Question

Не понимаю эту тему, кто знает помогите Если в основании пирамиды лежит ромб с острым углом α и стороной а и все боковые грани пирамиды наклонены к основанию под углом β, то боковая поверхность пирамиды равна

Answer 1

Площадь боковой поверхности пирамиды можно рассматривать как такую,  проекция которой на основание даёт само основание.

Тут можно применить свойство проекции: Sпр = S*cos β, отсюда

S = Sпр/cos β.

Для ромба площадь равна a²sinα.

Ответ: Sбок = (a²sinα)/cos β.

Answer 2

I will try to answer your question in detail. I understand that this topic is difficult, but I hope I can help you.

If the base of the pyramid is a rhombus with an acute angle α and a side a, and all the lateral faces of the pyramid are inclined to the base at an angle β, then the lateral surface area of the pyramid is equal to:

S = 2ab * sin(α) * sin(β)

This formula can be derived as follows:

Let O be the center of the rhombus, and let H be the height of the pyramid. Then, OH is perpendicular to the base, and the triangles OAB and OHC are similar. Therefore, we have:

OH / OA = HC / AB

Since OA = a * cos(α/2) and AB = a * sin(α), we can solve for HC:

HC = OH * AB / OA = OH * a * sin(α) / (a * cos(α/2)) = OH * tan(α/2)

Now, let S be the area of one of the lateral faces of the pyramid. Then, S is equal to the product of the base and the height of the face, divided by two. The base of the face is AB, and the height is HC * sin(β), since HC is inclined to the base at an angle β. Therefore, we have:

S = (AB * HC * sin(β)) / 2 = (a * sin(α) * OH * tan(α/2) * sin(β)) / 2

Since the pyramid has four lateral faces, the total lateral surface area is:

S = 4 * S = 2ab * sin(α) * sin(β)

I hope this explanation helps you understand the topic better. If you have any more questions, feel free to ask me.