which to prove, take the Segment A into pieces as B C and D, there would be 9 of them, and all of one length, but severall forms, viz. B a foot square taken from the middle of the Segment, then will there be 4 pieces like wedges a foot square at the base, and ending in a line at A, having no thick∣ness, as the four corner pieces are perfect Pyra∣mids, containing one foot square at the base, the other ending in a point, as D, each of whose di∣mensions, according to the last Probleme, must be 5 cubicall feet, being the length is 15, and con∣sequently 4 of these will contain 20 feet, and as for C, 2 such pieces turned end for end, will be equall to the figure B, containing 15 feet; then one of them is 7 ½ feet, and 4 of those will make 30, the totall of the 9 pieces in content is 65 cubical feet, equall as they are all together in the figure A, which is evidently proved, as was re∣quired.
But this Segment (according to common practise) is measur'd in the middle) by taking the Arithmeticall mean, that is by adding the sides of the two squares together, and taking the half of that for the common square, as in this, 1 and 3 makes 4, the half 2, whose square 4 multiplied into the length, 15, the product will be 60 feet, for the content, according to custome, which is apparently erroneous, and 5 feet too little in this piece, as before was demonstrated; divers other errours (in measuring of solid bodies) are crept in for want of Art, and having got possession of igno∣rant people, they plead prescription and custome