From the given bearings, the angles may be found as follows; = BCD 100° 00 = DC, N. 35° 20′ W. 123 20 180 00 ADC 56 40 Construction. Make AB = 6.90, and draw DA, CB, making the angle DAB 64°, and ABC = 139° 20′; produce DA and make the angle EAF 56° 40′ the given angle = ADC; lay off AF = 11.50 = the given side CD, and parallel to AD draw FC, meeting BC in C; lastly draw CD parallel to AF, meeting AD in D, then will ABCD be the trapezium.* Calculation. The angle E 180° 23°20'. ADC the sum of the angles BCD, * DEMONSTRATION. By construction FC is parallel to AD and CD to AF, therefore (34.1.) CD=AF and (29. 1.) the angle ADC = EAF; hence it is evident that the sides AB, CD, and the angles of the trapezium ABCD are re 2. In a trapezium ABCD, the angles are, A = 65°, B=81°, C=120°, and consequently D 94°; also the side AB = 20Ch. and CD = 11Ch.: required the area. Ans. 22A. 2R. 27P. 3. Required the area of a four-sided piece of land, bounded as follows; 1. N. 12° 30′ E. 2. N. 81 00 E. dist. 23.20Ch. 3. S. 36 00 W. 4. N. 89 00 W. dist. 12.90Ch. PROBLEM VIII. To find the area of a trapezium when three sides and the two included angles are given. As radius, RULE. Is to the sine of one of the given angles; So is the rectangle of the sides including this angle,. As radius, Is to the sine of the other given angle; So is the rectangle of the sides includidg this other angle, To a second quantity. Take the difference between the sum of the given angles and 180°; Then, As radius, Is to the sine of this difference; So is the rectangle of the opposite given sides, 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 to AB, they are (30.1.) parallel and equal to each other, and (33.1.) ECFD is a parallelogram; therefore ABFD - ABHGGHFD = (35.1.) ABCE + ECFD=(34.1) ABCE +2 ECD; to the first and last of these equals EXAMPLES. 1. In a trapezium ABCD, there are given AD = 23.32Ch., AB = 25.70Ch., and BC= 15.84Ch., the angle DAB 64° and ABC 82°: required the area. As rad. : sin. DAB, 64° = :: AD × AB, But Fig. 72, when the sum of the given angles DAB, ABC, is less than 180°, 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 180o, 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: hence the truth of the rule is manifest. N 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. 7 E. dist. 17.53Ch., the second, N. 87° E. dist. 10.80 Ch. and the third, N. 25 E. dist. 12.92Ch.: 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 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 |