Rhind Mathematical Papyrus (8 of 8)

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RMP 56

The base of a pyramid measures 360 royal cubits and its height 250 royal cubits. Beginners calculate the sekad: how much does the slope recede on 1 royal cubit of height:

half base / height = 180 : 250 = '2 '5 '50

sekad 1 royal cubit x '2 '5 '50 = 5 '25 palms

Advanced learners may solve more demanding problems:

a) imagine a circle in the square of the base and calculate the circumference

b) transform the base into a regular octagon of about the same area; imagine a circle

in the square of the grid and calculate the circumference of that circle

c) transform the volume of the pyramid into a cube

Let us again consult the following number sequence:

3 (plus 22) 25 47 69 ... 289 311 333 355 377

1 (plus 7) 8 15 22 ... 92 99 106 113 120

The diameter of our circle measures 360 = 3 x 120 royal cubits. How long is the circumference? 3 x 377 = 1131 royal cubits. The area of the base measures 360 x 360 = 129,600 square cubits. Octagon of about the same area:

side 164 royal cubits

grid 116+164+116 by 116+164+116 royal cubits

area (grid) 396x396 - 2x116x116 = 129,904 square cubits

side length of grid 396 rc diagonal 560 royal cubits

diameter of inscribed circle 4 x 99 = 396 royal cubits

circumference 4 x 311 = 1244 royal cubits

Finally, the cube. The volume of the pyramid measures

360 x 360 x 250 x '3 = 60 x 60 x 60 x 50 cubic cubits

while the edge of a cube of the same volume measures

60 royal cubits x cube root of 50

How do we approximate the cube root of 50? By using the equation

A x A x A = 50 x B x B x B plus/minus C

and looking out for numbers A and B that will keep C small:

4 x 4 x 4 = 50 x 1 x 1 x 1 plus 16

11 x 11 x 11 = 50 x 3 x 3 x 3 minus 19

70 x 70 x 70 = 50 x 19 x 19 x 19 plus 50

The value 4 is too great: 4 x 4 x 4 = 64

The value '3 x 11 is a little too small:

'3 x 11 x '3 x 11 x '3 x 11 = '27 x 1331 = 49 plus ...

The value '19 x 70 is again a little too great:

'19 x 70 x '19 x 70 x '19 x 70 = 50 plus ...

Pairs of such values generate further and better values:

4 (plus 11) 15 26 37 48 59 70 81

1 (plus 3) 4 7 10 13 16 19 22

11 (plus 70) 81 151 221 291 361 431 501 571

3 (plus 19) 22 41 60 79 98 117 136 155

641 711 781 851 921 991 1061 1131 1201

174 193 212 231 250 269 288 307 326

The value '307 x 1131 is the best approximation for the cube root of 50 (funny, we have already seen the number 1131), while the numbers 221 and 60 yield a simple solution to our problem:

60 x cube root of 50 = about 60 x 221 '60 = 221

A pyramid with a base of 360 royal cubits and a height of 250 royal cubits and a cube of the edge 221 royal cubits have nearly the same volume.

RMP 57 and 58

The base of a pyramid measures 140 royal cubits, the height 93 '3 royal cubits, and the sekad 5 palms 1 finger. Advanced learners may solve the following problems: Imagine a circle in the square of the base and another around the base: how long are the circumferences? Transform the area of the base into a circle: how long are the slopes of the pyramid? Imagine a hemisphere and a sphere in the frame of this pyramid: how long are the radii?

The base measures 140 royal cubits, the diagonal 198 royal cubits, and the average 169 royal cubits. If we regard these numbers as diameters of 3 circles, the circumferences measure:

140 royal cubits x '7 x 22 = 440 royal cubits

198 royal cubits x '99 x 311 = 622 royal cubits

169 royal cubits x '169 x 531 = 531 royal cubits

How can we transform the area of the base into a circle? We may use the value '7 of 22 and work with the following equation, looking out for good values of A and B which will keep C small:¨

22 x A x A = 7 x B x B plus minus C

22 x 22 x 22 = 7 x 39 x 39 plus 1

The numbers 22 and 39 provide a fine approximation for the square root of 're'. The pyramid base measures 140 royal cubits. I multiply this number by 22 and obtain 3080. Now I divide 3080 by 39 and obtain 79 minus '39 or practically 79. Hence, a circle of the radius 79 royal cubits and a square of the side 140 royal cubits have about the same area.

Now for the slope. The height measures '3 x 280 royal cubits, half the base measures '3 x 210 royal cubits, and the slope measures exactly '3 x 350 or 116 "3 royal cubits - according to the Sacred Triangle 3x70-4x70-5x70 = 210-280-350.

Now for the final answers: the radius of the imaginary sphere in the frame of this pyramid measures exactly 35 royal cubits, while the radius of the inscribed hemisphere measures exactly 56 royal cubits.

The imaginary sphere and hemisphere in the frame of the pyramid symbolize the sun and the sky enclosed in the Primeval Hill.

RMP 59

Base and height of a pyramid measure 12 and 8 royal cubits. Let us imagine a wooden model of this pyramid:

height 8 fingers base 12 fingers

half base 6 fingers or 3 x 2 fingers

height 8 fingers or 4 x 2 fingers

slope 10 fingers or 5 x 2 fingers

This pyramid is again defined by a Sacred Triangle, so let me call this type of a pyramid a 'Sacred Pyramid'.

The surface of the model (base and four faces) measures 384 square fingers. How much is the volume? 384 cubic fingers.

The radius of the inscribed sphere measures exactly 3 fingers, while the distance from the center of the sphere to a corner of the base measures exactly 9 fingers (quadruple 3-6-6-9).

Now let us calculate the volume of a sphere of the diameter 9 fingers. Using the value '81 x 256 for 're’ we again obtain 384 cubic fingers. This generates a pretty formula: A square of the side 8 units and a circle with a diameter of 9 units have nearly the same area; a Sacred Pyramid with a height of 8 units and a sphere with a diameter of 9 units have nearly the same volume

RMP 60

Imagine a cone. The diameter of the base measures 15 royal cubits, and its height measures 30 royal cubits. Calculate the volume of the cone. Then imagine a sphere of the same diameter, 15 royal cubits. Calculate the volume of the sphere, using the same value for re (e.g. '15 of 47). The cone and the sphere have exactly the same volume.

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