simple algebra, and how folding affects side lengths.

Given Data / Assumptions:

• The original sheet is a rectangle with length L and breadth B. • When folded along one set of sides, each half has perimeter 34 cm. • When folded along the other set of sides, each half has perimeter 38 cm. • Folding divides one side into two equal parts while keeping the other dimension unchanged. • We must find the area L * B of the original rectangle.

Concept / Approach:
When a rectangle is folded along its length, the length becomes L/2 and the breadth remains B for the folded part. When folded along its breadth, the breadth becomes B/2 and the length remains L. The perimeter of a rectangle is 2 * (length + breadth). By writing perimeter equations for both folding cases, we get two linear equations in L and B. Solving these gives us L and B, and we then multiply them to obtain the area.

Step-by-Step Solution:
Step 1: Fold along the length. The folded part has dimensions L/2 and B. Perimeter in this case is 2 * (L/2 + B) = 34. Simplify: L + 2B = 34. Call this Equation (1). Step 2: Fold along the breadth. The folded part now has dimensions L and B/2. Perimeter = 2 * (L + B/2) = 38. Simplify: 2L + B = 38. Call this Equation (2). Step 3: Solve the system of equations. Equation (1): L + 2B = 34. Equation (2): 2L + B = 38. Step 4: Multiply Equation (1) by 2 if needed or use elimination directly. From Equation (1): L = 34 - 2B. Substitute into Equation (2): 2(34 - 2B) + B = 38. 68 - 4B + B = 38. 68 - 3B = 38. 3B = 68 - 38 = 30. B = 10 cm. Step 5: Substitute B back to find L. L + 2 * 10 = 34 ⇒ L + 20 = 34 ⇒ L = 14 cm. Step 6: Compute the area of the original rectangle. Area = L * B = 14 * 10 = 140 sq cm.

Verification / Alternative check:
Check both folding scenarios. First scenario (fold along length): dimensions 14/2 and 10 give 7 cm and 10 cm, perimeter = 2 * (7 + 10) = 2 * 17 = 34 cm, which matches. Second scenario (fold along breadth): dimensions 14 and 10/2 give 14 cm and 5 cm, perimeter = 2 * (14 + 5) = 2 * 19 = 38 cm, also matching. Therefore, the dimensions 14 cm by 10 cm are consistent, and the area 140 sq cm is correct.

Why Other Options Are Wrong:
Option 150 sq cm, 160 sq cm, 170 sq cm, and 180 sq cm: These would correspond to different pairs of length and breadth. None of those pairs will simultaneously satisfy both perimeter conditions for the two different foldings. Only 14 cm by 10 cm meets both conditions exactly.

Common Pitfalls:
Students sometimes confuse which dimension gets halved in which folding scenario, leading to incorrect perimeter equations. Another source of error is algebraic manipulation when solving the two simultaneous equations. Mixing up L and B or substituting incorrectly can produce wrong dimensions and therefore wrong area. Carefully defining folding direction and writing explicit equations for each case helps prevent such errors.

Final Answer:
The area of the original rectangular sheet of paper is 140 square centimetres.