SECTION I. Containing rules for finding the areas of triangles, quadrilaterals, circles, and ellipses; also the method of protracting a survey and finding its area by dividing it into triangles and trapeziums. PROBLEM I. To find the Area of a Parallelogram, whether it be a Square, a Rectangle, a Rhombus, or a Rhomboides. RULE. Multiply the length by the height or perpendicular breadth, and the product will be the area.* Note. Because the length of a square is equal to its height, its area will be found by multiplying the side by itself. * DEMONSTRATION. Let ABCD (Fig. 68) be a rectangle; and let its length AB and CD, and its breadth AD and BC, be each divided into as many equal parts, as are expressed by the number of times they contain the lineal measuring unit; and let all the opposite points of division be connected by right lines. Then, it is evident that these lines divide the rectangle into a number of squares, each equal to the superficial measuring unit; and that the number of these squares is equal to the number of lineal measuring units in the length, as often repeated as there are lineal measuring units in the breadth, or height; that is, equal to the length drawn into the breadth. But the area is equal to the number of squares or superficial measuring units; and therefore the area of a rectangle is equal to the product of the length and breadth. Again, a rectangle is equal to any oblique parallelogram of an equal length and perpendicular height (36.1); therefore the area of 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 4|0)13610.0 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 perpendicular height are given. RULE. 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 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. Ch. 12.38 6.78 9904 8666 7428 2)83.9364 10)41.9682 Area, 4A. OR. 31P. 4.19682 4 .78728 40 31.49120 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. M PROBLEM III. To find the area of a triangle when two sides and their included angle are given. As radius, RULE. Is to the sine of the included angle; EXAMPLES. 1. 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 × AB: CD × AB; therefore (11.5.) rad.: sin. A:: AC X AB: CD × AB; but CD × AB is equal to twice the area |