rectangle E F G H. For the triangle A E K is equal to the triangle HD K, as may be found by applying one to the other, and also LFB equal to CGL. But the area of the rectangle EFGH is equal to the product of H G, the mean length of the parallel sides of the trapezium, and E H, the perpendicular distance. Hence the rule. Ex. The two parallel sides of a trapezoid are 3 ft. 6 in. and 4 ft. 10 in., and the perpendicular distance between them is 2 ft. 4 in.; find the area. = 3 ft. 6 in. 42 in.; 4 ft. 10 in. =58 in.; 2 ft. 4 in. 28 in. = =28×50=1400 sq. in.=9 sq. ft, 104 sq. in. Ex. XXX. Find the areas of the trapezoids which have the following dimensions: (1) Parallel sides 14 and 18; perpendicular distance 20. (2) Parallel sides 3 ft. and 7 ft.; perpendicular distance 10 ft. (3) Parallel sides 15 and 71; perpendicular distance 12. (4) Parallel sides 4-8 yds. and 2-6 yds.; perpendicular distance 7.5 yds. (5) Parallel sides 5 yds. 1 ft. and 4 yds. 2 ft.; perpendicular distance 3 yds. 1 ft. (6) Parallel sides 2 ft. 6 in. and 11 ft. 6 in.; perpendicular distance 5 ft. 4 in. (7) Parallel sides 5 yds. 2 ft. 9 in. and 2 yds. 1 ft. 7 in.; perpendicular distance 4 yds. 2 ft. 5 in. (8) Parallel sides 800 lks. and 450 lks.; perpendicular distance 160 lks. (9) If a field have two parallel sides, the sum of which is 2050 lks., and the perpendicular distance between them 375 lks., what is the acreage? (10) A field is bounded by four straight lines, of which two are parallel. If the sum of the parallel sides is 1875 lks., and the perpendicular distance between them is 240 lks., determine the area of the field. (11) A quadrilateral field has two parallel sides; one is 6·7 chains, the other 5.8 chains, and the perpendicular distance between them 7.4 chains; find the acreage of the field. (12) Two parallel sides of a quadrilateral field are 14 and 20 chains respectively, and the perpendicular distance between them is 12 chains; what is the area in ac. ro. po.? (13) Find the area of a field having two parallel sides 650 and 925 lks. long, the perpendicular distance between them being 850 lks. (14) Find the area (in square chains) of a quadrilateral figure, of which two parallel opposite sides are respectively 750 lks. and 1225 lks., and the perpendicular distance between them is 1550 lks. 75. To find the sum of the parallel sides when the area and the perpendicular distance are given. RULE. Divide double the area by the perpendicular distance, and the quotient will be the sum of the parallel sides. = = = For, half sum of parallel sides × perp. distance area (Art. 74). sum of parallel sides x perp. distance double the area. and, sum of parallel sides double the area÷perp. distance. Ex. The area of a trapezoid is 1728 sq. yds., and the perpendicular distance 24 yds. ; find the sum of the parallel sides. Double the area=1728×2=3456 sq. yds. Sum of parallel sides=3456÷24=144 yds. Ex. XXXI. (1) The area of a trapezoid is 7308, and the perpendicular distance 18; find the sum of the parallel sides. (2) A quadrilateral field which has two parallel sides contains 15 ac. 2 ro. 20 po., and the perpendicular distance is 625 links; find the sum of the parallel sides. (3) The area of a trapezoid is 25 ac. 2 ro. 15 po., and the perpendicular distance 3250 lks.; find the sum of the parallel sides. (4) Two of the four sides of a field are parallel, and differ by 4 feet in length; find them, their perpendicular distance being 19 ft., and the area of the field 475 sq. ft. [If the difference between the two parallel sides be 4 ft., then the shorter side is 2 ft. less, and the longer side 2 ft. more, than the mean length.] 