476. If a fourth proportional to lines representing p, q, r, be taken, it will represent ; and if p=1, it will represent gr; if gr also q and r be equal, it will represent q2. 477. If a mean proportional between lines representing a and b be taken, it will represent ab, which, when a=1, becomes √b. Hence it appears that any possible algebraical quantities may be represented by lines; and conversely, lines may be expressed algebraically; and if the relations of the algebraical quantities be known, the relations of the lines are known. 478. The area included within a rectangular parallelogram may be measured by the product of the two numbers which measure two adjacent sides. Let the sides AB, AC of the rectangular parallelogram AD be measured by the lineal quantities a, b, respectively; then axb will express the number of superficial units in the area, that is, the number of squares it contains, each of which is described upon a lineal unit. For instance, if the lineal unit be a foot, of which AB contains a, and AC contains b, the parallelogram AD contains axb square feet. For, 1st, if AB, AC be divided into lineal units, and straight lines be drawn through the points of division parallel to the sides, the whole figure is made up of squares, which are equal to each other, and to the square upon the lineal unit; and the number of them is evidently a taken b times, or axb. 2ndly. If AB and AC do not contain the lineal unit an exact number of times, that is, if a and b be fractional, let a=a+—, and b¬ß+1. 1 1 m n Then take another lineal unit which is th part of the former; and mn by what has been shewn the square described upon the larger unit contains mnxmn of that described upon the smaller. Again, the sides AB, AC respectively contain mna+n, mnß+m lineal units of the smaller kind, and therefore, by the first case, the whole figure contains (mna+ n)x (mnß+m) square units of the smaller kind; that is, the area 3dly. If the sides AB, AC be incommensurable with the lineal unit, a unit may be found which is commensurable with certain lines that approach as near as we please to AB, and AC, and therefore the product of such lines will represent the area of a rectangle differing from the rectangle AD by a quantity less than any that can be assigned, that is, we may, in this case also, without error express AD by AB× AC. (See Art. 260.) 479. COR. 1. Since, by Euclid, Book 1. Prop. 35, the area of an oblique-angled parallelogram is equal to that of the rectangular parallelogram upon the same base and between the same parallels, therefore area of any parallelogram=base-altitude. 480. COR. 2. Also, since by Euclid, Book 1. Prop. 41, the area of a triangle is half that of the parallelogram upon the same base and between the same parallels, therefore 481. COR. 3. Since any rectilineal figure may be divided into triangles, its area may be found by taking the sum of all the triangles. 482. The solid content or volume included within a rectangular parallelopiped may be measured by the product of the three numbers which measure its length, breadth, and height. Let the base of the parallelopiped be divided into its component squares, as in the preceding Proposition, and through each of the parallels suppose planes drawn at right angles to the base; and let the same thing. be done with one of the faces adjacent to the base. Then it is evident that the whole figure is divided into a certain number of equal cubes, each cube having for its face one of the squares described upon the lineal unit; (that is, if the lineal unit be a foot, each of these cubes will have its length, breadth, and height equal to a foot, and is called a cubic foot). Now the number of these cubes is manifestly equal to the number of squares in the base taken as many times as there are lineal units in the height; therefore content or volume=base-height, -length-breadth x height. (Art. 478.) COR. Any three of these quantities being given, the fourth may be found. Thus, if C be the content, I the length, b the breadth, and h the height, we have The following Examples will sufficiently illustrate the preceding Theory, for our present purpose: Ex. 1. If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. (EUCLID, Book 11. Prop. 7.) Let a and b represent the two parts into which the given line is divided; then a+b is the whole line, and (a+b)2+a2=the squares of the whole line, and of one of the parts; =a2+2ab+b2+a3, =2a2+2ab+b2, =2(a+b)a+b2, =twice the rectangle &c. together with the square of the other part. Q.E.D. Ex. 2. To find the radius of a circle inscribed in a given triangle. See Euclid's diagram, Book IV. Prop. 4; let r be the radius of the inscribed circle, and a, b, c the sides of the triangle respectively opposite to the angles A, B, C. Then (Art. 480) or 1 r(a+b+c) = axthe perpend' (p) upon a from the opposite angle ; To find p, let the segments into which a is divided by it be r and a-x; then (EUCLID, Book 1. Prop. 47) = 4a2 (a+b+c)(a+c—b)(a+b−c)(b+c−a) Ex. 3. To find the area of the square inscribed in a given circle; and also of the square circumscribed about a given circle. (1). Let r be the radius of the circle; then (see EUCLID, Book IV. Prop. 6) r2+r2=AD3, or 2r=inscribed square. (2). Again, + p2 +p2+g2 or 4r-circumscribed square, =twice the inscribed square. Ex. 4. To find the area of the equilateral and equiangular hexagon inscribed in a given circle. Let r be the radius of the given circle, then (EUCLID, Book Iv. Prop. 15) also r=the side of the inscribed hexagon; and the area of the hexagon the sum of the areas of the six equal equilateral triangles of which it is composed, Ex. 5. The depth of water in a cistern (whose form is a rectangular parallelopiped) is h feet, and the base contains a square feet. Find (1) the number of cubic feet of water; and (2) the depth of the same quantity of water in another cistern whose base contains b square inches. (1). The quantity of water-basexdepth=ah cubic feet.. (2). The depth for 2nd cistern- quantity of water (in cubic feet) =ah÷ base (in square feet) b 144ah 144 b = feet. It is not necessary here to multiply Examples, because the subject is now of sufficient importance to form a separate treatise, to which, in the regular course of reading, the student's attention will be hereafter directed. APPENDIX. EQUATIONS. In the infinite variety of Equations which ingenious persons may put together, it is not to be supposed that any general Rules can be laid down. for every operation necessary to their solution. Indeed the best method of solution is frequently that which lies under the least obligation to general Rules; as may be seen at once by comparing Exs. 1, 2, and 3 below with Art. 194. Numberless are the artifices by which algebraic calculation may be abridged, and their successful application can be learnt by practice only. There are, however, peculiar artifices of more frequent occurrence than others, which shall be exhibited in the following Examples. 4x-17 38-22x Ex. 1. =x ; find r. or x=x--+2, 10x-13 8x-30 5x-4 4x-17 Ex. 2. x-4 + ; find x. 2x-3 2x-7 x-1 4(x-4)-15(2x−3)+2 _ 4(2x-7)−2 ̧ 5(x-1)+1 = + x-1 |