 | William Enfield - Astronomy - 1811 - 476 pages
...inscribed and circumscribed figure is the sum of the parallelograms K/, Lm, M«, Do, that is, (because of the equality of all their bases) the rectangle under one of their bases K£, and the sum of their altitudes Aa ; that is, the rectangle AB/a. But this rectangle, because its breadth... | |
 | John Mason Good - 1819 - 788 pages
...ratios of equality.— For the difference of the inscribed and circumscribed figures is the sum of Hi'- parallelograms K /, L m, M n, D o; that is, (from...equality of all their bases), the rectangle under onp of their ba-es K i, and the sum of their altitudes A n, that is, the rectangle AB / a. But this... | |
 | Isaac Newton - Curves, Plane - 1850 - 184 pages
...circumscribed figure is the sum of the a parallelograms K /, L m, M n, K D o, that is, (because of the equality of all their bases,) the rectangle under...sum of their altitudes A a; that is, the rectangle AB / «. But this rectangle, because its breadth AB is diminished indefinitely, becomes less than any... | |
 | Florian Cajori - Mathematics - 1919 - 294 pages
...difference of the inscribed and circumscribed figures is the sum of the parallelograms K/, L;«, Mn, Do, that is (from the equality of all their bases), the rectangle under one of their bases K£ and the sum of their altitudes Aa, that is, the rectangle AB/«. But this rectangle, because its breadth... | |
 | Florian Cajori - Mathematics - 1919 - 320 pages
...difference of the inscribed and circumscribed figures is the sum of the parallelograms K/, L#z, M«, Do, that is (from the equality of all their bases), the rectangle under one of their bases Kb and the sum of their altitudes Aa, that is, the rectangle AB/«. But this rectangle, because its... | |
 | Michael Spivak - Mathematics - 2006 - 682 pages
...difference of the inscribed and circumscribed figures is the sum of the parallelograms Kl, Lm, Mn, Do, that is (from the equality of all their bases), the rectangle under one of their bases Kb and the sum of their altitudes Aa, that is, the rectangle ABla. But this rectangle, because its... | |
| |
But this rectangle, because its breadth AB is supposed diminished in infinitum, becomes less than any given space. And therefore (by Lem. I) the... 