Question

Диагональ основания правильной четырехугольной пирамиды в два раза больше высоты боковой грани, проведенной к стороне основания пирамиды. Найдите угол между плоскостями несмежных боковых граней пирамиды. Ответ дайте в градусах.

Answer 1

Решение.
1) Пусть сторона основания равна а. Боковая грань - правильный треугольник, так как плоский угол при вершине равен 60, Поэтому боковое ребро = а. Диагональ основания равна а*sqrt(2), а половина диагонали a*sqrt(2)/2. По теореме Пифагора a^2- a^2/2=16. Откуда a=4*sqrt(2). Высота боковой грани 2*sqrt(6).
S=4*4*sqrt(2)*2*sqrt(6)/2=32*sqrt(3)

Answer 2

Calculating the Diagonal of the Base of a Regular Quadrilateral Pyramid

To find the angle between the planes of the non-adjacent lateral faces of a pyramid, we first need to calculate the diagonal of the base of the pyramid and the height of the lateral face. Given that the diagonal of the base is twice the height of the lateral face, we can proceed with the calculations.

Let's denote: - \( d \) as the diagonal of the base of the pyramid - \( h \) as the height of the lateral face

We know that \( d = 2h \).

Finding the Angle Between the Planes of Non-Adjacent Lateral Faces

The angle between the planes of non-adjacent lateral faces of a pyramid can be calculated using the formula: \[ \cos(\theta) = \frac{h^2 + h^2 - d^2}{2h \cdot h} \] where \( \theta \) is the angle between the planes of the non-adjacent lateral faces.

Calculation

Given that the diagonal of the base of the pyramid is twice the height of the lateral face, we can substitute \( d = 2h \) into the formula to find the angle \( \theta \).

\[ \cos(\theta) = \frac{h^2 + h^2 - (2h)^2}{2h \cdot h} \] \[ \cos(\theta) = \frac{2h^2 - 4h^2}{2h^2} \] \[ \cos(\theta) = \frac{-2h^2}{2h^2} \] \[ \cos(\theta) = -1 \]

Since \( \cos(\theta) = -1 \), the angle \( \theta \) is \( 180^\circ \).

Therefore, the angle between the planes of the non-adjacent lateral faces of the pyramid is 180 degrees.