14. To find the area of an irregular figure by the method of equidistant ordinates. When a figure is bounded by an irregular outline A'B'C' D' E' F' G' on one side, this method will enable us to obtain a tolerably close approximation to the area enclosed by it and a straight line AG. Set of AB = BC = CD = A B G = FG, any number of equal parts. Denote each by h, and measure the several distances AA' BB', .... GG' perpendicular to AG; and denote them severally by a, b, c, g. Then the area of the space enclosed by the straight lines A'A, AG, GG' and the irregular boundary A'G' will be expressed by sa+b b+c +d d+e e + f + + + + 2 2 2 2 2 For these are the areas of A'ABB', B'BCC', ....; and if the distance h be so taken as to render the collected excesses of one set of parts above the figure, visibly equal to the collected defects of the others from the several trapezoïds, the area will be found to a visible degree of accuracy. When the figure is composed of boundaries along parts of which straight lines of considerable length can be drawn without varying much from the actual boundaries, a better method will be to take the extremities A', B', C' .... G' of these lines as the angles of the trapezoïds; to measure the lengths AB, BC, CD, .... FG of the intersections of the perpendiculars AA', BB', CC', Then the figure will be, calling these distances h1, h2, On these principles, the areas of fields bounded by irregular fences are estimated in land surveying: which will be illustrated under that head. II. PROBLEMS ON PLANE SURFACES. On the principles laid down in the preceding theorems the following series of problems admit of solution. The theorems for solution are referred to without specifying the rules in words, a step which is altogether unnecessary. PROBLEM I. To find the area of any parallelogram. Nos. 1, 2, 3, page 1. The length of a parallelogram is 12.25 and its breadth is 8.5: what is its area? Ans. 12.25.8.5 104 125 area. 2. The side of a square is 35 25 chs: required its area? Ans. 124ac 1r lp. 3. Find the area of a rectangular board whose length is 12ft, and whose breadth is 9in. Ans. 9 ft. 4. Find the content of a piece of land in the form of a rhombus, its length being 6.2 chs, and breadth 5:45. Ans. 3ac 1r 20p. 5. Required the number of sq. yds in a rhomboid whose length is 37ft. and height 5ft 3in. Ans. 21 sq. yds. 6. The two diagonals of a parallelogram are 185 5 and 137·9, and they intersect under an angle of 42° 10′ 18′′: what is the area ? Ans 8587.55. 7. The side of a rhombus is 183, and one diagonal is 23: find the other diagonal and the area. Ans. diag. 29-223278, area = 343.373. 8. If a, b, c be the distances of a tree from three of the angles of a square field, show that its area is expressed by a2 2ab cos (45° + ) + b2, where cos &= (a2 -c2) cos 45° + b2 sec 45° 2ab 9. The distances a, b, c, d of a point from the four angles of a rectangle are given to find its sides and area. 10. The stretching frame of a picture is 24in by 18; the frame of the picture (which is flat with a bevelled edge of in wide, and inclined to the picture in an angle of 45°) extends over 2in of the canvas; and the visible area of the picture is equal to the visible surface of the frame: it is required to find the width of the frame, including the bevelled edge. PROBLEM II. To find the area of a triangle. Rule I. No. 2, Cor. 2. page 471. 1. The base of a triangle is 625 and height 520 links: what is its area? Ans. 625.260 = 162500lks = lac 2r 20p. 2. Find the area in yds of the triangle whose base is 40 and perpendicular 30ft. Ans. 663 sq. yds. 3. Required the number of yds in a triangle whose base and height are 49 and 251ft respectively. Ans. 684 yds. 4. Required the area of a triangle whose base is 18ft 4in, and altitude 11ft Ans. 108ft. 53in. 10in. Rule II. No. 2, Cor. 3, and No. 7, pages 472, 473. 1. The containing sides are 30 and 40, and included angle 28° 57': what is the area? By natural numbers. By logarithms. t sin 28° 57' = 9 6848868 log 600 2.7781513 log 290 4277 = 2 4630381 2. How many sq. yds are contained in the triangle, one of whose angles is 45°, and the containing sides 25 and 211ft respectively? Ans. 20 86947. 