Squares — Third Square from Difference of Areas (Recovery Applied): The perimeters of two squares are 68 cm and 32 cm. Find the perimeter of a third square whose area equals the difference of their areas.

Introduction / Context:
The original stem was incomplete (“The perimeter of two squares are 68 cm.”) lacking the second perimeter. By the Recovery-First Policy, we minimally repair it to a standard, solvable form by specifying the second perimeter as 32 cm so the item remains meaningful and aligned with common exam patterns and the given options.

Given Data / Assumptions:

• Square 1: P1 = 68 cm ⇒ s1 = 17 cm ⇒ A1 = 289 cm^2
• Square 2: P2 = 32 cm ⇒ s2 = 8 cm ⇒ A2 = 64 cm^2
• Third square area A3 = A1 − A2 = 225 cm^2

Concept / Approach:
Compute areas from perimeters, take the difference, then find the square whose area equals that difference. From area, obtain side and then perimeter. This ties together square perimeter–area conversions consistently.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• 64 cm and 32 cm correspond to perimeters of the given squares, not the third.
• 8 cm and 40 cm do not match the difference area 225 cm^2.

Common Pitfalls:

• Treating “difference” as “sum” or mixing perimeters with areas.

Final Answer:
60 cm.