In a regular triangular pyramid, the side of the base is 12 dm, and the height of the pyramid is 10
In a regular triangular pyramid, the side of the base is 12 dm, and the height of the pyramid is 10 dm. Find the volume of the pyramid. Write down the formula for the lateral surface.
Knowing the side of the base of the equilateral triangle ABC, we calculate the area of the base of the prism.
Sbasn = AB ^ 2 * √3 / 4 = 144 * √3 / 4 = 36 * √3 dm2.
Then the volume of the pyramid is equal to: Vpir = Sosn * DO / 3 = 36 * √3 * 10/3 = 120 * √3 dm3.
In an equilateral triangle ABC, we calculate the height of CH.
CH = AB * √3 / 2 = 12 * √3 / 2 = 6 * √3 dm. Then, by the property of the medians, OH = CH / 3 = 6 * √3 / 3 = 2 * √3 dm.
In a right-angled triangle DOH, according to the Pythagorean theorem, DH ^ 2 = OH ^ 2 + DO ^ 2 = 12 + 100 = 112.
DН = 4 * √7 dm.
Then Sbok = 3 * Savd = 3 * AB * DH / 2 = 3 * 12 * 4 * √7 / 2 = 72 * √7 dm2.
Answer: The volume of the pyramid is 120 * √3 dm3, Side = 3 * AB * DH / 2 dm2.