PROBLEM XI. In a parabola IV H, the focus of which is F, any two of the three following parts, viz., the parameter PQ, the abscissa V G, and the ordinate G H being given, to find the third part. GH EXAMPLES. 1. The parameter PQ of a parabola is 50, and its ordinate 60 feet; required the abscissa V G. y2 602 Ans. By the first formula x = = p 50 =72. 2. The parameter of a parabola is 10, and its ordinate 4; required the abscissa. Ans. 1.6. 3. The abscissa of a parabola is 4, and its corresponding ordinate 10; required the parameter. PROBLEM XII. Ans. 25. To find the length of the arc of a parabola, its ordinate and abscissa being given. (See last figure.) FORMULA. Let x and y represent the same parts, as in the last Problem; then The arc V H = √x2 + y2 nearly. EXAMPLES. 1. Required the half arc V Q of a parabola, V F being feet, and F Q: 6. = Ans. V Q = √ 32 + 62 = 6 feet 11 inches. 2. The abscissa is 2, and the ordinate 6; required the length of the half arc of the parabola. Ans. 6.4291. NOTE. 1. The parabola is the path of projectiles in vacuo; it is also used in the astronomical theory of comets. NOTE 2. The student who wishes for further information concerning this curve, as well as concerning the ellipse and hyperbola, may consult the various works on conic sections.-(See Hann's Analytical Geometry, in Weale's Series.) 2* PART III. MENSURATION OF SUPERFICIES OR SURFACES. The area of any surface is estimated by the number of squares in that surface, without regard to its thickness; the side of those squares being one inch, one foot, one yard, &c. Hence the area is said to be so many square inches, or square feet, or square yards, &c. A TABLE OF SQUARE MEASURE. To find the area of a parallelogram; whether it be a square, a rectangle, a rhombus, or a rhomboid. RULE.-Multiply the length by the breadth or perpendicular height, and the product will be the area. 1. The length of a rectangular board is 7 feet, and its breadth 4 feet; required its area in square feet. (See first figure.) By the Rule. 7 × 4 = 28 square feet, the area required. 2. The side of a square is 18 inches; required its area in feet. (See last figure.) 24 square feet, the area required. 3. Find the area of a rhombus, the length of which is 6.2 feet, and its perpendicular breadth 5·45. (See second figure.) 334 square feet nearly. Ans. 33.79 = 4. The length of a table is 7 feet 8 inches, and its breadth 3 feet 10 inches; required its area. Here the operation is performed by duodecimals, and the area is found to be 29 square feet, 4 inches or 12ths, and 8 parts or ths. 5. What length must be cut off a rectangular board, the breadth of which is 9 inches, to make a square yard? A square yard contains 1296 square inches, whence by the second formula. 9)1296 = 144 inches 12 feet, the length required. 6. How many square feet of deal will make a box 6 feet long, 5 broad, and 2 feet 8 inches deep? Ans. 118 square feet 8'. 7. How many square yards are contained in a floor 23 feet long 14 feet wide? Ans. 37 square yards. 8. The base of the largest Egyptian pyramid is a square, the side of which is 693 feet; required the number of acres it occupies. Ans. 11a. Or. 4p. 9. A square court yard is 42 feet long, and 23 feet 10 inches broad; what did it cost paving at 48. 10d. per square yard? Ans. £26 18s. 6d. 10. Required the side of a square, the area of which is 500 square feet. = ✔ A, that is By the fourth formula s = side of the square S=500 · = 22.3607 feet 22 feet 4 inches nearly. 11. What is the side of a square the acre? = area of which is an Ans. 69.6 yards nearly. 12. A square in a city contains 64 acres of ground, required the side of the square. Ans. 173.92 yards. PROBLEM II. To find the area of a triangle. RULE I.-Multiply the base by the perpendicular height, and take half the product for the area. RULE II. When the three sides only are given: add the three sides altogether, and take half the sum; from the half sum subtract each side separately; multiply the half sum and the three remainders continually together; and take the square root of the last product for the area of the triangle. A: When all the three sides of the triangle are given, let them be reD presented by a, b and c, and their half sum by s; then A√8 (8a) (s—b) (s—c). 1. Let the base AD = EXAMPLES. 42 feet, and the perpendicular CB=33 feet; required the area in square yards. By Rule I. 42 × 33÷2=693, and 693÷9=77 square yards. 2. To find the number of square yards in a triangle, the sides of which are 13, 14, and 15 feet. 3. The base of a triangle is 40, and its perpendicular 30 feet; required the area in square yards. Ans. 66 square yards. 4. Find the area of a triangle, the three sides of which are 20, 30 and 40 feet. Ans. 32.27 square yards. 5. The base of a triangle is 49 and its height 25 feet, how many square yards does it contain? Ans. 68.736 square yards. 6. The base of a triangle is 18 feet 4 inches, and its height 11 feet 10 inches; required the area. Ans. 108 feet 5' 8". 7. The hypothenuse of a right angled triangle is 102 feet, and its base 100; required the area in square yards. Ans. 125 square yards. 8. The side of an equilateral triangle is 5.1 feet, required the Ans. 11.2626 square feet. area. 9. The base of a triangle is 121 yards; required its perpendicular, when it contains an acre of land. Ans. 80 yards. 10. The equal sides of an isosceles triangle are each 50 feet, and its base 28; how many square yards does it contain? Ans. 743 square yards. PROBLEM III. To find the area of a trapezoid. Add together the two parallel sides; multiply that sum by the perpendicular distance between them, and take half the product for the area. EXAMPLES. 1. In a trapezoid the parallel lines are A B 7.5, and DC 12.25, also the perpendicular distance A P is 15.4 feet; required the |