User:Kotnog/pyramid building

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PYRAMID BUILDING using BRICKS

PHASE A:

Since pyramid is 1/3 of base times height of a cube, find the quantity of bricks that forms a cube.

THEORY NO 1: The volume of a brick is similar to the quantity of bricks on the cube but not equal. They may not equal in value in but similar in numerical. S3 = V(xyz)

cube out of bricks

cube out of bricks

THEORY NO 2: The volume of a brick divided by its dimension is equal to the quantity of base area of a cube. Since cube has all edges of equal length, also the base and height but not the same to the number of quantities of piled bricks on each dimension.

Therefore:

S3 = [((V/l)l) ((V/w)w) ((V/h)h)]

File:G:tn02.jpg

Finding quantity x, y, z.

(x) quantity of bricks piled horizontally = V / l

(y)quantity of bricks piled diagonally = V / w

(z)quantity of bricks piled vertically = V / h

To find the corresponding quantities of bricks to be piled on each direction, multiply x to the length for horizontally, y to width for vertically, and z to height of a brick for diagonally. Thus this result will determine that each side is equal forming a cube.
S3 = [(xl)(yw)(zh)] S3 = V(xyz)

File:G:/tn012.jpg

Sample Problem:
Dimension of a brick is 3 x 4 x 5 ft. Find the quantity x, y, and z.

UNIT	QUANTITY
Length = 5 ft.	x = V/l
Width = 4 ft.	y = V/w
Height = 3 ft.	z = V/h

Formula: S3 = [((V/l)l) ((V/w)w) ((V/h)h)]

   = [((60/5)l) ((60/4)w) ((60/3)h)]
   = [(12.5) (15.4) (20.3)]
   = [3600.60]
   = 3√216000

S = 60 cu. ft. Conclusion: A 3600 bricks with a dimension of 3 x 4 x 5 ft. each can be formed into a cube by layering 12 times horizontally, 15 times vertically and 20 times diagonally.

PHASE B: To find their quantity Area by level of the pyramid, (x) subtracted by height of bricks (h) times level of Pyramid (p) divided by the length of brick (l) until the equation is equal to zero (0), and same as the (y). As shown below: x0 = x-(hp)/l y0 = y-(hp)/w Cubes Dimension Quantity of Bricks Horizontal (x) x0 = x-(hp)/l  x0 = 0 Diagonal (y) y0 = y-(hp)/w  y0 = 0 Level of Pyramid (p) base  top And the equation must be: A0 = x-(hp)/l * y-(hp)/w To find their unit Area by level, multiply the x, y by length (l), width(w). A0 = x0y0 (lw) A1 = x1y1 (lw) A2 = x2y2 (lw) A3 = x3y3 (lw)

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