SCHOLIUM. It is evident that the method employed in this proposition, for finding the limits of the ratio of the circumference of the diameter, may be carried to a greater degree of exactness, by finding the perimeter of an inscribed and of a circumscribed polygon of a greater number of sides than 96. The manner in which the perimeters of such polygons approach nearer to one another, as the number of their sides increases, may be seen from the following Table, which is constructed on the principles explained in the fore going Proposition, and in which the radius is supposed =1. The part that is wanting in the numbers of the second column, to make up the entire perimeter of any of the inscribed polygons, is less than unit in the sixth decimal place; and in like manner, the part by which the numbers in the last column exceed the perimeter of any of the circumscribed polygons is less than a unit in the sixth decimal place, that is, than 1 of the radius. Also, as the numbers in the second column are 1000000 less than the perimeters of the inscribed polygons, they are each of them less than the circumference of the circle; and for the same reason, each of those in the third column is greater than the circumference. But when 1 1 1000000 the arc of of the circumference is bisected ten times, the number of sides 6 in the polygon is 6144, and the numbers in the Table differ from one another only by part of the radius, and therefore the perimeters of the polygons differ by less than that quantity; and consequently the circumference of the circle, which is greater than the least, and less than the greatest of these numbers, is determined within less than the millionth part of the radius. Hence also, if R be the radius of any circle, the circumference is greater than Rx 6.283185, or than 2R × 3.141592, but less than 2R ×3.141593; and these numbers differ from one another only by a millionth part of the radius. So also R2+3.141592 is less, and R2×3.141593 greater than the area of the circle; and these numbers differ from one another only by a millionth part of the square of the radius. In this way, also, the circumference and the area of the circle may be found still nearer to the truth; but neither by this, nor by any other method yet known to geometers, can they be exactly determined, though the errors of both may be reduced to a less quantity than any that can be assigned. A STRAIGHT line is perpendicular or at right angles to a plane, when it makes right angles with every straight line which it meets in that plane. 2. A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common section of the two planes, are perpendicular to the other plane. 3. The inclination of a straight line to a plane is the acute angle contained by that straight line, and another drawn from the point in which the first line meets the plane, to the point in which a perpendicular to the plane, drawn from any point of the first line, meets the same plane. 4. The angle made by two planes which cut one another, is the angle contained by two straight lines drawn from any, the same point in the line of their common section, at right angles to that line, the one, in the one plane, and the other, in the other. Of the two adjacent angles made by two lines drawn in this manner, that which is acute is also called the inclination of the planes to one another. 5. Two planes are said to have the same, or a like inclination to one another, which two other planes have, when the angles of inclination above defined are equal to one another. 6. A straight line is said to be parallel to a plane, when it does not meet the plane, though produced ever so far. 7. Planes are said to be parallel to one another, which do not meet, though produced ever so far. 8. A solid angle is an angle made by the meeting of more than two plane angles, which are not in the same plane in one point. PRCP. I. THEOR. One part of a straight line cannot be in a plane and another part above it. If it be possible let AB, part of the straight line ABC, be in the plane, and the part BC above it: and since the straight line AB is in the plane, it can be produced in that plane (2. Post. 1.); let it be produced to D: Then ABC and ABD are two straight lines, and they have the common segment AB, which is impossible (Cor. def. 3. 1.). Therefore ABC is not a straight line. PROP. II. THEOR B D Any three straight lines which meet one another, not in the same point, are în one plane. Let the three straight lines AB, CD, CB meet one another in the points B, C and E; AB, CD, CB are in one plane. Let any plane pass through the straight line EB, and let the plane be turned about EB, produced, if necessary, until it pass through the point C Then, because the points E, C are in this plane, the straight line EC is in it (def. 5. 1.): for the same reason, the straight line BC is in the same; and, by the hypothesis, EB is in it; therefore the three straight lines EC, CB, BE are in one plane: but the whole of the lines DC, AB, and BC produced, are in the same plane with the parts of them EC, EB, BC (1. 2. Sup.) Therefore AB, CD, CB, are all in one plane. E B COR. It is manifest, that any two straight lines which cut one another are in one plane: Also, that any three points whatever are in one plane PROP. III. THEOR. If two planes cut one another, their common section is a straight line. Let two planes AB, BC cut one another, and let B and D be two points in the line of their common section. From B to D draw the straight line BD; and because the points B and are in the plane AB, the straight line BD is in that plane (def. 5. 1.): for the same reason it is in the plane CB; the straight line BD is therefore common to the planes AB and BC, or it is the common section of these planes. B D A. PROP. IV. THEOR. If a straight line stand at right angles to each of two straight lines in the point of their intersection, it will also be at right angles to the plane in which these lines are. Let the straight line AB stand at right angles to each of the straight lines EF, CD in A, the point of their intersection: AB is also at right angles to the plane passing through EF, CD. And B H Through A draw any line AG in the plane in which are EF and CD; let G be any point in that line; draw GH parallel to AD; and make HF=HA, join FG; and when produced let it meet CA in D; join BD, BG, BF. Because GH is parallel to AD, and FH=HA: therefore FG GD, so that the line DF is bisected in G. because BAD is a right angle, BD2=AB2 +AD2 (47. 1.); and for the same reason, BF2AB2+AF2, therefore BD2+BF2= 2AB2 + AD2 + AF2; and because DF is bisected in G (A. 2.), AD2+AF2=2AG2+ 2GF2, therefore BD2+BF2=2AB2+2AG2 +2GF2. But BD2+ BF2 (A. 2.) 2BG2+2GF2, therefore 2BG2+ 2GF2=2AB2+2AG2+2GF2; and taking 2GF2 from both, 2BG2=2AB2 +2AG2, or BG2=AB2+AG2; whence BAG (48. 1.) is a right angle. Now AG is any straight line drawn in the plane of the lines AD, AF; and when a straight line is at right angles to any straight line which it meets with in a plane, it is at right angles to the plane itself (def. 1. 2. Sup.). AB is therefore at right angles to the plane of the lines AF, AD. = E |