To solve the problem of finding the area of the overlapping part when a rectangular paper is folded along its diagonal, follow these steps:

Step 1: Define the Rectangle and Folding

Let the rectangle be (ABCD) with (AB=10\,\text{cm}) (length) and (BC=6\,\text{cm}) (width). Fold along diagonal (AC), so point (B) maps to (B'). The overlapping part is (\triangle AEC) (where (E) is the intersection of (AB') and (CD)).

Step 2: Key Observations

• (\angle EAC = \angle ACD) (alternate interior angles, since (AB\parallel CD)).
• Folding preserves angles: (\angle BAC = \angle B'AC), so (\angle ACD = \angle B'AC), making (\triangle AEC) sceles ((AE=EC)).
• (\triangle ECB') is right-angled at (B') ((\angle AB'C = 90^\circ)).

Step 3: Calculate (AE)

Let (AE=x), then (EB'=10-x) and (EC=x). By Pythagoras in (\triangle ECB'):
[x^2 = (10-x)^2 + 6^2]
[x^2 = 100 - 20x + x^2 + 36]
[20x = 136 \implies x=\frac{34}{5}=6.8\,\text{cm}]

Step 4: Compute the Area of (\triangle AEC)

Using coordinates or shoelace formula for (\triangle AEC) (points (A(0,0)), (E(\frac{16}{5},6)), (C(10,6))):
[Area = \frac{1}{2}\left|0(6-6) + \frac{16}{5}(6-0) + 10(0-6)\right|]
[= \frac{1}{2}\left|\frac{96}{5} - 60\right| = \frac{102}{5}=20.4\,\text{cm}^2]

Answer: (\boxed{20.4}) (or (\boxed{\frac{102}{5}}) as an exact fraction).

(\boxed{20.4})

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