Here the content of the triangle by the [18 e xij], is found to be 62.44/125 for the base of the pyramis. The altitude is 9.15/19: Because by the [12 e xviij], the side is of treble power to the ray. But if from 144, the quadrate of 12 the side, you take the subtriple i. 48, the remainder 96, by the [9 e xij], shall be the square of the heighth. And the side of the quadrate shall be 9.15/19. Now the third part of 9.15/19 is 3.5/19. And the plaine of 62.44/125 and 3.5/19, shall be 203.1103/2375 for the solidity of the pyramis.
So in the example following, Let 36, the quadrate of 6 the ray, be taken out of 292.9/1156 the quadrate of the side 17.3/34 the side 16.3/34 of 256.9/1156 the remainder shall be the height, whose third part is 5.37/102; the plaine of which by the base 72.1/4 shall be 387.11/24 for the solidity of the pyramis given.
If the pyramis be unperfit, first measure the whole, and then that part which is wanting: Lastly from the whole
subtract that which was wanting, and the remaine shall be the solidity of the unperfect pyramis given: As here, let ao, the side of the whole be 16.5/12, eo the side of the particular be 8.1/16. Therefore the perpendicular of the whole ou, shall be 15.5/32: Whose third part is 5.5/96: Of which, and the base 93.3/11 the plaine shall be 471.134/1056 for the whole pyramis. But in the lesser pyramis, 9 the square of the ray 3, taken out of 65.1/256 the quadrate of the side 8.1/16 the remaine shall be 56.1/256; whose side is almost 7-1/2 for the heighth. The third part of which is 2-1/2. The base likewise is almost 22. The plaine of which two is 55, for the solidity of the lesser pyramis: And 471 - 55 is 416, for the imperfect pyramis.
After this manner you may measure an imperfect Prisma.
8. Homogeneall Prisma's of equall heighth are one to another as their bases are one to another, 29, 30, 31, 32 p xj.
The reason is, because they consist equally of like number