# In a regular quadrangular pyramid, the side edges are 10 cm and form an angle

**In a regular quadrangular pyramid, the side edges are 10 cm and form an angle of 60 degrees with the base plane. Find the surface area of the pyramid.**

The POC triangle is rectangular with an acute angle OCP = 600 and a hypotenuse CP = 10 cm, then:

Cos60 = OC / CP.

OC = CP * Cos60 = 10 * 1/2 = 5 cm.

The OC segment is half the length of the AC diagonal, then AC = 2 * OC = 2 * 5 = 10 cm.

Determine the area of the square at the base of the pyramid. Sosn = AC ^ 2/2 = 100/2 = 50 cm.

Then AC = √Sbase = √50 = 5 * √2 cm.

The height of the PH of the lateral face is the median of the HRV triangle, then CH = BC / 2 = 5 * √2 / 2 cm. Let us determine the leg of the PH from the right-angled triangle CHP.

PH ^ 2 = CP ^ 2 – CH ^ 2 = 100 – 25/2 = 175/2.

PH = 5 * √7 / √2.

Let’s define the area of the ВСН triangle.

Svsr = ВС * РH / 2 = (5 * √2) * (5 * √7 / √2) / 2 = 25 * √7 / 2.

Then Sпов = Sсн + 4 * Sbok = 50 + 4 * 25 * √7 / 2 = 50 * (1 + √7) cm2.

Answer: The total surface area is 50 * (1 + √7) cm2.