The apothem of a regular quadrangular pyramid is L, and the flat angle at the apex is alpha. Find the total surface area of the pyramid.
The lateral faces of a regular pyramid are isosceles triangles, which have an apothem with the height, median, and bisector of the triangle. Then in the triangle KCD, CH = DH, the angle DНK = α / 2, and the triangles KНD and KНС are rectangular.
In a right-angled triangle КНD, tg (α / 2) = DH / КН = DH / L.
DН = L * tg (α / 2), then СD = 2 * L * tg (α / 2) see.
Determine the area of the base of the pyramid.
Since there is a square at the base, then Sbn = DH ^ 2 = 4 * L ^ 2 * tan2 (α / 2) cm2.
The area of the side face of the pyramid is: Sksd = KН * CD / 2 = L * 2 * L * tan (α / 2) / 2 = L ^ 2 * tan (α / 2), then Side = 4 * L2 * tan (α / 2) cm2.
Spov = Sbok + Sbn = 4 * L ^ 2 * tan (α / 2) + 4 * L ^ 2 * tan2 (α / 2) = 4 * L ^ 2 * tan (α / 2) * (1 + tan ( α / 2) cm2.
Answer: The total surface area is 4 * L ^ 2 * tan (α / 2) * (1 + tan (α / 2) cm2.
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