2. With a little attention to the course pursued in the preceding questions, it will be evident that we have always taken for the unknown quantity a line, which being once known, serves, by observing the conditions of the question, to determine all the others. This is the course to be pursued in all cases, but there is a choice with regard to the line to be used; there are often several, each of which has the property of determining all the others, if once known. Among these some would lead to more simple equations than others. The following rule is given to aid in such cases. 3. If among the lines or quantities, which would, when taken each for the unknown quantity, serve to determine all the other quantities, there are two which would in the same way answer this purpose, and it would be foreseen that such would lead to the same equation, (the signs + and -excepted); then we ought to employ neither of these, but take for the unknown quantity one which depends equally on both; that is, their half sum, or their half difference, or a mean proportional between them, or &c., and we shall always arrive at an equation more simple than by employing either the one or the other. 4. The question we have resolved, (Prob. VI,) may be used to illustrate what is here said. In this question there is no reason for taking AD rather than AE, for the unknown quantity; by taking AD for the unknown quantity x, we have x+c for AE; and, by taking AE for the unknown quantity x, we should have x c for AD; and, as to the rest, the mode of proceeding is the same for each case; so that the equations differ only in the signs. If, therefore, instead of taking either for the unknown quantity, we take their half sum, and desig→ nate it by x, since their half difference DE=c is given, we shall have AE=x+1; c, and AD=x − 1 cx, whence, according to the proposition adopted in the first solution, a more simple equation than the former, and which gives x= √1 c3+ab; and, since AE=x+ c, we have immediately and as before found. AE=c+√ c2+ab, PROBLEM VIII. an Let it be required to draw a right line BFE from one of the gles B of a given square BC, so that the part FE intercepted by DE and DC, shall be of a given length. Draw EG perpendicular A to BE to meet BC produced in G, and from the angle E draw EH perpendicular to BG. Let BC or DC=a,FE=b, BF=y, and CG=x. Since the triangle EHG is similar to the triangle BCF B and the side EH=the side BC, hence the hypothenuse EG= the hypothenuse BF. But or BE2+EG2=BG ̧, and because the triangle BCF and BEG are similar, or hence, BF: BC:: BG: BE yaa+xy+b y2+by=a2+ax multiplying this equation by 2, we have 2y2+2by=2a+2ax. Subtracting the last from the former equation, we have or hence, b2=— a2+x2, x=√b2+a2 having the value of x, y may be found in the equation y2+by=a2+ax completing the square y2+by+b=a2+ax+1b hence Let DE=%, then and hence, Scholium. This problem is susceptible of several modes of solution, but perhaps none more simple than the one here given, for most of the modes of which it is susceptible, involve powers and equations higher than quadratics. EXAMPLES FOR PRACTICE. Ex. 1. To find the side of a square, inscribed in a given semicircle, whose diameter is d. Ans. d√5 1 Ex. 2. To find the side of an equilateral triangle inscribed in a circle whose diameter is d; and that of another circumscribed about the same circle. Ans. d√3, and d√3 Ex. 3. To find the sides of a rectangle, the perimeter of which shall be equal to that of a square, whose side is a, and its area half that of a square. Ans. a+a2 and a — a√2 Ex. 4. Having given the perimeter (12) of a rhombus, and the sum (8) of its two diagonals, to find the diagonals. Ans. √2+ √5 4+ √2 and 4--√2 Ex, 5. Required the area of a right angled triangle, whose hypothenuse is x and the base and perpendicular ** and ** Ans. 1.029085 Ex. 6. Having given the two contiguous sides (a, b) of a parallelogram, and one of its diagonals (d), to find the other diagonal. Ans. (2a+2b3 — d') Ex. 7. Given the base (194) of a plane triangle, the line that bisects the vertical angle (66), and the diameter (200) of the circumscribing circle, to find the other two sides. Ans. 81.36587 and 157, 43865 Ex. 8. The lengths of two lines that bisect the acute angles of a right angled plane triangle, being 40 and 50 respectively, it is required to determine the three sides of the triangle. Ans. 35.80737, 47.40728, and 59.41143 Ex. 9. Given the hypothenuse (10) of a right angled triangle, and the difference of two lines drawn from its extremities to the centre of the inscribed circle (2), to determine the base and perpendicular. Ans. 8.08004 and 5.87447 Ex. 10. Having given the lengths (a, b) of two chords, cutting each other at right angles, in a circle, and the distance (c) of their point of intersection from the centre, to determine the diameter of a circle. Ans. †√ { †(a2+b2)+2c2 } Ex. 11. Two trees, standing on a horizontal plane, are 120 feet asunder; the height of the highest of which is 100 feet, and that of the shortest 80; where in the plane must a person place himself, so that his distance from the top of each tree, and the distance of the tops themselves, shall be all equal to each other? Ans. 2021 feet from the bottom of the shortest and 403 feet from the bottom of the other Ex. 12. Having given the sides of a trapezium, inscribed in a circle, equal to 6, 4, 5, and 3, respectively, to determine the diameter of the circle. Ans. (130x133) or 6.574572 Ex. 13. Supposing the town a to be 30 miles from B, в 25 miles from c, and c 20 miles from A; where must a house be erected that it shall be at an equal distance from each of them? Ans. 15.118578 miles from each, viz: in the centre of a circle whose circumference passes through each of the three towns. Ex. 14. In a plane triangle, having given the perpendicular (p), and the radii (r R) of its inscribed and circumscribing circles, to determine the triangle. DETERMINATION OF ALGEBRAICAL EXPRESSIONS FOR SURFACES AND SOLIDS. 7. We have seen in the Elements of Geometry, that surfaces depend upon the product of two dimensions, and solids upon the product of three dimensions; so that, if the several dimensions of one or two solids, or two surfaces, which we would compare, have to the several dimensions of the other, each the same ratio, the two surfaces will be to each other as the squares, and the two solids as the cubes, of the homologous dimensions; and more generally still, if any two quantities of the same nature are expressed each by the same number of factors, and if the several factors of the one have to the several factors of the other, each the same ratio, the two quantities will be to each other as their homologous factors, raised to a power whose exponent is equal to the number of factors. If, for example, the two quantities were a b c d, a' b' c' d', and we had What is here said is true not only of simple quantities; the same may be shown with respect to compound quantities. Let the quantities whose dimensions are proportional be a" b a' c d a b + c d : a' b' + c' d' : : a b + c d : + a a2 It follows, from what is here proved, that the surfaces of similar figures are as the squares of their homologous dimensions, and that the solidities of similar solids are as the cubes of their homologous dimensions; for, whatever these figures and these solids may be, the former may always be considered as composed of similar triangles, having their altitudes and bases proportional, (Prop. XXIII. B. IV. El. Geom.,) and the latter as composed of similar pyramids, having their three dimensions also proportional. (Prop. XXXII, Cor. 5, B. II, El. Sol. Geom.) It will hence be perceived, that quantities may be readily compared, when they are expressed algebraically; and this may be done, whether the quantities be of the same or of a different species, as a cone and a sphere, a prism and a cylinder, provided only that they are of the same nature, that is, both solids, or both surfaces. Let it be required to investigate the properties of a pyramid and and also of a frustum of a pyramid. Leth the altitude, s= the greater base and s' the smaller base, and h' the altitude of the vertical pyramid taken from the top of the frustum, = then we shall have √s': √s::h': h + h' or the altitude of the whole pyramid, and consequently, and (h + h') √s' = h' √s = h √s' + h' √s' h'√s-h' √s' = h √s' |