Through D, F, C, draw perpendiculars gd, Ff, Cc, to the base, and produce GH both ways to meet Dd and Cc in g and h. Then the area of ABCD = area of ABHG = AB×Fƒ. Ff is the average height of the trapezium; hence the area= base x average height. For 2Ff=dg+ch, =dD+ Dg+ch, =dD+Cc, since Dg=Ch; .. Ff=¿(Dd+Cc), the sum of the greatest and least heights. If the ABC be a right angle, the greatest and least heights become the parallel sides; and then the area = base x sum of the parallel sides. α In the annexed particular example of a symmetrical trapezium, where ab bisects the parallel sides, and cd, meeting ab in e at right angles, bisects the c other sides, we may consider the figure to consist of two trapeziums, having e a common base ab, and average height ec, or ed; then the area = ab× cd = breadth × average length; or= breadth × sum of the parallel sides. D C NOTE. If DC were a curved line, a straight line might be so drawn, terminated in AD and BC, or in those lines produced, partly above and partly below the curve, that the included and excluded areas should nearly balance; then the area of the irregular surface will approximately be represented by that of the trapezium, found as before. A B PROB. 10. Find the area of a flat oblong frame of uniform width, as of a picture, or a walk round a rectangular grass-plot. Let ABCD be the outer, and EFGH the inner, boundary. Produce GH to meet AD in h, The area required may be obtained in several ways. 1st. It is equal to the area of the rectangle BD-area of rectangle FH, = AB× AD-EF × EH. 2nd. It is equal to twice the rectangle AH + twice the rectangle DG, =2(Ah+Gh) x breadth of frame. 3rd. Or, by joining the corresponding corners, by the lines BF, CG, &c., the frame is divided into four trapeziums, of the form exhibited in the 2nd. case of Prob. 9. And since the area of each breadth × sum of parallel sides, .. the area of the four= = (breadth × sum of outer and inner boundaries of the frame. EXERCISES G. (1) Each side of a square is 75'4 feet; what is the distance between the two opposite corners ? Ans. 106-63068 ft. (2) A roof of 10 ft. span, and formed by two equal rafters makes a right angle at the ridge; of the slope from the ridge to the eave. find the length Ans. 7.071...ft. (3) How many plots, each 11 yards square, can be obtained from 50 acres? Ans. 2000. (4) The foot of a mast 120 ft. high is 25 ft. from the ship's side; find the length of a rope reaching from the side to a point in the mast at two-thirds of its height. Ans. 83-815...ft. (5) A ten acre field is in the form of a square; find the cost of laying down a diagonal drain at 15d. per linear yard. Ans. £19. 8s. 10gd. (6) The diagonal of a square board is 10 yards; what is the side of the square, and its area? (1) Ans. 7071...yds. (2) Ans. 50 sq. yds. (7) The sides of a rectangular plot are 108 ft. and 144 ft.; if the former dimension be shortened by 12 ft., how much must the latter be increased, so that the area may remain unaltered? Ans. 18 ft. (8) In the preceding example, if the longer side be shortened by 12 ft., how much must the creased? other be inAns. 9 ft. (9) Compute the lengths of the outer boundaries of each of the three rectangular plots in (7) and (8). Ans. 504; 516; 499. (10) The sides of a triangle are 4, 5, and 6; alter the last dimension so that the triangle shall become rightangled. Ans. Add 403...to it. (11) Out of a piece of metal, 15 inches square, as many circular portions as possible are cut, each 1 inch in diameter; how many will there be? Ans. 225. (12) Find the areas of the trapeziums of which the dimensions are as follows: (13) The hypothenuse of a right-angled triangular field is 50 yds., and the other sides are in the ratio of 3 to 4; find its area, and cost at 80 guineas per acre. (1) Ans. 600 sq. yds. (2) Ans. £10. 8s. Sd. (14) Find the side of a square which cost £27. 1s. 6d. paving, at 8d. per square yard. Ans. 28 yds. (15) How many square feet of flooring can be covered by a board whose length is 10 ft. 5 in. and the breadths of the two ends 24 ft. and 12 ft.? Ans. 22 ft. 19 in. (16) Find the areas of the several lozenge-shaped parallelograms whose diagonals are as follows: (17) The diagonal of a square is 20 ft.; find the side approximately to 3 places of decimals. Ans. 14.142 ft. (18) Two vessels sail from the same point, one due North at 9 knots per hour, and the other East at 11 knots; find how far they are apart in 12 hours. Ans. 170:5512...miles. (19) A ladder 40 feet high reaches to 3rds of the height of a building, when placed across a street 8 yds. wide; how much must the ladder be lengthened, so that it may reach the top of the building without changing its resting-place in the street? Ans. 13.665 ft. (20) Prove that the rectangle, whose sides are 18 units and 8 units, has a longer diagonal than the square of the same area. 2 (21) Shew that the area of the square, whereof the perimeter is 40 units, is greater than that of any other oblong of the same perimeter also, shew that with the same perimeter, the greater the inequality of the sides, the less the area. (22) In measuring a narrow rectangle, if there be any liability of error, shew that it is more important to be accurate in measuring the breadth than the length. (23) Find how many feet of planking are required to form a single shelf, round a room 24 ft. by 16 ft., if the shelf is 1 ft. broad, and it is interrupted by a door and two windows, that are respectively 3 feet and 4 feet wide. Ans. 933 sq. ft. (24) A room whose floor is 36 ft. by 27 ft. is surrounded by desks placed 1 foot from the wall; how much of the floor is enclosed within their inner boundary if they be a yard broad? Ans. 532 sq. ft. (25) A picture, each of whose sides is 1 foot, is surrounded by a flat frame containing 14 sq. ft.; find the outer circumference of the frame. Ans. 6 ft. (26) A copy-book is ruled for writing, with lines 1 inches apart, and with sloping transverse lines, which intersect the others at equal intervals of of an inch. Find the area of each of the equal parallelograms, into which the surface is divided. Ans. of a square inch. (27) A hole, a yard square, is to be diminished to half its size, and still to be a square. What is the length of the side of the diminished square, and what of its diagonal? (1) Ans. 6.364 ft. nearly. (2) Ans. 1 yard. (28) A square, and another square, one-fourth of its size, are to be formed into one square; find the proportion of its side to the side of the first square. Ans. 1.183: 1. (29) A trench is cut round a camp 4 ft. deep and 5 feet wide; and the earth is formed into a rampart 3 ft. in perpendicular height, the face of which slopes so that its upper edge is one foot from the vertical side of the trench; find the shortest ladder that will reach |