1. Required the area of the quadrilateral C ABCD, of which the sides are AC 236, BD 348, AB 392, and DC 427 feet, and the liagonal AD 473. √(606·5) × (606·5-348) × (606·5—392) A x (606.5-473) 67003.90 DAC √(568) × (568—236) × (568-427) × (568-473)=50259-08 ABD 117262.98 square feet. Ans. 2 acres 2 roods 30 perches 21 yards 6 feet. 2. Required the area of the trapeze ABCD, the sides AB 218, BC 194, CD 166 yards, and the perpendiculars from A upon BC 136, and upon CD 152 yards. Ans. 25808 yards, 5 acres 1 rood 13 perches 4 yards. E 3. Required the area of a trapeze ABCD, the sides AB 842, BC 938, CD 753, AD 826 links, and the angle A 78° 28'. By trigonometry BD = 1055·05. Ans. Area 683885 square links, 6 ac. 3 ro. 14 per. 64 yds. 4. Required the area of a trapeze ABCD, three sides AB 543, BC 428, CD 634 links, and the angles B 74° 40′ and C 84° 20'. By trigonometry BD = 729-077. Ans. Area 185392.38 links, = 1 ac. 3 ro. 16 per. 19 yds. 5. Required the area of a trapeze, the four sides 328, 456, 572, and 298, and the diagonal from the angle between the first and second 598 feet. Ans. 3 ac. 1 ro. 31 per. 29 yds. 3.8 ft. 6. Required the area of a trapeze, the diagonal 1268 links, the perpendiculars from one of its extremities upon the opposite sides 784 and 672, and the length of these sides 856 and 548 links. Ans. 5 acres 31 perches 14 yards 6-858 feet. PROB. IX. Given a diagonal of a quadrilateral, and the perpendiculars upon it from the opposite angles; to find the area. RULE. Add the perpendiculars together, and multiply half the sum by the diagonal. 1. Required the area of the quadrilateral ABCD, of which the sides are AB 68 and BC 54 yards, the diagonal BD 133, and the perpendiculars AF 37 and CE 44 yards. B F 37+4481 5386.5 square yards. Ans. 1 acre 18 perches 2 yards. 2. Required the area of the trapeze ABCD, the sides AB 672, BC 834, the diagonal BD 1296, and the perpendiculars AE 418 and CF 550 links. Ans. 6 ac. 1 ro. 3 per. 18 yds. 3. Required the area of a parallelogram, of which one of the diagonals is 486 feet, and each of the perpendiculars upon it from the opposite angle 126. Ans. 486 x 126 = 61236 feet area, = 1 acre 1 rood 24 perches 28 yards. 4. Required the area of a trapeze, the diagonal 1356, the angles at one of its extremities 57° and 42°, and the perpendiculars on it 568 and 724 links. Ans. 8 acres 3 roods 21 perches 2 yards 6 feet. 5. Required the area of a quadrilateral, of which the diagonals cut one another at right angles, the segments of the one are 328 and 523 feet, and of the other 498 and 672. Ans. 11 acres 1 rood 28 perches 18 yards. PROB. X. Given the diagonals of a quadrilateral, and the angle at their intersection; to find the area. RULE. Multiply half the product of the diagonals by the natural sine of the angle. Or add the logarithms of one diagonal, half the other, and the log. sine of the angle: the sum, lessened by 10 in the index, will be the logarithm of the area. NOTE 1.. If the angle made by the diagonals be a right angle, half the product of the diagonals is the area. The triangle ACD = AED+ DEC = {AE × EDx sin. E+ECX ED × sin. E = ACX ED x sin. E; and ABC ACX EBX sin. E. = 1. Required the area of the quadrilateral ABCD, of which the diagonals are AC 674 and BD 398 feet, and the acute angle at E 67° 30'. Nat. sine of 67° 30′ 92388 674 B E 622.69512 Ans. Area 123916-32888 square feet, 2 acres 3 roods 15 perches 4 yards. 2. Required the area of a parallelogram, the diagonals 436 nd 324 yards, and their angle 48° 38'. Ans. 53009 yards, = 10 acres 3 roods 32 perches 11 yards. 3. Required the area of a trapeze, the sides 856 and 643, he diagonal joining their extremities 1154, and the other 345 links, and the angle made by the diagonals 57° 30'. Ans. 6 acres 2 roods 7 perches 7 yards." 4. Required the area of a quadrilateral, the diagonals 72 and 18 feet, and containing a right angle. Ans. 192 yards. 