OF THE POLYGRAPHIC LINES, AND HOW BY THESE we can describe regular polygons; that is, figures of many equal sides and angles.

Operation XXVI.

urning the Instrument over we see the innermost Lines on its other face, called Polygraphic from their principal use, which is to describe on a given line figures having as many equal sides (and angles) as required. This is easily done by taking with a compass the length of the given line and fitting it to the points marked 6-6; next, without changing the Instrument, we take the distance between the points marked with the same number as the number of sides of the figure we wish to draw. For instance, to describe a [regular] figure of 7 sides we take the distance between points 7-7, which will be the radius of the circle containing the heptagon to be drawn. Place one leg of the compass first at one end and then at the other end of the given line and trace with it a short intersection of arcs; then taking this as center we shall describe with the same compass-opening a faint circle. Passing through the ends of the given line, this will contain that line exactly 7 times inside its circumference, whereby the heptagon is drawn.

DIVISION OF A CIRCLE INTO AS MANY PARTS as is desired.

Operation XXVII.

ith these lines the circumference can be divided into many parts, working by the converse of the preceding operation. Take the radius of the given circle and fit it to the number designating the parts into which the circle is to be divided; then take always the distance between points 6-6 [on the Polygraphic Lines so separated] and this will divide the circumference into the desired [number of] parts.

EXPLANATION OF THE TETRAGONIC LINES, AND HOW BY THEIR means one may square the Circle, or any other regular figure, and may also transform all these, one into another.

Operation XXVIII.

hese Tetragonic Lines are so called from their principal use, which is to square all regular areas and the circle as well, which is done by an easy operation. For when we wish to construct a square equal to a given circle, all we have to do is to take its radius with a compass, and opening the Instrument fit this to the two points along the Tetragonic Lines marked by two tiny circles; then, not moving the Instrument, if you take on the compass the distance between the points marked 4-4 on those Lines, you will have the side of a square equal [in area] to the given circle. Likewise if you want the side of the pentagon, or the hexagon, equal to the same circle, you will take the distance between points 5-5, or 6-6; for they are respectively the sides of the pentagon and the hexagon equal to that same circle.
      Moreover if we want the converse (given a square or other regular polygon, to find a circle equal to it), we take one side of the given polygons and fit it to the points of the Tetragonic Lines corresponding to the number of sides of the given figure; then, without moving the Instrument, take the distance between the little circle marks, and by making this the radius you will describe the circle equal to the given polygon. In conclusion, you can in this way find the side of any regular figure that is equal to any other. For example, we wish to construct an octagon equal to a given pentagon [both regular]. The Instrument is set so that the side of the given pentagon fits at points 5-5, and without changing the Instrument the distance between points 8-8 will give the side of the octagon sought.

GIVEN DIFFERENT REGULAR FIGURES, HOWEVER dissimilar to one another, how to construct a single one equal to all of them together.

Operation XXIX.

olution of the present problem depends on the preceding one and on Operation 10, explained before. Let there be given to us a circle, a triangle, a pentagon, and a hexagon; we are asked to find one square equal to all the said figures [combined]. First, by the preceding operation we find separately four squares, equal to the given four figures; then, by Operation 10, we find a single square equal to those four, which will doubtless be equal to all four given figures.

HOW ONE MAY MAKE ANY DESIRED REGULAR figure that equals any given irregular (but rectilineal) figure.

Operation XXX.

he present operation is no less useful than interesting, showing us not just the way to square all irregular [rectilineal] figures, but how to reduce them either to a circle or to any other regular figure you like. Since every rectilineal figure is resolvable into triangles," once we can make a square equal to any triangle we shall make separately individual squares, equal to each of the triangles into which the given rectilineal figure is resolved, and then by Operation 10 we reduce all these squares into a single square, and obviously the square will have been found that is equal to the given rectilineal figure. By means of the Tetragonic Lines we can at will convert that square in turn into a circle, a pentagon, or any other regular rectilineal figure. Thus the solution of the present question is reduced to our having to find a square equal to any given triangle, which you will have very easily from the ensuing lemma.

