The lateral surface of a regular triangular pyramid is 30420mm2, and its lateral edge is 169mm.

| January 10, 2021 | education

The lateral surface of a regular triangular pyramid is 30420mm2, and its lateral edge is 169mm. Find the area of the base of the pyramid.

The side faces of a regular pyramid are equal-sized isosceles triangles, then Svsd = Sside / 3 = 30420/3 = 10140 mm2.
Let the angle of the internal combustion engine = α, then Svsd = ВD * СD * Sinα / 2.
Sinα = 2 * Svsd / СD * СD = 2 * 10140/169 * 169 = 120/169.
Determine the cosine of the angle of the BCD.
Cos2α = 1 – Sin2α.
Cosα = √ (1 – Sin2α) = √ (1 – (120/169) 2) = 119/169.
Let’s apply the cosine theorem for a triangle.
СV2 = DS2 + DВ2 – 2 * DС * DВ * Cosvsd = 28561 + 28561 – 2 * 28561 * 119/169 = 57122 – 40222 = 16900.
CB = √16900 = 130 mm.
The ABC triangle is equilateral, then AH = CB * √3 / 2 = 130 * √3 / 2 = 65 * √3 mm.
Determine the area of ​​the base.
Sbn = ВС * АН / 2 = 130 * 65 * √3 / 2 = 4225 * √3 cm2.
Answer: The base area is 4225 * √3 cm2.



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