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

| July 24, 2021 | education

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.



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