The sides of the base of a regular quadrangular pyramid are 10 dm and 2 dm, and its height is 2 dm.

The sides of the base of a regular quadrangular pyramid are 10 dm and 2 dm, and its height is 2 dm. find the area of the diagonal section of this pyramid.

Since the truncated pyramid is correct, there are squares at its bases.
By the Pythagorean theorem, we determine the lengths of the diagonals at the base of the pyramid.
AC^2 = AD^2 + CD^2 = 100 + 200 = 200.
AC = 10 * √2 dm.
A1C1^2 = A1D1^2 + C1D1^2 = 4 + 4 = 8.
A1C1 = 2 * √2 dm.
The diagonal section of the truncated pyramid is an isosceles trapezoid АА1С1С, the height of which is equal to the height of the pyramid.
Then Ssec = (AC + A1C1) * OO1 / 2 = (10 * √2 + 2 * √2) * 2/2 = 12 * √2 dm2.
Answer: The area of the diagonal section is 12 * √2 dm2.