In a regular quadrangular pyramid, the diagonal of the base is 8 roots of 2 cm, and the height is 3 cm. Find the lateral surface area.

To solve the problem, consider the figure.

The area of ​​the side surface of the pyramid is equal to half the product of the perimeter of the pyramid base and the apothem.

S = (Pabcd * SK) / 2.

Consider a right-angled triangle ACD with legs AD = CD and hypotenuse AC = 8 * √2.

Find the sides of the base by the Pythagorean theorem.

AD^2 + CD^2 = (8 * √2) ^2.

2 * AD^2 = 64 * 2

AD = 8 = CD = AB = AC.

The perimeter of the base is: P = 4 * AD = 32 cm.

Consider a right-angled triangle SOK with OK = AD / 2 = 8/2 = 4 cm.

Then by the Pythagorean theorem apothem SK2 = OK2 + SO2 = 16 + 9 = 25.

SK = 5 cm.

Then the side surface of the pyramid is equal to:

S = (32 * 5) / 2 = 80 cm2.

Answer: The lateral surface area is 80 cm2.