Two identical square sheets of paper each have a diagonal of 18 cm. They are laid on top of each other so that the overlapping region is itself a square with side AB. If the total area covered by the combined figure is 246 square cm, find the length of AB.

Step-by-step solution

1. 1

   Get each square's area straight from its diagonal. For a square of side $s$, the diagonal is $d=s\sqrt2$, so $s^2=\dfrac{d^2}{2}$. The area is therefore half the square of the diagonal — no need to find the side first:

   $\text{Area}=\frac{1}{2}d^2=\frac{1}{2}(18)^2=\frac{324}{2}=162\ \text{cm}^2$

2. 2

   Set up the inclusion-exclusion relation. When two regions overlap, the area actually covered counts the shared part only once:

   $\text{Covered}=\text{Area}_1+\text{Area}_2-\text{Overlap}$

   This identity is the whole engine of the problem.

3. 3

   Solve for the overlap area. The two sheets together have $162+162=324\ \text{cm}^2$ of paper, but they cover only $246\ \text{cm}^2$:

   $\text{Overlap}=324-246=78\ \text{cm}^2$

   The missing $78\ \text{cm}^2$ is exactly the doubly covered region.

4. 4

   Convert the overlap area to a side length. The problem states that the overlap is itself a square, so its side is the square root of its area:

   $AB=\sqrt{78}\ \text{cm}$

5. 5

   Simplify and evaluate. $78=2\times3\times13$ has no square factors, so $\sqrt{78}$ cannot be simplified:

   $AB=\sqrt{78}\approx8.83\ \text{cm}$

6. 6

   Check that the answer is geometrically possible. Each sheet has side $\sqrt{162}\approx12.73$ cm, and the overlap side $8.83$ cm is smaller than that ✓ — an overlap can never exceed the sheet itself. Also $78<162$, so the overlap is under half of each sheet, consistent with a partial overlap ✓.