MENSURATION OF SUPERFICIES. THE area of any figure is the measure of its surface, or the space contained within the bounds of the surface, without any regard to thickness. The area is estimated by the number of squares contained in the surface, the side of those squares being either an inch, a foot, a yard, &e. And hence the area is said to be so many square inches, or square feet, or square yards, &c. Our ordinary lineal measures, or measures of length, are as in the first of the following tables; and the annexed table of square measures is taken from it by squaring the several numbers. LINEAL MEASURES. To find the area of a parallelogram; whether it be a square, a rectangle, a rhombus, or a rhomboid. RULE.* Multiply the length by the breadth, or a perpendicular height, and the product will be the area. Take any rectangle ABCD, and divide each of ts sides into as many equal parts as is expressed by the number of times they contain the linear 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, or the area of the figure, is equal to the number of linear measuring units in the length as often repeated, as there are linear measuring units in the breadth or height, that is, equal to the length multiplied by the height, which is the rule. And since a rectangle is equal to an oblique parallelogram standing upon the same base, and between the same parallels; (Euc. I. 35) therefore the rule is true for any parallelogram in general. Q. E. D. EXAMPLES. 1. To find the area of a square, whose side is 6 inches, or 6 feet, &c. 2. To find the area of a rectangle, whose length is 9, and breadth 4 inches, or feet, &c. RULE 2. If any two sides of a parallelogram be multiplied together, and the product again by the natural sine of their included angle, the last product will give the area of the parallelo❤ gram. DEMONSTRATION. For having drawn the perpendicular AP, the area, by the first rule, is APXBC; but as rad. 1 (s. ≤P) :s. ZB:: AB: AP = s. <B×AB; therefore AP × BC=BC × s. ZB X AB is the area. A 3. To find the area of a rhombus, whose length is 6.20 chains, and perpendicular height 5.45. 4. To find the area of the rhomboid, whose length is 12 feet 3 inches, and breadth 5 feet 4 inches. 5. To find the area of a rectangular board, whose length is 12.5 feet, and breadth 9 inches. Ans. 93 ft. 6. To find the square yards of painting in a rhomboid, whose length is 37 feet, and breadth 54 feet. Ans. 217 square yards, 1 2 PROBLEM II. To find the area of a triangle. RULE 1.* Multiply the base by the perpendicular height, and half the product will be the area. RULE 2.† When the three sides only are given;-add the three sides together, and take half the sum; from the half sum subtract each side separately; multiply the half sum and the three remainders continually together; and the square root of the last product will be the area of the triangle. * A triangle is half a parallelogram of the same base and altitude,) Euc. I. 41) and therefore the truth of the rule is evident. Now, as Cp Cm, Ap-An, and Bm Bn, it is evident, that CH, or CL, will half the perimeter of the triangle, and that HA, Ap, and C will be the differences between the half perimeter and each side respectively. And since CH-CL, CG common, and the angle HCG angle LCG, therefore GLGH, and the angle GLC angle GHC a right angle. Also, = |