The height of a regular quadrangular pyramid is 10 cm and forms an angle of 45 degrees with the side edge. Find S total.

The AOK triangle is rectangular, since the OK is the height. The angle of the OAK, by condition, is equal to 45, then the legs of the OAK triangle are equal.

OK = OA = 10 cm.

Since there is a square at the base of the pyramid, the diagonal AC, at point O, is divided in half, then AC = AO * 2 = 20 cm.

From the right-angled triangle ACD, we determine the length of the legs AD and CD.

AC ^ 2 = 2 * CD ^ 2.

CD ^ 2 = AC ^ 2/2 = 400/2 = 200.

СD = 10 * √2 cm.

Sbn = СD ^ 2 = 200 cm2.

Let’s build the height of the KH, which will also be the median of the triangle, then DH = CH = CD / 2 = 10 * √2 / 2 = 5 * √2 cm.

Point O is the point of intersection of the diagonals, which divides them in half, then the segment OH is the middle line of the triangle ACD. Then OH = AD / 2 = 10 * √2 / 2 = 5 * √2 cm.

From a right-angled triangle KOH, KH ^ 2 = OK ^ 2 + OH ^ 2 = 100 + 50 = 150.

KN = 5 * √6 cm.

Then Ssdk = CD * KН / 2 = 10 * √2 * 5 * √6 / 2 = 25 * √12 = 50 * √3 cm2.

Side = 4 * Sdk = 4 * 50 * √3 = 200 * √3 cm2.

Spov = Sb + S side = 200 + 200 * √3 = 200 * (1 + √3) cm2.

Answer: The total surface area is 200 * (1 + √3) cm2.