x = √ac b 8 times the linear unit. 20 9 √36 6 times the linear unit. We can then draw the line corresponding to the particular value of x. This is the most practical method. Again, we may obtain the required line from the algebraical result, by means of geometrical theorems; this method is called the Construction of Quantities'; it is often elegant, and is, moreover, useful to those who wish to obtain a complete knowledge of Algebraical Geometry. THE CONSTRUCTION OF QUANTITIES. 9. Let x = a + b. A D B In the straight line A X, let A be the point from whence the value of x is to be measured; take A B = a, and BC≈ b, then A CAB+ BC= a + b is the value of x. Χ b, in B A take B Db, then AD=AB - BD = a-b. Let x- then a :: bc, с and x is a fourth proportional to the three given quantities c, b, and a; hence the line whose length is expressed by x, is a fourth proportional to three lines, whose respective lengths are c, b, and a. From A draw two lines AC D, ABE, forming any angle at A; take B AB = c, BE = a, and AC=b, join B ̊C, and draw DE parallel to BC; then, AB:AC::BE: CD, or cb:: a: CD .. CD is the required value of r. The same property of right-angled triangles may be advantageously em a2 ployed in the construction of the equation x = ; for, take AC = b, and draw C E perpendicular to A C and equal to a, join A E and draw E B per Again, x is a line, the square upon which is equal to the sum of the rectangles ab, cd. This sum may be reduced to a single rectangle, and the rectangle converted into a square, the base of which is the required value of x.-Euclid, i. 45, and ii. 14; or Geometry, i. 57, 58. Let x= √a2 + b2; take a straight line A Ba, from B draw BC (= b) perpendicular to AB; AC is the value of x. Let x = a + b2 + c2, from C draw CD(c) perpendicular to AC, A D is the required value of x. Let x = √a2-b2 = √(a+b)(a−b); B (in the b2. a is a mean proportional between a + b and ab; or by taking last figure but one) A B = a, and AE= b, we have B E = √a2 √{a2 + b2 — c2 —d2}, find y2 = a2 + b2 and 2o = c2 + d2 and then x. Let x = 11. Of course the preceding methods will equally apply, when instead of the letters we have the original numbers, the linear unit being understood as usual. Thus x= √12 = √3.4 is a mean proportional between 3 and 4; hence (see last figure but one) take A C equal four times the unit, and C B equal three times the unit, CE is the value of x; or since √12— √164 √√4o 22, by constructing a right-angled triangle of which the hypothenuse is four times the linear unit, and one side twice that unit: the remaining side = √12. = Similarly x = √7=√4+4 − 1 = √2a + 2a − 1, which is of the form a + b2 - c2. Let x 52+ 1. In the last figure let A B, B C, and C D = Let x= then x is the hypothenuse of a right-angled triangle, each of whose sides is half the unit. this may be constructed as the last. 3 = and so on for all numbers, since any finite number can be decomposed into a series of numbers representing the squares upon lines. If the letter a be prefixed to any of the above quantities, it must be introduced under the root. 12. In constructing compound quantities, it is best to unite the several parts of the construction in one figure. This construction fails when b is greater than a, for then the circle never cuts the line A X; this is inferred also from the impossibility of the roots. 13. Since theorems in geometry relate either to lines, areas, or solids, the corresponding equations must in each case be homogeneous, and will remain so through all the algebraic operations. If, however, one of the lines in a figure be taken as the linear unit and be therefore represented by unity, we shall find resulting expressions, such as x = x = √ a, " b a α x = √√a2 + b,&c., in which, prior to construction, the numerical unit must be expressed; thus these quantities must be written x 1, √a × 1, √a3 +b × 1, and then constructed as above. b CHAPTER II. DETERMINATE PROBLEMS. 14. GEOMETRICAL Problems may be divided into two classes, Determinate and Indeterminate, according as they admit of a finite or an infinite number of solutions. If A B be the diameter of the semicircle A E B, and it be required to find a point C in A B such, that draw ing CE perpendicular to A B to meet the circumference in E, CE shall be equal to half the radius of the circle, this is a determinate problem, because there are only two such points in AB, E C B each at an equal distance from the centre. Again, if it be required to find a point E out of the line A B such, that joining EA, E B, the included angle A E B shall be a right angle, this is an indeterminate problem, for there are an infinite number of such points, all lying in the circumference A E B. The determinate class is by no means so important as the indeterminate, but the investigation of a few of the former will lead us to the easier comprehension of the latter; and therefore we proceed to the discussion of determinate problems. 15. In the consideration of a problem, the following rules are useful. 1. Draw a figure representing the conditions of the question. 2. Draw other lines, if necessary, generally parallel or perpendicular to those of the figure. 3. Call the known lines by the letters a, b, c, &c., and some of the unknown lines by the letters x, y, z, &c. 4. Consider all the lines in the figure as equally known, and from the geometrical properties of figures deduce one, two, or more equations, each containing unknown and given quantities. 5. From these equations find the value of the unknown quantities. 6. Construct these values, and endeavour to unite the construction to the original figure. 16. To describe a square in a given triangle AB C. Let D E F G be the required square CHK the altitude of the triangle. The question is resolved into finding the point H, because then the position of DE, and therefore of the square, is determined. Let CK = a, A Bb, CH = x; then by the question, DE HK, and DE AB:: CH: CK, or DE b :: Ꮖ : a, G H E a + b Thus x is a third proportional to the quantities (a+b) and a. In CA take C L = a, produce CA to M so that LM = b, join M K, and draw LH parallel to M K; CH is the required value of x. 17. In a right-angled triangle the lines drawn from the acute angles to the points of bisection of the opposite sides are given, to find the triangle. Let CE a, BD b, AD = CD=x, AE= EB = y. Then the square upon CE square upon CA + square upon A E, or a2 = 4x2 + y2 a the sides take A F = 2 with centre F and radii b and 2 a, describe circles cutting the other side produced in G and H, respectively; draw GI parallel to FH; then 2 AI is the required value of y. Hence AD, and therefore AC and A B are found, and the triangle is determined. 18. To divide a straight line, so that the rectangle contained by the two parts may be equal to the square upon a given line b. Let A B = a AP x |