EXAMPLES. 1. Required the area of a square field, a side of which measures 7.29 four-pole chains. 2. Required the area of a rectangular field whose length is 13.75 chains, and breadth 9.5 chains. 3. Required the area of a field, in the form of a rhomboides, whose length AB is 42.5 perches, and perpendicular breadth CD is 32 perches. Fig. 15. P. 42.5 32 850 1275 40)1360.0 4)34 8A. 2R. 4. What is the area of a square tract of land, whose side measures 176.4 perches? Ans. 194A. 1R. 36.96P. 5. What is the area of a rectangular plantation whose length is 52.25 chains, and breadth 38.24 chains? Ans. 199A. 3R. 8.6P. 6. The length of a field, in the form of a rhombus, measures 16.54 chains, and the perpendicular breadth 12.37 chains: required the area. Ans. 20A. 1R. 33.6P. 7. Required the area of a field in the form of a rhomboides, whose length is 21.16 chains, and perpendicular breadth 11.32 chains. Ans. 23A. 3R. 32.5P. PROBLEM II. To find the area of a triangle when the base and pendicular height are given. RULE. per Multiply the base by the perpendicular height, and half the product will be the area.* • DEMONSTRATION. A triangle is half a parallelogram of the same base EXAMPLES. 1. The base AB of a triangular piece of ground, measures 12.38 chains, and the perpendicular CD 6.78 chains required the area. Fig. 49. 2. Required the area of a triangular field, one side of which measures 18.37 chains, and the distance from this side to the opposite angle 13.44 chains. Ans. 12A. 1R. 15P. 3. What is the area of a triangle whose base is 49 perches and height 34 perches? Ans. 5A. OR. 33P. PROBLEM III. To find the area of a triangle when two sides and their included angle are given. RULE. As radius, Is to the sine of the included angle ; EXAMPLES. 4. In a triangular lot of ground ABC, the side AB measures 64 perches, the side AC 40.5 perches, and their contained angle CAB 30°: required the area. Fig. * DEMONSTRATION. In the triangle ABC, Fig. 49. let AB and AC be the given sides, including the given angle A, and let CD be perpendicular on AB. Then by trig. rad. : sin. A :: AC: CD; but (cor. 1.6.) AC : CD :: AC X AB : CD X AB; therefore (11.5.) rad. : sin. A :: AC × AB: CD × AB; but CD X AB is equal to twice the area of the triangle: hence the truth of the rule is 2. What is the area of a triangle two sides of which measure 15.36 chains and 11.46 chains respectively, and their included angle 47° 30′? Ans. 6A. 1R. 38P. 3. One side of a triangular field bears N. 12° E. distance 18.23 chains, and at the same station the other adjacent side bears N. 78° 30′ E. distance 13.84 chains : required the area. Ans. 11A. 2R. 11P. 4. Required the area of a triangular piece of ground, one side of which bears N. 82° 30′ W. dist. 19.74 chains and at the same station, the other adjacent side S. 24° 15' E. dist. 17.34 chains. Ans. 14A. 2R. 8P. PROBLEM IV. To find the area of a triangle when one side and the two adjacent angles are given. RULE. Subtract the sum of the two given angles from 180°, the remainder will be the angle opposite the given side. Then, As the rectangle of radius and the sine of the angle opposite the given side, Is to the rectangle of the sines of the other angles, To double the area.* * DEMONSTRATION. Let AB, Fig. 49, be the given side of the triangle ABC, and A and B the given angles; also let CD be perpendicular on AB: Then by trig. Therefore (23.6.) rad. × sin. ACB : sin. A × sin. B :: AB X AC : CD × AC :: (cor. 1.6.) AB: CD :: AB2: AB X CD; but AB X CD is equal to double the area of the triangle ABC; therefore (11.5.) rad. X sin. ACB: sin. A X |