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If the apex of the pyramid is directly above the center of the square, it is a right square pyramid with four isosceles triangles; otherwise, it is an oblique square pyramid. When all of the pyramid's edges are equal in length, its triangles are all equilateral. It is called an equilateral square pyramid, an example of a Johnson solid.
The base regularity of a pyramid's base may be classified based on the type of polygon: one example is the star pyramid in which its base is the regular star polygon. [24] The truncated pyramid is a pyramid cut off by a plane; if the truncation plane is parallel to the base of a pyramid, it is called a frustum.
Its dihedral angle can be obtained by adding the angle of an equilateral square pyramid and a cube: [6] The dihedral angle of an elongated square bipyramid between two adjacent triangles is the dihedral angle of an equilateral triangle between its lateral faces, arccos ( − 1 / 3 ) ≈ 109.47 ∘ {\displaystyle \arccos(-1/3)\approx 109.47 ...
The Egyptians knew the correct formula for the volume of such a truncated square pyramid, but no proof of this equation is given in the Moscow papyrus. The volume of a conical or pyramidal frustum is the volume of the solid before slicing its "apex" off, minus the volume of this "apex":
The most famous example of a seked slope is of the Great Pyramid of Giza in Egypt built around 2550 BC. Based on modern surveys, the faces of this monument had a seked of 5 + 1 / 2 , or 5 palms and 2 digits, in modern terms equivalent to a slope of 1.27, a gradient of 127%, and an elevation of 51.84° from the horizontal (in our 360 ...
A triangular-pyramid version of the cannonball problem, which is to yield a perfect square from the N th Tetrahedral number, would have N = 48. That means that the (24 × 2 = ) 48th tetrahedral number equals to (70 2 × 2 2 = 140 2 = ) 19600. This is comparable with the 24th square pyramid having a total of 70 2 cannonballs. [5]
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The cosine of the larger of the two non-right angles is the ratio of the adjacent side (the shorter of the two sides) to the hypotenuse, , from which it follows that the two non-right angles are [1] θ = sin − 1 1 φ ≈ 38.1727 ∘ {\displaystyle \theta =\sin ^{-1}{\frac {1}{\varphi }}\approx 38.1727^{\circ }} and θ = cos − 1 1 φ ...