3. Given the base of a triangle equal to 476 25 yds, and the angles at the base 27° 10′ 15′′ and 35° 10′ 18, to find the area and the three perpendiculars from the angles to the opposite sides. Ans. area 33678, and perpendiculars 141 43, 217-47, 274.33. 4. The base of a triangle is 27.25 chains, the vertical angle is 57° 15′, and the difference of the angles at the base is 18° 18'. What is the area of the triangle? Ans. 328.972. 5. The difference of the segments into which the perpendicular divides the base is 10, the base itself is 50, and the vertical angle is 100°. What is the area of the triangle? Ans. 511 938. Rule III. No. 6, page 473. b = 15, s - c = 5; 1. To find the area of a triangle whose sides are 20, 30, 40. Here 2s 20+30 + 40, or s = 45, s — a = 25, s and hence the area is 45.25.15.5 75/15290°4737. 2. How many yards are there in a triangle whose sides are 30, 40, 50ft respectively? Ans. 663. 3. Find the area of a field whose sides are 2569, 4900, and 5025 links respectively. Ans 61ac 1r 39p. 4. The area of a triangle is 6, and two of its sides 3 and 5: find the third side and all the angles. Ans. 4 or 2/13. 5. The sides of a triangle are in arithmetical progression, their sum is 27 linear feet, and the sum of their squares is 261 sq. ft: find the area in yds. Ans. 15. 6. The area of a triangle is 1000 and its sides in the ratio, 1, §: find the sides. Ans. 40'074, 50·093, 66.791. 7. Find the side of an equilateral triangle whose area is 100. Ans. 15.197. 8. The sides of a triangle are √3, √4, √5: what is the area? Ans. 11. 9. A triangle has its sides each equal to 3√3/3: what is its area? Ans. 204. 10. Find the area of the triangle whose sides are 4, 4 + √3 and 4 √3. Ans. 2/3. 11. On the perpendicular of an equilateral triangle whose side is a, another equilateral triangle is described, and on the perpendicular of this another, and so on ad inf.: it is required to find the sum of the areas of all the triangles so described. Ans. 2√3. 12. The sides a, b, and the area A, are given to show that 1. In a trapezoïd the parallel sides are 750 and 1225, and their distance 1540 links what is the area? Ans. 15ac Or 33p. 2. The greater and less ends of a plank are 15 and 11in, and its length 12ft 6in: what is its area? Ans. 13 ft. 3. A quadrangular field ACDB has perpendiculars CP, DQ drawn to the side AB; and the following measures were taken from which to find its area: AP = 110, AQ= 745, AB = 1110, CP = 352, and DQ 595 links respectively. Ans. 4ac Ir 5 792p. = 4. Find the area of the trapezoïd whose opposite parallel sides are 35 and 19; and the angles made by the oblique sides with the parallel sides are 42° 10′ 15′′ and 73° 6′ 20′′ respectively. Ans. 306.861. 5. From the triangle ABC, whose base AB is 40 and the perpendicular upon it is 60, to cut off a trapezoïd by a line parallel to AB which shall have the area 480. Ans. The breadth OP of the trapezoïd is 131⁄2 nearly. 6. What length must be cut from the broader end of a board which is 13 ft long, and whose ends are respectively 18in and 14in, to make 10ft square? Ans. 7.1 nearly. 7. The breadth of a ditch at the top was 72ft, at the bottom 38, and the sloping side 26 and 20ft; and the top and bottom horizontal: find the area of the vertical section. Ans. 885 ft. 8. The area of the section of the ditch being 154ft, and its depth 5ft; also the breadths at the top and bottom are as 9 to 5: what are those breadths? Ans. 36 and 20ft. 1. To find the area of a trapezium, one of whose diagonals is 42 and the perpendiculars upon it 16 and 18. Ans. (1618) 42 = 714. 2. A diagonal is 65 ft, and the perpendiculars on it are 28 and 33 ft: how many yds does it contain? Ans. 222 yds. 3. In a quadrangular field ABCD, owing to obstructions there could only be taken the following measures; BC= 265, AD = 220, AC = 378, AE 100, CF 70 yds, when DE and BF are perpendicular to AC: it is required to construct the figure and compute the area. Ans. 17ac 2r 21p. 4. The two diagonals of a trapezium are 31-2956 and 62.1598, and they intersect under an angle of 105° 18′ 25": what is the area? 5. One angle of a trapezium is 107° 18′ 10′′, and the diagonal divides it in the ratio of 7 to 5, whilst the sides containing it are 123-456 and 654321, and the greater side makes the less angle with the diagonal, the diagonal itself being 1000: it is, from these data, required to find the areas of the two triangles into which the other diagonal divides the trapezium. 