FI=KH=HT-SG=GL, which let be represented by x. GO is a parallelogram, of which one side is a+2x, and the altitude: therefore, the area (Art. 513.)=(a+2x)×x =ax+2x2. PH is a parallelogram equal to GO; there. fore its area ax+2x2. CP is a parallelogram, whose base 18=b, and whose altitude=x; hence its area=bx. SA is a parallelogram whose area is equal to the area of CP. By the Figure, the sum of the areas of these four parallelo. grams the area of the walk, which=2(ax+2x2)+2× (bx)=2ax+4x2+2bx-by substituting figures, 24x+4x2+ 36x=4x2+60x: therefore, 1. By the given conditions, 4x2+60x=12×18 3. Comp. the square, x2+15x+225÷4=225÷4+54 The solution of this problem may be much abridged by considering the parallelogram LKHG=(2x+18)×x, and the parallelogram_LNCD=12×x. By multiplying each of these expressions by 2, we have 4x2 +60x, which=12x 18-216, as above, and when reduced the same result is obtained. XIX. See figure 2. Let AB=6x, and BC=5x; then, 1. By Art. 513, 6x × 5x=30x2=the area 2. By the conditions of the problem, 30x2÷6=125 3. Clearing of fractions, 5x2=125 4. Dividing by 5, x2=25, and x=5. Therefore, AB=6x=30, BC=5x=25. XX. See Figure 10. Let the right angled triangle be represented by ABC, and the parallelogram by BCHE. Let x=AC-EC, and y=EB. By the problem, BC=60. By Art. 513, 60y=the area of the parallelogram, and 60x 2=30x=the area of the triangle; then, 1. By the given conditions, 30x60y::5:8 2. By Euclid I. 47, 602+y2x2 3. Multiplying extremes and means in the 1st, 240x=300y 4. Dividing by 240, x=5y÷4 5. Substituting for y in the 2d, 3600+y2=25y3÷16 6. Multiplying by 16, 57600+16y2 =25y2 7. Transposing and uniting, 57600=9y2 8. Extracting the square root, 240=3y, and y=80 9. Substi. the value of y in the 4th, x=5×80÷4=100. Therefore the base of the parallelogram 80, and the base of the triangle 100. By Art. 513, 1×60×100=3000= the area of the triangle, and 60 × 80=4800=the area of the parallelogram. Second method. Let x=AB, the base of the triangle; then by Art. 518, 60×x÷2=30x=the area of the triangle. By the problem, AB-EC, and by Euclid I. 47, EC2=60o +the square of a side of the parallelogram; therefore a side of the parallelogram=√x2 −3600, and the area of the parallelogram=60× √x2−3600: hence, 1. By the condition of the problem, 30x: 60√x2 -3600:: 5:8 2. Multiplying extremes and means, 240x=300√x2 — 3600 3. Dividing by 60, 4x=5√x2 - 3600 6. Clearing of fractions, 16x2=25x2-90000 7. Transposing and uniting, 9x2=90000 8. Dividing by 9, x2=10000, and a=100; therefore, 30×100=3000=the area of the triangle, and 60 × √x2-3600=60√10000-3600=60 √6400=60×80 =4800. XXI. See Figure 12, Let x=AB, a side of the base of the smaller vat=the hight of the larger vat. Also. let y= FE, a side of the base of the larger vat=the hight of the smaller vat; then the area of the base of the smaller vat, and y2= the area of the base of the larger vat. Moreover, on the principle that the contents of a solid are found by multiplying together its three dimensions, a3y= the solidity of the smaller vat, and y3r=the solidity of the larger vat: Hence, 1. By the given conditions, x2y: yax::4;5 y3x-20=x3y 3. Multiplying extremes and means in the 1st, 5x2y=4y2x 4. Dividing by xy, 5x=4y 5. Dividing by 5, x=4y÷5 6. Substi. a's value in the 2d, ya ×1— X 20= 16y3 25 8. Clearing of fractions, 100y3 — 2500=80y3 9. Transposing and uniting, 20y3=2500 10. Dividing by 20, y3-125, and y=5=side of the base of the larger vat the depth of the smaller vat 11. Substituting y's value in the 4th, 5x=4×5, and x=4= a side of the base of the smaller vat=the depth of the larger. Therefore the depth of the larger vat=4 feet, and the depth of the smaller 5 feet. XXII. See figure 8. The equilateral triangle ABC, is divided by the lines AP, CP, BP, into the triangles APC, APB, BPC, whose several altitudes are PE, PD, PF. Let PD=a, PE=b, and PF=c: Also let x=of one of the equal sides of the triangle; then, by Art. 518, the area of APC=ax, the area of APB=bx, and the area of BPC =cx; hence the area of the equilateral triangle ABC=ax +bx+cx. Moreover, in the triangle ABC, the perpendicular from Cupon AB, bisects the base AB; hence, by Euclid I. 47, AC2=AB2+the square of the perpendicular. By substituting, 42-2 the square of the perpendicular; therefore √3a2 the perpendicular. By Art. 518, x× √3x2 the area of ABC: But it has before been shown that ax+bx+cx=the area of 4BC: therefore, = 1. xx√3x2=(a+b+c)xx 2. Dividing by x, √3x2=a+b+c 3. Removing a from under the radical sign, (Art. 271, √3×x=a+b+c 4. Dividing by √3, x= a+b+c. XXIII. See Figure 12. Let x=the breadth of the street then by the problem, 9x-3=a side of the square. As in problem 18th, (9x-3+2x)×2x+(9x-3)×2x=40x2 - 12x = the area of the street. Also by the problem, 4x (9x-3) the perimeter of the square: hence, 1. By the given conditions, 40x2-12x-288-36x-12 4. Completing the square, 100x2-120x+36=540+36 XXIV. See Figure 3. Let x= BE, then 2x=AB. let y=BD, then 2y=BC. Let AD=b, and CE=a. 1. By Euclid I. 47, 4x2 +y2= b2 2. 66 66 4y2+x2=a2 4. Transposing in the 2d, x2=a2 — 4y2 5. Substituting in the 1st, 4(a2-4y2 )+y2=b2 6. Expanding, 4a2-16y2+y2=b2 7. Transposing and uniting, -15y2=-4a2+b2 8. Changing signs, 15y2=4a2-b2 4a2-b2 9. Dividing by 15, y2 : 15 10. Extracting the square root, y= 4a2-b2 11. Substituting y's value in the 1st, 4x2 + Also, 15 12. Clearing of fractions, 60x2+4a2-b2-1562 13. Transposing and uniting, 60x2 = 16b2 — 4a2 14. Dividing by 4, 15x2=4b2-α2 General Principles.-1. Algebra is useful in investigating the nature and relation of curves. 2. The positions of the several points in a curve drawn on a plane, are determined, by taking the distance of each from two right lines perpendi. cular to each other. 3. When two lines are drawn at right angles to each other, from any point in the same curve, they are together called the co-ordinates belonging to that point. For distinction's sake, one is called the ordinate, and the other the abscissa. 4. To compensate for the impossibility of drawing lines to every point in a curve, in order to express the relations of the several abscissas to their corresponding ordinates by an equation, and then determine each point, and the nature of the curve, the equation is made to depend on some property which is common to every pair of co-ordi nates. 5. Each abscissa is equal to twice the correspond. ing ordinate. 6. If one of the co-ordinates be taken of any particular length, the other will be given by the equation. 7. When ordinates are drawn on both sides of the axis to which they are applied, those on one side will be positive, while those on the other side will be negative. The abscissas also are positive or negative, according as they are on one side or the other of the point from which they are measured. 8. An abscissa vanishes at the point where the curve meets the axis from which the abscissas are measured; and an ordinate vanishes at the point where the curve meets the axis from which the ordinates are measured. 9. When the two axes meet the curve at the same points, the two co-ordinates vanish together. 10. An abscissa or an ordinate changes from positive to negative,or the contrary, by passing through the point where it is equal to 0. 11. In geometry, lines are supposed to be produced by the motion of a point. It might be ad. ded, that a surface is produced by the motion of a line, and a solid is produced by the motion of a surface. 12. The cube of the abscissa is equal to the square of the ordinate, multiplied by the difference between the diameter of the circle and the abscissa. 13. The curve is described, by taking ab. scissas of different lengths, and applying an ordinate to each. 14. If a point is conceived to move in such a manner, as to pass through the extremities of all the ordinates assigned by an equation; the line which it describes is called the locus of the point. The line is also called the locus of the equation by which successive positions of the point are determined. 15. The different orders of lines are distinguished, by the greatest index, or sum of the indices of the co-ordinates, |