In a regular quadrangular pyramid, the side of the base is 8 cm, and the side edge

| April 25, 2021 | education

In a regular quadrangular pyramid, the side of the base is 8 cm, and the side edge is inclined to the plane of the base at an angle of 45. Find the total surface area.

Determine the length of the AC diagonal at the base of the pyramid.

Since there is a square at the base of the pyramid, its diagonal is equal to: AC = AB * √2 = 8 * √2 cm.

Point O divides the diagonals of the square in half, then OS = AC / 2 = 8 * √2 / 2 = 4 * √2 cm.

Determine the length of the height of the pyramid, from the right-angled triangle POC. Since in a right-angled triangle one of the angles is 45, the legs of this triangle are equal.

RO = OC = 4 * √2 cm.

In a right-angled triangle PОН, we define the length of the hypotenuse PН by the Pythagorean theorem.

PH ^ 2 = PO ^ 2 + OH ^ 2 = 32 + 16 = 48.

PH = 4 * √3 cm.

Then the area of ​​the side face BСP is equal to: Svav = ВС * РН / 2 = 8 * 4 * √3 / 2 = 16 * √3 cm2.

Then Side = 4 * Svsr = 4 * 16 * √3 = 64 * √3 cm2.

Sbn = AB^2 = 64 cm2.

Spov = Sb + S side = 64 + 64 * √3 = 64 * (1 + √3) cm2.

Answer: The total surface area of ​​the pyramid is 64 * (1 + √3) cm2.



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