6. The four sides of a field taken in order are 25, 35, 31, and 19 poles, and the diagonals are equal: required the area of the field. 7. ABCD is a quadrangular field, whose sides taken in succession are AB = 15 ch 24 1, BC = 18 ch 86 1, CD = 9 ch 90 1, and DA = 11 ch 141; also the angles at A and C are 105° 28′ and 89° 54′: find its area. Ans. 17.5169ac. 8. The side AD of a quadrangular field ABCD was 311yds, and the angles were BAC 44° 20′, CAD = 41° 19′, ADB = 24° 10′, and DBC = 37° 4' : required its area. = Ans. 8 6531ac. 9. The sides of a quadrangular field taken in order are 26, 20, 16, and 10 poles, and the angle contained by the longest sides is 56°: what is its area? Ans. 1ac 127 676p. 10. ABCD is a quadrilateral, whose angles B and C are right angles, as are likewise those formed by the diagonals BD, AC: the diagonals themselves are given, d and d1, and it is required to find the area, the sides, and the remaining angles formed by the several lines of the figure. 11. The two diagonals AC, BD, of a trapezium bisect each other; one of the angles, B, is 30°, the side BC is 100, and twice the product of the diagonals is equal to the square of their difference: required the sides, angles, and area of the figure. PROBLEM V. To find the area of any polygon. This will be effected if we divide it into triangles or quadrilaterals, or both, and take sufficient measures either of lines or lines and angles for the trigonometrical determination of all the parts requisite for the purpose. One or two examples of the manner of dividing the polygon will suffice to show the nature of the problem, whilst the actual computation is left for the student. 1. Suppose a regular pentagon whose side is 170 fathoms, to be fortified, and that the salient angle of the bastion is 71°, and its face 47 fathoms ; required the flank and curtain, supposing the line of defence to be perpendicular to the flank determine, also, how many acres would be contained within the boundaries of the fortification, supposing the work completed. 64 57, area 26 051. VOL. I. I i 2. The polygon in the annexed figure had the following parts measured, from which to determine the area : viz. AC 55, FD= 52, GC = 44, Gm = 13, Bn = 18, Go 12, Ep = 8, and Dq = 23. Ans. area 1878.5. 3. The sides FE, AB, BC, were found to be 15.726, 25.182, and 23 629, and the angles were taken as follows: FED = 152° 10′, EFG = 65° 18', FEG : 66° 28', EGD = 31° 15', CGD = 32° 18', GCD = 58° 40', AGC = 100° 5′, GAC = 41° 50': it is required to find the area. PROBLEM VI.-To find the area of a regular polygon.* 1. Required the area of a regular pentagon, each side being 25 ft. Ans. 1075 298356. 2. Find the area of an equilateral triangle whose side is 20 ft. 3. Required the area of a hexagon whose side is 20. Ans. 173 2058. Ans. 1039 23048. Ans. 1931 37084. Ans. 3077 68352 PROB. VII-To find the circumference and diameter of a circle from one another. No. 11, page 475. The approximation 1: 314159 is the number used for general scientific purposes; but for mere round numbers 7: 22 or 113: 335 are used as the ratio of the diameter to the circumference. In the same manner, circumference: diameter 1: 318309. 1. Find the circumference of a circle whose diameter is 20. Ans. 7 22: 20: 629 as the roughest approximation. 2. If the circumference of the earth be 24877.4 miles, what is its diameter ? 1 :: 24877-4: 7918.72 3.14159: 113 :: 24877-4: 7918.72 1: 318309: 24877-4: 7918-72 3. Required the circumference of a circle whose radius is 22. Ans. 139 01547. π * It is sometimes convenient to possess the values n cot in a table; and likewise the value of the circumscribed radius. In the latter case the tabulated value is multiplied by the side of the polygon, and in the former by its square. The following is such a table to the duodecagon inclusive. To show its use we may take Ex. 1, in which we have 252.1·7204774 = 1075-29837, the area. |