User:Canisandfelis/sandbox

objetivos

O processo de dedução da formula do tronco de pirâmide tradicional é por demais que complicado, tendo-se em vista que ha de se extrair uma pirâmide secundaria de uma pirâmide inicial, para se ter um determinado tronco com uma altura h. Esta altura tem de ser determinada pela lei de semelhança de triângulos, para que a progressão do comprimentos dos quadrados à essa altura determine o volume da pirâmide inicial, da qual se obterá um tronco de altura especifica. Assim um processo de visualização de geometria espacial interna se estabelece no começo e um processo algébrico conduz a formula final, mas o processo algébrico deixa para traz, a visualização de geometria espacial interna do começo, entende-se o processo, através da visualização de geometria espacial interna de começo, com o processo algébrico sequente, mas não o processo geométrico espacial como um todo! O processo a seguir mostrara o processo como um todo, através da construção geométrica espacial e, em seguida a decomposição em sólidos, para a obtenção da visualização geométrica espacial completa.

Obijetives

The process of deduction of traditional pyramid trunk formula is too complicated, having in mind that ha to extract a secondary pyramid a pyramid, to have a particular trunk with a height h, this time must be determined by the law of similarity of triangles, so that the progression of square lengths at this point to determine the volume of the original pyramid, which will get a tall trunk specifies. So a process of visualization of internal space geometry is established at the beginning and an algebraic process leads the final formula, but the algebraic process leaves behind, the visualization of internal space geometry from the beginning, the process, through the internal spatial geometry preview beginning, with the process of algebraic, but not spatial geometric process as a whole. The following process will show the process as a whole, through the spatial and geometric construction, then the decomposition in solids, for obtaining the full spatial geometric visualization.

Frustum of pyramid

First tends to imagine a frustum of pyramid

You begin with the orthographic projection of the square base of the top in the area of the square base and larger base subtraction base area smaller, you get a square edge.

Tracing-if straight segments 45° angles, the largest square edge square to the corners of the lower court, if you have 4 trapezes, first has to deduce the areas of trapezoids, needs only the height:

(L-l)/2

Has the height, you need larger bases and now less:

(L-l)/2.(L + l)

You need then to share everything by two to get the area of trapezoids

(L-l)/2.(L + l)/2

(L-l)/2.(L + l) 1/2

Developing if you have:

(L-l).(L + l)/4

By adding the height if you have:

(L-l).(L + l)/4.h

Now the square edge has a three-dimensional spatial dimension, as the areas of bases are trapezoids, cutting with the base segments that determined the trapezoids, has four trapezoidal base solid.

Original trunk if taken up base square as s so the trunk volume of pyramid was deducted from the decomposition of Polyhedra, in case the decomposition of a trunk of height h, the geometric visualization process-see space as a whole, not only to deduce the height of a pyramid «by similarity of triangles and the process of simple algébrico.ao square development is at the base of the topThis fits perfectly to the solid square edge three-dimensional.

Works now with the four trapezoidal bases solids, from the base to top of solido trapezoidal bases with the smaller base of top area, construct a diagonal plan uniting, the smaller base off the larger base below, if you have two Polyhedra of equal trapezoidal bases.

(L-l).(L + l)/4.h/2

(L-l).(L + l)/4.h.1/2

Developing if has eight equal solids

(L-l).(L + l)/8.h

Are four trapezoidal bases solids

(L-l).(L + l)/8.h.4

Developing

(L-l).(L + l)/2.h

If you have the square of the sum for the difference

(L ^ 2-l ^ 2) h/2

L ^ 2 and l ^ 2 areas are the foundations of the original stem, then:

(A-a). h/2

Added the base polyhedron blocks equal the area of the top of the stem and if you have:

(A-a). h/2 + a. h

The volume of the trunk looks like this:

V = (A-a). h/2 + a. h

This demonstrates that a frustum of pyramid decomposes in the Union of these Polyhedra in sequence determines the volume of the same wood, without the determination of the height of the original pyramid predicted by similarity of triangles, to subtract a pyramid of another, and the whole process internal space geometric, is apparent.

References: Euclid's elements