# The height of a regular triangular prism is equal to H. The straight line passing through the centroid of the base

**The height of a regular triangular prism is equal to H. The straight line passing through the centroid of the base and the middle of the side of the lower base, forms an angle a (alpha) with the base plane. Find the full surface of the prism.**

From the right-angled triangle OO1H we define the segment OH.

OH = OO1 * Ctgα = H * Ctgα.

Since the triangle ABC is equilateral, point O divides the height of the ВН into segments ВO and HO, related as 2/1, then the height of the ВН = 3 * H * Ctgα.

The height of an equilateral triangle is determined by the formula: BH = AB * √3 / 2, then

AC = 2 * ВН / √3 = 2 * 3 * H * Ctgα / √3 = 2 * √3 * H * Ctgα.

Determine the area of the base.

Sbn = ВН * AC / 2 = 3 * H * Ctgα * 2 * √3 * H * Ctgα / 2 = 3 * √3 * H2 * Ctg2α.

Determine the area of the side face.

Side = АА1 * АС = Н * 2 * √3 * H * Ctgα = 2 * √3 * H2 * Ctgα.

Determine the total area of the prism.

S floor = 2 * S main + 3 * S side = 2 * 3 * √3 * H ^ 2 * Ctg2α + 3 * 2 * √3 * H ^ 2 * Ctgα = 6 * √3 * H ^ 2 * Ctg2α + 6 * √ 3 * H ^ 2 * Ctgα = 6 * √3 * H ^ 2 * Ctgα * (Ctgα + 1) cm2.

Answer: The total surface area of the prism is 6 * √3 * H ^ 2 * Ctgα * (Ctgα + 1) cm2.