The base of the pyramid is a rhombus, the diagonals of which are equal to 6 in 8 in.

The base of the pyramid is a rhombus, the diagonals of which are equal to 6 in 8 in. The height of the pyramid passes through the intersection of the diagonals of the rhombus. Find the side surface of the pyramid if the apothem is 8 dm.

By the property of a rhombus, its diagonals are mutually perpendicular and the point of intersection is divided in half. Consider one of the right-angled triangles obtained by intersecting the diagonals of a rhombus: its two legs will be equal, respectively
a = 1/2 * 6 = 3 dm
and
b = 1/2 * 8 = 4 dm.
Then, in accordance with the Pythagorean theorem, its third side, and accordingly the side of the rhombus, is
c = √ (a² + b²) = √ (3² + 4²) = 5 dm.
Since the base of the pyramid is a rhombus, its lateral surface is made up of four equal triangles. By the known one of the sides of the triangle c and the height h dropped on it (apothem of the pyramid), we find the area of ​​one triangle that makes up the lateral surface:
S1 = 1/2 * c * h.
Then the total area of ​​the side surface of the pyramid will be equal to
S = 4 * S1 = 4 * 1/2 * c * h = 2 * c * h;
S = 2 * 5 * 8 = 80 dm².