Consider a square sheet of side 1 unit. The sheet is first folded along the main diagonal. This is followed by a fold along its line of symmetry. The resulting folded shape is again folded along its line of symmetry. The area of each face of the final folded shape, in square units, equal to _______

The correct option is 1/8.

Let us analyze the folding sequence step-by-step as shown in the provided illustration:

Step 1: Original Square Sheet
We begin with a square sheet of side length 1 unit.
The area of the square sheet is:
$$\text{Area}=1\times 1=1\text{ square unit}$$ The diagonal of the square has a length of:
$$d=\sqrt{1^2+1^2}=\sqrt{2}\text{ units}$$

Step 2: First Fold (Along the Main Diagonal)
The sheet is folded along its main diagonal of length $\sqrt{2}$.
This forms a right-angled isosceles triangle with two perpendicular sides of length 1 unit each, and a hypotenuse of length $\sqrt{2}$ units.
The area of this triangle (the face of the folded shape) is:
$${\text{Area}}_1=\frac{1}{2}\times 1\times 1=\frac{1}{2}\text{ square unit}$$

Step 3: Second Fold (Along the Line of Symmetry)
Next, the triangle is folded along its line of symmetry (the perpendicular line from the right angle to the hypotenuse of length $\sqrt{2}$).
This line of symmetry bisects the hypotenuse, resulting in two smaller right-angled isosceles triangles.
The hypotenuse of the original triangle ($\sqrt{2}$) is halved to:
$$\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\text{ units}$$ This folded shape is a right-angled triangle with perpendicular sides of length $\frac{1}{\sqrt{2}}$ units and a hypotenuse of length 1 unit.
The area of each face of this folded shape is:
$${\text{Area}}_2=\frac{1}{2}\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{1}{4}\text{ square unit}$$

Step 4: Third Fold (Along the Line of Symmetry again)
The resulting triangle has perpendicular sides of length $\frac{1}{\sqrt{2}}$ and a hypotenuse of length 1.
It is folded again along its line of symmetry, which is the altitude from the right angle to the hypotenuse (denoted as $AB$ in the diagram).
Let the final folded right-angled triangle be $AOB$, where:
- The base $OB$ is half of the hypotenuse of the previous shape:
$$OB=\frac{1}{2}\text{ unit}$$ - The hypotenuse $AB$ is:
$$AB=\frac{1}{\sqrt{2}}\text{ unit}$$ - The height $AO$ can be calculated using the Pythagorean theorem:
$$AO=\sqrt{A{B}^2-O{B}^2}$$
$$AO=\sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2-{\left(\frac{1}{2}\right)}^2}=\sqrt{\frac{1}{2}-\frac{1}{4}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\text{ unit}$$

Now, we calculate the area of the face of the final folded shape ($\Delta AOB$):
$$\text{Final Area}=\frac{1}{2}\times \text{base}\times \text{height}$$
$$\text{Final Area}=\frac{1}{2}\times OB\times AO$$
$$\text{Final Area}=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}\text{ square unit}$$

Therefore, the area of each face of the final folded shape is 1/8 square units.