76. To find the perpendicular distance when the area and the sum of the parallel sides are given. RULE. Divide double the area by the sum of the parallel sides, and the quotient will be the perpendicular distance between them. For, half sum of parallel sides perp. distance area (Art. 74). sum of parallel sides x perp. distance = double the area. and, perp. distance = double the area ÷ sum of parallel sides. Ex. The area of a trapezoid is 1008, and the sum of the parallel sides is 28; find the perpendicular distance between them. (1) The area of a trapezoid is 6 sq. yds. 5 sq. ft. 18 sq. in., and the sum of the parallel sides is 16 ft. 6 in.; find the perpendicular distance. (2) Find the perpendicular distance between the parallel sides of a trapezoid, whose area is 32 acres, the sum of the parallel sides being 62 ch. 50 lks. (3) A trapezoid contains 11 ac. 0 ro. 4 po., and its two sides are respectively 48 po. and 36 po.; find its breadth. (4) What is the perpendicular distance of a trapezoid whose area is 52-36, and parallel sides 7·25 and 8·15 respectively? THE TRAPEZIUM. 77. A Trapezium is a quadrilateral or four-sided figure, of which no two sides are parallel. B Every trapezium may be divided into two triangles by drawing a straight line joining two of its opposite angles. Thus the trapezium ABCD is divided into the two triangles ABC, ADC, by A C. AC is called the diagonal. BE and D F are the perpen E F diculars upon it. The area of the trapezium will be the sum of the areas of the two triangles. 78. To find the area of a trapezium. RULE. Divide the trapezium into two triangles; find the area of each triangle, and the sum of both areas will be the area of the trapezium. Ex. 1. The diagonal A C of a trapezium ABCD is 16 ft., the perpendicular D E 8 ft. and BF 6 ft.; what is its area? Area of triangle AD C-16x8÷2-64 sq. ft. (Art. 66). Area of triangle A B C=16×6÷2-48 sq. ft. .. Area of trapezium A B C D=64+48=112 sq. ft. Ex. 2. In the trapezium A B CD, AB=14, BC=13, CD=12, DA 9, and the diagonal C A = 15; find its area. Area of triangle A B C=84. Area of triangle AD C=54. .. Area of trapezium A B C D = 84+54=138. E A Ex. XXXIII. (1) Find the area of a trapezium whose diagonal is 640 yds. and the two perpendiculars 180 and 120 respectively. (2) How many sq. yds. of paving are in the trapezium A B C D, whose diagonal A C=20 yds. and the perpendiculars DE 7.2 yds. and BF=5·8 yds. (3) In surveying a four-sided field A B C D, I measured A C =190 ft., and perpendiculars from the other corners upon A C 45 ft. and 60 ft. respectively. Find its area in sq. ft. (4) One diagonal of a quadrilateral field is 20 ch. 25 lks., and the perpendiculars on it from the opposite angles are 6 ch. 25 Iks. and 7 ch. 25 lks. respectively; find the area in ac. ro. po. (5) A diagonal of a four-sided field is 18 ch. 75 lks., and the perpendiculars on it from two corners are 6 ch. 25 lks. and 7 ch. 75 lks.; find its rent, at 12s. a rood. (6) ABCD is a quadrilateral field. A B measures 48 ch., BC=20 ch., A C=52 ch., and the perpendicular from D on A C-30 ch. Find the acreage of the field. [Find the area of the triangle A B C by Art. 69, and of A D C by Art. 66.] (7) Required, in sq. lks., the area of the trapezium A B C D, whose sides A B, BC, CD, and D A, are 30, 40, 120, and 130 links respectively, and the diagonal A C 50 links. (8) What is the area, in sq. ch., sq. lks., of a four-sided field, whose west side measures 195 lks., south side 260 lks., east side 264 lks., north side 269 lks., and the diagonal from N.W. to S.E. 325 lks. (9) A B C D is a trapezium of which the diagonal A C is 325 yds., and the sides A B, B C, CD, DA, are 123, 208, 116, 231 yds. respectively; find the area in sq. yds. (10) The four sides of a field are 75, 100, 125, and 200 links long respectively; the first two sides form a right-angle; find the area of the field in square poles. [B C=75, CD=100, D A=125, A B=200. Á Find the length of B D by Art. 71.] D (11) Find the cost of paving the courtyard ABCD (see fig. to Quest. 10) at 1s. 6d. per sq. yd., when A B=48 yds., BC-15 yds., CD=20 yds., and AD=D B, the angle B C D being a right angle. (12) A B C D is a quadrilateral field; D A B 850 yds.; BC=557 yds.; CD= 685 yds.; A B is parallel to CD, and the angle at A is a right angle; find the area. [AE DC, E B=AB-AE. Given EB and BC, find EC by Art. 72.] POLYGONS. с THE IRREGULAR POLYGON. 79. A Polygon is a figure which has more than four sides. 80. Polygons are regular or irregular. 81. A regular polygon is one whose sides are all equal. 82. An irregular polygon is one whose sides are not equal. 83. To find the area of an irregular polygon. |