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Ar. Co. 0.00000 As rad. x sin. E,
sin. E 230 20'. Ar. Co. 0.40222
sin. ECD 100° 00' 9.99335 : sin. ECD x sin. EDC, Х
sin. EDC 56 40. 9.92194 SCD 11.50
1.06070 :: CD,
: 2 EDC 274.731 2 EAB 70.105
2 ABCD 204.326
ABCD = 102.163 Ch. = 10A. OR. 34.6P.
2. In a trapezium ABCD, the angles are, A= 65°, B=81°, C = 120°, and consequently D= 94°; also the side AB= 20 Ch. and CD= 11 Ch.: required the
Ans. 22A. 2R. 27P.
3. Required the area of a four-sided piece of land, bounded as follows:
1. N. 12° 30' E.
PROBLEM VIII. To find the area of a trapezium when three sides and the. ***,
two included angles are given.
Take the difference between the sum of the given angles and 180°; Then,
If the sum of the given angles be less than 180°, subtract the third quantity from the sum of the other two, and half the difference will be the area of the trapezium. But if the sum of the given angles exceed 180°, add all the three quantities together and half the sum will be the area.*
• DEMONSTRATION. Let ABCD (Fig. 72 or 73) be the trapezium, having the given sides, AD, AB, BC, and given angles DAB, ABC. Complete the parallelograms ABCE, ABFD, and join ED, CF; then because EC, DF, are each parallel and equal to AB, they are (30.1.), parallel and equal to each other, and (33.1.) ECFD is a parallelogram; therefore ABFD ABHG + GHFD (35.1.) ABCE + ECFD
(34.1.) ABCE + 2 ECD; to the first and last of these equals add ABCE, then ABFD + ABCE 2 ABCE + 2 ECD = 2 ABCDE.
EXAMPLES. 1. In a trapezium ABCD, there are given AD 23.32 Ch., AB = 25.70 Ch., and BC = 15.84 Ch., the angle DAB = 64° and ABC = 82o: required the area.
But Fig. 72, when the sum of the given angles DAB, ABC, is less than 1800, 2 ABCDE = 2 ABCD + 2 EAD; therefore in this case ABFD + ABCE - 2 ABCD + 2 EAD; or ABFD + ABCE 2 EAD 2 ABCD.
And, Fig. 73, when the sum of the given angles DAB, ABC, exceeds 180°, 2 ABCDE = 2 ABCD - 2 EAD; therefore ABFD + ABCE = 2 ABCD 2 EAD; or ABFD + ABCE + 2 EAD = 2 ABCD.
But by prob. 3. one of the first two proportions gives 2 BAD (= ABFD), and the other gives 2 ABC (= ABCE); also because the angle EAD is the difference between the sum of the given angles and 180°, and the side EA BC, the third proportion gives 2 EAD:
Ar. Co. 0.00000 : sin. difference 340.
1.36773 BC 15.84
: : AD BC, SAD 23.32
2. What is the area of a four-sided lot of ground, three sides of which, taken in order, measure 6.15, 8.46 and 7.00 chains, respectively, the angle contained by the first and second sides 56°, and that contained by the second and third sides 98° 30'? Ans. 4A. OR. 25P.
3. One side of a quadrilateral piece of land bears S. 7E. dist. 17.53 Ch., the second, N. 87 E. dist. 10.80 Ch. and the third, N. 25 E. dist. 12.92 Ch.: what is the area? Ans. 21A. 3R. 2P.
Note. As in triangles any three parts, except the three angles, being given, the area may be found, so in trapeziums, unless two sides are parallel, any five parts except the four angles and one side, being given, the area may be found. Several other problems might therefore be introduced for finding the areas of triangles and trapeziums, depending on the different parts, sufficient to limit them, that may be given; but as they seldom occur in practice, and when they do may readily be solved by means of trigonometry and the preceding problems, they are omitted.
To find the area of a trapezoid.
Multiply the sum of the parallel sides by their perpendicular distance, and half the product will be the
1. Required the area of a trapezoid ABCD, of which the parallel sides AD, BC measure 6.14 and 9.48 chains respectively, and their perpendicular distance BF or DE, 7.80 chains.
Ch. 6.14 9.48
60.9180 Ch. = 6A. OR. 15P.
* DEMONSTRATION. The trapezoid ABCD, Fig. 74, = the triangle
AD X BF BC X DE, ABD + BDC by prob. 2),