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area CA=X CA2,
or (P. 15)
Scholium 1. Let CA-R, and area CAA: then, AR, making CA=1; we shall have
In common calculations, we take =3.1416.
But we have found the area of the circle whose radius is 1 to be 3.1415926 (P. 12): therefore, we have
Scholium 2. The problem of the quadrature of the circle, as it is called, consists in finding a square equivalent in surface to a circle, the radius of which is known. Now it has just been proved, that a circle is equivalent to the rectangle contained by its circumference and half its radius (P. 15); and this rectangle may be changed into an equivalent square, by finding a mean proportional between its length and its breadth (B. IV., PROB. 3). To square the circle, therefore, is to find the circumference when the radius is given; and for effecting this, it is enough to know the ratio of the diameter to the circumference.
Hitherto the ratio in question has never been determined except approximatively; but the approximation has been carried so far, that a knowledge of the exact ratio would afford no real advantage whatever beyond that of the approximate ratio. Accordingly, this problem, which engaged geometers so deeply, when their methods of approximation were less perfect, is now degraded to the rank of those idle questions, with which no one possessing the slightest tincture of geometrical science, will occupy any portion of his time.
Archimedes showed that the ratio of the diameter to the circumference is included between 310 and 31; hence, 31 or 22 affords at once a pretty accurate approximation to the number above designated by ; and the simplicity of this first approximation has brought it into very general
use. Metius, for the same quantity, found the much more accurate value 5. At last, the value of , developed to a certain order of decimals, was found by other calculators to be 3.1415926535897932 &c.: and some have had patience enough to continue these decimals to the hundred and twenty-seventh, or even to the hundred and fortieth place. Such an approximation is practically equivalent to perfect accuracy: the root of an imperfect power is in no case more accurately known.
PROPOSITION XVII. THEOREM.
If the circumferences of two circles intersect each other, the arc of the common chord in the less circle will be longer than the corresponding arc of the greater.*
Let A and B be the centres of two circles, AC, BC, their radii, C and D the points in which their circumferences intersect and CD their common chord: then will the arc DEC described with the radius BC, be longer than the arc DFC described with the greater radius AC.
Join the centres A and B,
Bisect the arcs CE, ED, and also the arcs CF, FD, and draw chords subtending the new arcs there will thus be inscribed in the two segments DEC, DFC, portions of two polygons, having the same number of sides in each.
Now, since the point F is within the triangle DEC,
*The arc considered in this demonstration is the one which is less than a semicircle.
EC plus ED is greater than CF plus FD (B. I., P. 8): hence, the half, CE is greater than the half, CF. If now, with C as a centre, and CE as a radius, we describe an arc EH, the chord CE being greater than CF, the arc CFH will be greater than the arc CF (B. III., P. 5). If we suppose the arc CKE to move with the chord CE then, when the chord CE becomes the chord CH, the arc CKE will pass through the points C and H, and will have with CFH, the common chord CH.
If, now, we bisect the arc which is equal to CKE, and also the arc CFH, we know from what has already been shown, that the chord of half the outer arc will be greater than the chord of half the inner arc CFH, much more will it be greater than the chord of CL, which is half the arc CF; that is, the chord of the arc CK, one-half of CE, will always be greater than the chord of the arc CL, onehalf of CF. Hence, the perimeter of that portion of the polygon inscribed in the segment CED, will be greater than the perimeter of the corresponding polygon inscribed in the segment CFD. If, then, we continue the operations indefinitely, the limit of the outer perimeter will be the arc CED, and of the inner, the arc CFD: hence, the arc CED is greater than the arc CFD.
Cor. If equal chords be taken in unequal circles, the arc of the chord in the greatest circle will be the shortest; for, the circles may always be placed as in the figure.
PLANES AND POLYEDRAL ANGLES.
1. A straight line is perpendicular to a plane, when it is perpendicular to every straight line of the plane which passes through its foot: conversely, the plane is perpendicular to the line. The point at which the perpendicular meets the plane, is called the foot of the perpendicular.
2. A line is parallel to a plane, when it cannot meet that plane, to what distance soever both be produced. Conversely, the plane is parallel to the line.
3. Two planes are parallel to each other, when they cannot meet, to what distance soever both be produced.
4. The indefinite space included between two planes which intersect each other, is called a diedral angle: the planes are called the faces of the angle, and their line of common intersection, the edge of the angle.
A diedral angle is measured by the angle contained between two lines, one drawn in each face, and both perpendicular to the common intersection at the same point. This angle may be acute, obtuse, or a right angle. If it is a right angle, the two faces are perpendicular to each other.
5. A POLYEDRAL angle is the indefinite space included by several planes meeting at a common point. Each plane is called a face: the line in which any two faces intersect, is called an edge: and the common point of meeting of all the planes, is called the vertex of the polyedral angle.
Thus, the polyedral angle whose vertex is S is bounded by the four faces, ASB, BSC, CSD, DSA. Three planes, at least, are necessary to form a polyedral angle.
A polyedral angle bounded by three planes, is called a triedral angle.
1. Let it be granted, that from a given point of a plane, a line may be drawn perpendicular to that plane.
PROPOSITION II. THEOREM.
2. Let it be granted, that from a given point without a plane, a perpendicular may be let fall on the plane.
PROPOSITION I. THEOREM.
A straight line cannot be partly in a plane, and partly out of it.
For, by the definition of a plane (B. I., D. 9), when a straight line has two points common with it, the line lies wholly in the plane.
Scholium. To discover whether a surface is plane, apply a straight line in different ways to that surface, and ascertain if it coincides with the surface throughout its whole
Two straight lines which intersect each other, lie in the same plane, and determine its position.
Let AB, AC, be two straight lines which intersect each other in A; a plane may be conceived in which the straight line AB is found; if this plane be turned round AB, until it pass through the point C, then the line AC, which has two of its points A and C, in this plane, lies wholly in it; hence, the position of the plane is determined by the single condition of containing the two straight lines AB, AC.