The diagonal of a square is 9√2 cm. The square is reshaped into an equilateral triangle using the same perimeter. What is the area (in sq cm) of the largest incircle that can be drawn inside this triangle?

Introduction / Context:
This geometry question combines properties of a square, an equilateral triangle and an incircle. The key idea is that a wire shaped as a square is bent to form an equilateral triangle while keeping the same perimeter. Then we use the relation between side length of an equilateral triangle and the radius of its incircle. This type of problem is popular in aptitude tests because it checks multiple geometry concepts in a single scenario.

Given Data / Assumptions:

• Diagonal of the square is 9√2 cm.
• The square is reshaped into an equilateral triangle with the same perimeter.
• We need the area of the largest possible incircle of that equilateral triangle.
• Use the standard formulas for diagonal of a square and properties of an equilateral triangle.
• Assume no loss of material while reshaping, so perimeter remains constant.

Concept / Approach:
For a square of side s, the diagonal is s√2. Once we know s, we find the perimeter of the square. This perimeter is then used as the perimeter of the equilateral triangle. So if the triangle has side length a, then 3a equals the original perimeter. For an equilateral triangle, the inradius r is given by r = (a√3) / 6. The area of the incircle is then π * r^2. We apply these formulas step by step.

Step-by-Step Solution:
Step 1: Let the side of the square be s. Given diagonal = s√2 = 9√2, so s = 9 cm. Step 2: Perimeter of the square = 4s = 4 * 9 = 36 cm. Step 3: This perimeter becomes the perimeter of the equilateral triangle, so 3a = 36 which gives a = 12 cm. Step 4: For an equilateral triangle, inradius r = (a√3) / 6 = (12√3) / 6 = 2√3 cm. Step 5: Area of the incircle = π * r^2 = π * (2√3)^2 = π * 4 * 3 = 12π sq cm.

Verification / Alternative check:
We can also note that the area of an equilateral triangle of side 12 cm is (√3 / 4) * a^2 = (√3 / 4) * 144 = 36√3 sq cm. The relation between area, inradius r and semiperimeter s of any triangle is Area = r * s. Here semiperimeter s = 36 / 2 = 18. So r = Area / s = 36√3 / 18 = 2√3 cm, matching our earlier computation. Substituting r into π * r^2 again gives 12π sq cm, confirming the result.

Why Other Options Are Wrong:
The values 6π and 9π correspond to smaller radii that do not satisfy the triangle and incircle relations. The value 15π would require a larger radius than allowed by the equilateral triangle with side 12 cm. The extra option 18π would similarly exceed the maximum possible incircle for a fixed perimeter. Only 12π is consistent with both the perimeter transformation and known formulas for equilateral triangles and incircles.

Common Pitfalls:
A frequent mistake is assuming the square and triangle have equal areas instead of equal perimeter. Another common error is misusing the formula for inradius or mixing up circumradius and inradius. Learners may also forget the relation between diagonal and side of a square. Carefully identifying which quantity stays constant and then applying the correct formulas in the correct sequence is essential for success in such mixed shape problems.

Final Answer:
The area of the largest incircle is 12π sq cm.