5. The diagonals of a quadrilateral are 567 and 743 links, and they contain an angle of 73° 30′; the side joining their extremities opposite to this angle 324. Ans. 2 roods 3 perches 4 yards 33 feet. 6. Required the area of a quadrilateral, the diagonals 924 links and 1256, and they bisect one another in an angle of 52° 30'. Area 4 acres 2 roods 16 perches 17 yards 44 feet. NOTE 2. If the sides be given instead of the diagonals, Add the squares of each pair of opposite sides, and subtract the less sum from the greater: one-fourth of the remainder, multiplied by the natural tangent of the angle contained by the diagonals, will be the area.* See Appendix, Prop. 42. NOTE 3. When the quadrilateral is in a circle, or its opposite angles are together 180°, From half the perimeter subtract each side separately; multiply the four remainders successively, and the square root of the product will be the area. See Appendix, Prop. 44. 7. Required the area of a quadrilateral, of which the sides are 7, 8, 9, and 10 yards, and the angle contained by the diagonals 80°. 102+82164 4 34 8.5 Nat. tan. 80° = 5.67128 Ans. 48-20588 square yards. 8. Required the area of a trapeze in a circle, the sides 326, 438, 247, and 392 feet. Ans. 117976 square feet, = 2 ac. 2 ro. 33 per. 103 yds. * If a table of natural tangents be not at hand, multiply by the natural sine, and divide by the natural cosine. Or add the log. of half the remainder to the log. tangent: the sum is the log. of the area. 9. Required the area of a quadrilateral in a circle, the sides 24, 26, 28, 30 yards. Ans. 723-98895 yards, = 23 perches 281 yards. 10. Required the area of a quadrilateral, of which the opposite angles are together 180°, the sides 40, 55, 60, 75 chains. Ans. 3146-427 ch. 314 ac. 2 ro. 22 per. 25 yds. 1·532 ft. POLYGONS. PROB. XI. To find the area of any rectilineal figure. RULE. Draw diagonals so as to divide the figure into quadrilaterals and triangles, and find the areas of these figures separately, and add them: the sum is the area of the whole. 1. Required the area of the pentagon ABCDE, of which the sides are AB 354, BC 432, CD 518, DE 465, and EA 397 feet; and the diagonals A AC 574, and AD 612 feet. B E Whole figure, 7 ac. 5 per. 16 yds. 6.35 feet, 2. In order to obtain the area of the field ABCDE, I measured along the diagonal AC; and at b, 326 links from A, 306431-6 took the perpendicular 6E, 97 links: then I measured to c, 543 links from A, where I took the offset cB 354 links; and measuring on to d, 749 links from A, I took the links. The whole diagonal AC is 987 links. area. By Prob. VII. EbdD=4(Eb+Dd) x bd = E B offset dD 158 Required the 53932.5 links. 15811-0 DdCdc xdD= 18802.0 Area of whole, 2 ac. 2 ro. 21 per. 5.78 yds. 263244.5 = 3. Required the area of the field B E y Prob. IV. . BFC4CFX LB = 359520.0 links. y Prob. IX. ABFG=&(AH+FK)XBG= 479053.5 FCDE= (FM+DN) × CE = 848815·0 Ans. 16 ac. 3 ro. 19 per. 24-85 yds. = 1687388.5 4. Measured along a diagonal from east west, at 230 from its east extremity, a erpendicular to it on the south side, of #56 links, reached to an angle, and at 380 rom the same extremity a perpendicular n the north side, of 428, reached an angle. At 673, a perpendicular of 560 reached an ngle on the south side; at 812, a perpen Hicular of 230 reached an angle on the north; at 1140, a perpendicular of 340 reached an angle on the south; and at he west extremity 1270, there was a perpendicular of 530 on the north side. Ans. 7 ac. 3 ro. 11 per. 4 yds. 61 feet. 5. In a hexagon are given the sides AB 536, BC 498, CD 620, DE 580, EF 398, and AF 492 links, and the diagonals AC 918, CE 1048, and AE 652 links. Ans. 6 ac. 2 ro. 9 per. 23 yds. 8.413 feet. 6. In a heptagon are given the sides AB 294, BC 456, CD 572, DE 640, EF 612, FG 498, and GA 386, and the diagonals AC 540, AD 864, AE 630, and AF 490 links. Ans. 6 ac. 1 ro. 34 per. 18.3 yds. B D A F E |