Question

ПОМОГИТЕ ПОЖАЛУЙСТА !!!! В правильной шестиугольной пирамиде SABCDEF сторона основания AF=6,высота So-корень из 22.Найдите площадь боковой поверхности пирамиды

Answer 1

Большие диагонали пересекаются в точке О и делят правильный шестиугольник на шесть правильных треугольников.
Тр-ник АВО - правильный, ОМ - его высота. ОМ=а√3/2=6√3/2=3√3.
В тр-ке SМО SМ=√(SO²+OM²)=√(22+27)=7.
Площадь боковой поверхности: S=P·l/2=6АВ·SM/2=6·6·7/2=126 (ед²).

Answer 2

Problem Analysis

We are given a regular hexagonal pyramid with base side AF measuring 6 units and a height of √22 units. We need to find the lateral surface area of the pyramid.

Solution

To find the lateral surface area of the pyramid, we need to calculate the area of each triangular face and then sum them up.

The area of a triangle can be calculated using the formula: Area = (1/2) * base * height.

In this case, the base of each triangular face is the side length of the hexagon, which is 6 units. The height of each triangular face is the height of the pyramid, which is √22 units.

Let's calculate the area of each triangular face and then sum them up to find the lateral surface area of the pyramid.

Calculation

Area of each triangular face = (1/2) * base * height = (1/2) * 6 * √22 = 3√22 square units.

Since there are six triangular faces in a regular hexagonal pyramid, the total lateral surface area of the pyramid is 6 * 3√22 = 18√22 square units.

Therefore, the lateral surface area of the given pyramid is 18√22 square units.

I hope this helps! Let me know if you have any further questions.