LEMMA FOR [CARRYING OUT] THE THINGS said above.

Operation XXXI.

et it be proposed to make a square equal to the given triangle ABC. Draw to one side two lines at right angles, DE and FG; then take a proportional compass of which one end opens to double the other and hold one of the longer legs at corner A; open the other leg until, when rotated, it just grazes the opposite line BC. Now invert the [proportional] compass and mark with the shorter legs the distance FH, which will be half the perpendicular that falls from angle A on the opposite side BC. This done, take line BC with the longer legs and carry this to FI; holding one leg at point I, move the other out to point H. Again inverting the compass, without further widening or narrowing it, mark with the short legs the distance IK, and holding one of their points at K let the other cut the vertical FG at point L. Thus we now have line LF, which is the side of the square that is equal to triangle ABC. Notice that although we have set forth this operation as performed linearly without the Instrument, that is not because this could not also be easily found on the Instrument. For if we wished to reduce some triangle to a square, as for instance triangle ABC, then taking the vertical that falls from angle A upon the opposite side BC we would see how many graduations this contains along the Arithmetic scale, as say 45, and fit that distance crosswise to points 45-45 on the Geometric Lines; then taking half the line BC we would see likewise how many graduations along the Arithmetic Lines it contains; finding it to contain, say, 37, we then take crosswise on the Geometric Lines the distance between points 37-37, which would give us line LF, the square on which will equal triangle ABC.

OF THE ADDED LINES, FOR THE QUADRATURE of segments of a circle and of figures containing parts of circumferences, or straight and curved lines together.

Operation XXXII.

here remain finally the Added Lines, so called because they add to the Tetragonic Lines what with those was left to be desired-that is, a way of squaring segments of circles and the more specifically explained below. These Added Lines are other figures mentioned in the heading above, which will be marked with two series of numbers, of which the outer series begins at this mark D followed by the numbers 1, 2, 3, 4, and so on out to 18. The inner series begins from this mark going on then to 1, 2, 3, 4, and so on, also out to 18. By means of these Lines we can (for the first time) square any given segment of a circle not more than a semicircle. In order that this may be better understood, their use will be explained by example.
      We wish, for instance, to find the square equal to the segment of a circle, ABC. Bisect its chord AC at point D, and take the distance DC with a compass. Opening the Instrument, fit this across the points marked DD, and leaving the Instrument at this setting, take next the altitude of the segment (that is, line DB) and see at which points of the outer scale this fits crosswise. Let this be, say, at points 2-2. This done, we must next take with the compass the the distance between points 2-2 of the inner scale and form our square on a line of that length, which [square] will be equal to the segment ABC. Now if we have an area contained by two segments of circles with a common chord, as in the next diagram ABCD, we could easily reduce this to a square by drawing the chord AC which divides this figure into two circular segments, whereupon by the rule already given there are found two squares equal to the two separate segments which by recourse to Operation 10 are reduced to a single square, completing the task.
      By a not dissimilar operation we can also square any sector of a circle; for drawing the chord of its arc cuts that into one circular segment and one triangle, two parts which can easily be reduced to two squares by the things already taught, and those two then [reduced] to a single square.
      Finally it remains for us to show how the same Lines can serve us to square a segment larger than a semicircle, or a trapezium contained by two straight lines and two arcs (like that in the next figure, ABCD), or the lune similar to X; all of which operations have the same resolution. As to the segment larger than a semicircle, if we square the excluded smaller segment in the foregoing manner, and subtract such a square from the square equal to the entire circle, the square equal to the difference will obviously be equal to the larger segment of the circle. Likewise, having found the square equal to the whole segment BAFDC and subtracted from this the square equal to the segment AFD, the remainder will equal the trapezium [ABCD]. And proceeding similarly for the lune X, draw the common chord of the two segments of circles and take separately the squares equal to those two segments; their difference will be the square equal to the lime. As to finding the difference between two given squares and reducing this to another square, that was explained by use of the Geometric Lines in Operation 11.