| ME | = 2000000 | = | diameter. | |||
| AB | = 1902113 | = | AD ÷ DB | |||
| MB | = 1618034 | = | MC + MH | = | MP + PB | |
| AS | = 1538841·5 | |||||
| EP | = 1453086 | = | AG + FB | |||
| AF | = 1175570 | = | AE = GB | |||
| MC | = 1000000 | = | radius = CD + DX | = | CH + CX | |
| AD | = 951056·5 | = | DB = DS | |||
| PB | = 854102 | |||||
| QS | = 812298·5 | |||||
| MP | = 763932 | = | CH × 2 | = | base of Cheops. | |
| AG | = 726543 | = | GH = XH | = | HN = PF = FB = Slant edge of Cheops. = slant edge of Pent. Pyr. | |
| DE | = 690983 | = | DH = XD | = | apothem of Pentagonal Pyramid. | |
| MH | = 618034 | = | MN = XC = | { | apothem of Cheops. altitude of Pentagonal Pyramid. side of decagon inscr'd in circle. | |
| MS | = 500000 | |||||
| 485868 | = | { | mean proportional between MH and HC altitude of Cheops. | |||
| OP | = 449027 | = | GF = GD + DF | |||
| HC | = 381966 | = | half base of Cheops. | |||
| SO | = 363271·5 | = | HS | |||
| CD | = 309017 | = | half MH | |||
| PR | = 277516 | |||||
| GD | = 224513·5 | |||||
| SP | = 263932 | |||||
The triangle DXH represents a vertical section of the pentagonal pyramid; the edge HX is equal to HN, and the apothem DX is equal to DE. Let DH be a hinge attaching the plane DXH to the base, now lift the plane DXH until the point X is vertical above the centre C. Then the points A, E, B, O, N of the five slant slides, when closed up, will all meet at the point X over the centre C.
We have now built a pyramid out of the pentangle, whose slope is 2 to 1, altitude CX being to CD as 2 to 1.
| Apothem | DX = DE |
| Altitude | CX = HM or MN |
| Altitude | CX + CH = CM radius. |
| Apothem | DX + CD = CM radius. |
| Edge | HX = HN or PF |
Note also that
| (MP) | = CH |
| 2 | |
| OP | = HR |
Let us now consider the Pentangle as the symbol of the Great Pyramid Cheops.
The line MP = the base of Cheops.
The line CH = half base of Cheops.
The line HM = apothem of Cheops.