We have here an example of a quantity capable of a minimum, or least value; and, in order to find this minimum, it is obvious that we have only to solve the quadratic and put the part under the radical equal to zero. Thus gives m2 - 2pa = 0, m2 = 2pa, for the minimum of m2. If we inquire what is the greatest value of m2 in this problem, we shall find it infinite, or that m2 has no maximum. [See the value of x and fig. 373.] If now we subtract from a by insensible degrees, ma Fig. 373. or, as it is commonly expressed, diminish a according to the Law OF CONTINUITY, a will at length become = 0, Fig. 374. where the first value only is applicable. Continuing the same motion, a will evidently become minus, and the point P will at the same time take up its position on the left of the line AC, or without the angle, and we shall have 叉 AX Fig. 375. where both values of x are applicable, only that the second, being minus, requires the area m2 to be laid off on the left. If, in like manner, we make p pass through the value 0 it will change its sign, and P will take up a position be low the line AB, where both values are applicable, as indicated in the figure. If, at the same time, we make a and p both minus, there will result where we have the same values as in the original solution, only changing into, as we evidently ought to do, since x is measured from A in an opposite direction. This is obvious also from the comparison of the figures 37, and 37, where all the parts of the one correspond to all the parts of the other. The figure 37, will have the same double construction that 37 has in 372. What would be the result of making p = 0? what, when p = 0, a=0? The problem we have just been discussing is admirably adapted to show the correlation of algebraical signs and geometrical figures. This problem will also enable us to divide any polygon into any required parts. 28°. The parallel sides of a trapezoid are a, b, and the altitude, h, it is required to cut off an area m2 adjacent to b by a line parallel thereto. What will be its breadth? = 29°. Given the sides of a triangle, ABC, viz., AB = 10 chains, AC 8 chains 43 links, BC = 4'70; also the position of a point, P, viz., distant from AB by 1'80, and from AC by one chain. Required to draw a line through the point P, that shall divide the triangle into two equal parts. The student will solve and verify. 30°. What is the greatest rectangle with a given perimeter ? may We denote the sides by a +x and a meter be = 4a, and we shall find m2 denoting the area; whence it is obvious that ma cannot exceed a2; and therefore that x = 0, when the area m2 is a maximum, or that among rectangles of the same perimeter, the square is the maximum. 31°. What is the maximum triangle that can be constructed on a given base, and of a given perimeter ? Ans. The triangle must be isosceles. 32°. Given the area and diagonal of a rectangle, to determine its sides. 33°. Let ABCD be a quadrilateral field, of which the sides AB, BC, CD, DA, are, respectively, 16, 34, 30, 29 rods in length; also, let the diagonal BD = 37 rods. It is required to divide the field into two equal parts, by a line cutting the opposite sides, AB, CD, so that the ratio of the segments of the one shall be equal to that of the corresponding segments of the other. Ans. AX= ? DY=? [Produce AB, CD, until they meet, and consult (140), (141), (132), (159).] BOOK THIRD. PLANE GEOMETRY DEPENDING ON THE CIRCLE, SECTION FIRST. The Circle, Definition 1. The circle is a plane figure described by the revolution of a straight line of invariable length about one of its extremities as a fixed point. Def. 2. The describing line is called the Radius [rod] of the circle, the fixed point the Centre, and the curve line that bounds it its Circumference. Cor. All radii, or lines drawn from the centre to the cir- (162) cumference, are equal to each other. PROPOSITION I. Angles at the centre of the same or equal circles, are to (163) each other as their subtending arcs; and the corresponding sectors have a like ratio. In the first place, let the angles be commensurable. For example, suppose the angle AOB to contain the angle BOC twice; then it is manifest that in applying the angle BOC twice to the angle AOB, the point C will fall on C', the middle point of the arc AB, since (162) OC' = OC. Therefore the arc AB contains the arc BC twice; and, in the same way, the sector AOB is double the sector BOC. Fig. 38. .. ZAOB: BOC :: arc AB : arc BC :: sec. AOB: sec. BOC, each being as 2 to 1. KL So in general, if a, b, be any commensurable angles, or such that when a is divided into any number, m, of equal parts, b shall also be exactly divisible into some number, n, of the same equal parts; then it will follow, by superposing one of these equal angles m times upon a and n times upon b, that the corresponding arcs A and B will be divided into m and n equal The same of the sectors K, L. arcs. Fig. 382. a : b :: A : B :: K: L, being as m to n. Finally, let us take the most general case, or that of incommensurability, where a and b having no common measure, are incapable of being divided into the same equal parts. Let b be increased by the angle x, so that b+x shall be commensurable with a, then will the corresponding arc B+X be commensurable with A; and, from what has just been proved, there results the proportion Fig. 383. a or or a : b :: A: B, B Q. E. D. and the same is equally true of the sectors. Cor. 1. In the same or equal circles, arcs may be taken (164) as the measures of their angles at the centre. Thus, if b be taken for the unit of angles and B for Scholium. As the right angle seems the most suitable for comparing angles, so its measure, the Quadrant, or quarter circumference, would appear to be the appropriate unit of arcs, and for this purpose the French have sometimes employed it, dividing the quadrant into a hundred equal parts, which they called degrees: but custom has established a different unit, theth part of the quad |