The height of a regular quadrangular pyramid is 4 cm and its apothem forms with a height of 45

| April 26, 2021 | education

The height of a regular quadrangular pyramid is 4 cm and its apothem forms with a height of 45 find the area of the base of the pyramid and the side surface of the pyramid.

Since the angle between height and apothem is 45, the HOY triangle is rectangular and isosceles, PO = HO = 4 cm.Then PH ^ 2 = 2 * HO ^ 2 = 2 * 16 = 32. PH = 4 * √2 cm.

At the base of the pyramid is the square ABCD, then AO = CO = BO = DO, since the diagonals of the square are divided in half at point O. AH = BH, since PH is the median of the APB triangle, then OH is the middle line of the ABC triangle, then AB = BC = 2 * OH = 2 * 4 = 8 cm.

Determine the area of ​​the base. Savsd = AB ^ 2 = 8 ^ 2 = 64 cm2.

Determine the area of ​​the triangle PAB.

Sarv = AB * РН / 2 = 8 * 4 * √2 / 2 = 16 * √2 cm2.

Side = Sarv * 4 = 4 * 16 * √2 = 64 * √2 cm2.

Answer: The base area is 64 cm2, the lateral surface area is 64 * √2 cm2.



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