This figure is cut out from a sheet of paper. Folding the sides upwards along the dashed lines, one gets a (non-equilateral) pyramid with a square base. Calculate the area of the base.
[img]https://1.bp.blogspot.com/-lPpfHqfMMRY/XzcBIiF-n2I/AAAAAAAAMW8/nPs_mLe5C8srcxNz45Wg-_SqHlRAsAmigCLcBGAsYHQ/s0/2005%2BMohr%2Bp1.png[/img]
1. **Identify the given information and the goal:**
   - We are given a figure that, when folded, forms a pyramid with a square base.
   - We need to calculate the area of the base of this pyramid.

2. **Define the variables:**
   - Let the horizontal side of the triangle be \( a \).
   - Let the vertical side of the triangle be \( b \).

3. **Use the Pythagorean theorem:**
   - The slant height of the pyramid (the hypotenuse of the right triangle) is given as 5 units.
   - The height of the pyramid (the vertical distance from the apex to the base) is given as 7 units.

4. **Set up the equations using the Pythagorean theorem:**
   - For the right triangle with sides \( a \) and 5, and hypotenuse 7:
     \[
     a^2 + 5^2 = 7^2
     \]
     \[
     a^2 + 25 = 49
     \]
     \[
     a^2 = 24
     \]

   - For the right triangle with sides \( b \) and 5, and hypotenuse 7:
     \[
     b^2 + 5^2 = 7^2
     \]
     \[
     b^2 + 25 = 49
     \]
     \[
     b^2 = 24
     \]

5. **Verify the consistency of the equations:**
   - Both \( a^2 \) and \( b^2 \) are equal to 24, which is consistent with the given conditions.

6. **Calculate the area of the base:**
   - Since the base is a square and each side of the square is equal to \( \sqrt{a^2} = \sqrt{24} \):
     \[
     \text{Side length of the square} = \sqrt{24} = 2\sqrt{6}
     \]
   - The area of the square base is:
     \[
     \text{Area} = (\text{side length})^2 = (2\sqrt{6})^2 = 4 \times 6 = 24
     \]

The final answer is \(\boxed{24}\).