In a regular quadrangular pyramid, the side of its base is 10 and the height is 12, find the area of the side surface.

The side faces of the pyramid are isosceles triangles.

Let’s construct the height of the PH, which will also be the median of the HRV triangle. Point O divides the diagonal AC in half, then OH is the middle line of the triangle ABC, and then OH = AB / 2 = 10/2 = 5 cm.

From the right-angled triangle ОPН, by the Pythagorean theorem, we determine the length of the hypotenuse PH.

PH ^ 2 = PO ^ 2 + OH ^ 2 = 144 + 25 = 169.

PH = 13 cm.

Let’s calculate the area of the side face of HPB.

Svsr = BC * PH / 2 = 10 * 13/2 = 45 cm2.

Since all the side faces are equal, then Sside = 4 * Svav = 4 * 45 = 180 cm2.

Answer: The lateral surface area is 180 cm2.