A takes 15 seconds to cross a rectangular field diagonally while walking at 52 m/min. B takes the same 15 seconds to cross the same field along its sides (from one corner to the opposite corner via two adjacent sides) while walking at 68 m/min. What is the area of the field (in sq m)?

Introduction / Context:
This problem combines speed-time-distance with rectangle geometry. Person A walks along the diagonal, so the distance covered equals the diagonal length. Person B walks along the sides from one corner to the opposite corner, so the distance covered equals length + breadth. Since both take the same time, we can calculate these distances from their speeds. With diagonal and (length + breadth) known, we use the identity: (length + breadth)^2 = length^2 + breadth^2 + 2*length*breadth. Here length^2 + breadth^2 equals diagonal^2. This allows us to solve for length*breadth, which is the area.

Given Data / Assumptions:

• Time for both = 15 seconds
• A speed = 52 m/min
• B speed = 68 m/min
• Diagonal distance = A speed * time
• Length + breadth distance = B speed * time

Concept / Approach:
Convert speeds to m/s or convert time to minutes. Compute diagonal d and sum S = L + B. Use S^2 = d^2 + 2LB to find LB (area).

Step-by-Step Solution:
Time = 15 sec = 15/60 = 0.25 min Diagonal d = 52 * 0.25 = 13 m Sum of sides S = 68 * 0.25 = 17 m S^2 = (L + B)^2 = 17^2 = 289 d^2 = L^2 + B^2 = 13^2 = 169 (L + B)^2 = L^2 + B^2 + 2LB => 289 = 169 + 2LB 2LB = 120 => LB = 60 Area = L*B = 60 sq m

Verification / Alternative check:
If area is 60, one valid pair is L=12 and B=5 (since 12*5=60). Then L+B=17 and diagonal sqrt(12^2+5^2)=sqrt(169)=13. Both distances match the computed S and d, confirming the area.

Why Other Options Are Wrong:
50, 70, 80, 90 do not satisfy the derived equation LB = (S^2 - d^2)/2. They would require different S or d than what the speeds and time produce.

Common Pitfalls:
Not converting seconds to minutes, assuming B's path is perimeter instead of L+B, or trying to solve L and B separately when only area is needed.

Final Answer:
The area of the field is 60 sq m.