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)

Sample Problem:

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