The lateral surface of a regular quadrangular prism has an area of 16 dm2. The diagonal of the base is 4√2.

The lateral surface of a regular quadrangular prism has an area of 16 dm2. The diagonal of the base is 4√2. Find the cross-sectional area of the prism passing through the diagonals of two adjacent side faces that have a common vertex.

The section of the prism passing through the diagonals of two adjacent lateral faces that have a common vertex is an isosceles triangle in which the lateral faces are the diagonals of the lateral faces of the prism, the base is the diagonal of the base of the prism.
The base of a regular quadrangular prism is a square, knowing its diagonal we can find its side: d ^ 2 = a ^ 2 + a ^ 2;
d ^ 2 = 2 * a ^ 2;
a ^ 2 = (d ^ 2) / 2 = 16;
Base side a = √16 = 4 in.
Side surface area – the sum of the areas of the four side faces, which are equal rectangles. The area of ​​the side face is equal to the product of the side of the base and the height of the prism.
Side = 4 * a * h;
h = S side / 4a = 16 / (4 * 4) = 1 dm.
The square of the diagonal of the side face can be found as the sum of the squares of the side of the base and the height:
D ^ 2 = h ^ 2 + a ^ 2 = 1 + 16 = 17;
Side face diagonal D = √17 dm.
Let us find the height of the prism section as a leg of a right-angled triangle, in which the diagonal of the side face is the hypotenuse, and half of the diagonal of the base is the second leg. H ^ 2 = D ^ 2- (0.5 * d) ^ 2 = 17- (2√2) ^ 2 = 17-8 = 9; H = √9 = 3 dm.
Let us find the area of ​​the given section of the prism as half of the product of the height of the section and the diagonal of the base: 0.5 * 3 * 4√2 = 6√2 dm2, which is approximately equal to 8.485 dm2.