MENSURATION OF PLANES. THE Area of any plane figure, is the measure of the space contained within its extremes or bounds ; without any regard to thickness. This area, or the content of the plane figure, is estimated by the number of little squares that may be contained in it; the side of those little measuring squares being an inch, a foot, a yard, or any other fixed quantity. And hence the area or content is said to be so many square inches, or square feet, or square yards, &c. To find the Area of any Parallelogram ; whether it be a Square, a Rectangle, a Rhombus, or a Rhomboid. MULTIPLY the length by the perpendicular breadth, or height, and the product will be the area. EXAMPLES # The truth of this rule is proved in the Geom. theor. 81, cor. 2. The same is otherwise proved thus : Let the foregoing rectangle be the figure proposed : and let the length and breadth be divided into several parts each equal to the linear measuring unit, being here 4 for the length, and 3 for the breadth ; and let the opposite points of division be connected by right lines. Then it is evident that these lines divide the rectangle into a number of little squares, each equal to the square measuring unit E ; and further, that the number of these little squares, or the area of the figure, is equal to the number of linear measuring units in the length, repeated as often as there are linear measuring EXAMPLES. Ex. 1. To find the area of a parallelogram, the length being 12:25, and height 8.5. 12.25 length 8.5 breadth 6125 9800 104.125 area Ex. 2. To find the area of a square, whose side is 35.25 chains. Ans. 124 acres, 1 rood, I perch Ex. 3. To find the area of a rectangular board, whose length is 121 feet, and breadth 9 inches. Ans of feet Ex. 4. To find the content of a piece of land, in form of a rhombus, its length being 6•20 chains, and perpendicular height 5 45 Ans. 3 acres, 1 rood, 20 perches, Ex. 5 To find the number of square yards of painting in a rhomboid, whose length is 37 feet, and breadth 5 feet 3 inches. Ans. 21, square yards PROBLEM ņ. t To find the Area of a Triangle. RULE I. MULTIPLY the base by the perpendicular height, and take half the product for the area*. Or, multiply the one of these dimensions by half the other. measuring units in the breadth, or height ; that is, equal to the length drawn into the height ; which here is 4 x 3 or 12. And it is proved. (Geom. theor. 25, cor. 2), that any oblique parallelogram is equal to a rectangle, of equal length and perpendicular breadth. Therefore the rule is general for all parallelograms whatever. • The truth of this rule is evident, because any triangle is the half of a parallelogram of equal base and altitude, by Geom. theor. 26. EXAMPLES. EXAMPLES. Ex. 1. To find the area of a triangle, whose base is 625, and perpendicular height 520 links? Here 625 X 260 = 162500 square links, or equal 1 acre, 2 roods, 20 perches, the answer. Ex. 2. How many square yards contains the triangle, whose base is 40, and perpendicular 30 leet? Ans. 66 square yards. Ex. 3. To find the number of square yards in a triangle, whose base is 49 feet, and height 254 feet? Ans 684, or 68.7361. Ex. 4. To find the area of a triangle, whose base is 18 feet 4 inches, and height 11 feel 10 inches ? Ans. 108 feet, 54 inches. Rule || When two sides and their contained angle are given : Multiply the two given sides together, and take half their product : Then say, as radius is to the sine of the given angle, so is that half product, to the area of the triangle. Or, multipiy that half product by the natural sine of the said angle, for the area". Ex. 1. What is the area of a triangle, whose two sides are 30 and 40, and their contained angle 28° 57' ? By Logarithnis. By Natural Numbers. First { X 40 X 30 = 600, then, 1 : 600 : : .484046 sin. 28° 57' 600 log. 9•684887 2 778151 Answer 290.4276 the area answering 2:463038 Ex 2. How many square yards contains the triangle, of which one angle is 45°, and its containing sides 25 and 214 feet? Ans. 20.86947 For, let AB, AC, be the two given sides, including the given angle A. Now | AB X CP is the area, by the first rule, Cp being the perpendicular. But, by trigonomi try, as sin. P, or rad us : AC:; sin. Z A: CP, which is therefore = AC X sin. LA.taking radius=1. Therefore the area JAB X cpis= } AB X AC xsin LA to rad u: 1; or, as radius : sin. < A:: | AB X AC : the area. RULE III. When the three sides are given : Add all the three sides together, and take half that sum. Next, subtract each side severally from the said half sum, obtaining three remainders. Then multiply the said half sum and those three remainders all together, and extract the square root of the last product, for the area of the triangle*. * For let abc be the given triangle Draw the parallels AE, BD, meeting the two sides AC, CB, produced, in D and E, and making CD CB, and DA B H I Ac in H, and AE produced in 1. Lastly, with centre h, and radi. K US HF, describe a circle meeting AC produced in K ; which will pass through g, because G is a right angle, and through 1, because, by means of the parallels, al = FB DF, therefore hd = HA, and AF = HI - JAB. Hence na or hp is half the difference of the sides Ac, cs, and нс half their sum or = {AC + CB ; also hk = HI = - 17 or AB; conseq. CK = - DAC + CB + JAB half the sum of all the three sides of the triangle ABC, or ck = $s, calling s the sum of those three sides. Again uk=H= fir='AB, OP KL=AB; theref. ct=ckKL = {s AB, and AK = CK is – AC, and al = DK = CK-CD= js-CB. Now, by the first rule, AG.cc = the A ACE, and AG, FG = the A ABE, theref. AG.CF = A ACB. Aise by the parallels, AG: CG :: DF or IA : CF, ther-f. Ag.cp=(A ACB =) CG.IA = CG.DF, conseq. AG.CF.CG • DF = APACB CA E But CG CF = CK . CL = {s. fs- AB, and AG, DF = AK. AL - BC; theref AG .cf .cg . DF = A ACB = is. is – AB - AC. js BC is the square of the area of the triangle ABC. Q: E, D. Otherwise, Because the rectangle AG , CF = the A ABC, and since cg : AG CF : Dr. drawing the first and second terms into CF, and the third and fourth into AG, the propor. becomes cg .ch: AG, CF :: AGGF: AG. DF, Or CG.CF: A ABC :: A ABO: BG.DF, that is, the A ABC is a mean proportional between cg .cf and AG. DF, or between 1s. As - AB and ds - Ac. js - BC. Q. E. D. Ex. 1. To find the area of the triangle whose three sides are 20, 30, 40. 20 45 45 45 40 15 2d rem. 5 3d rem. 2)90 45 half sum Then 45 x 25 x 15 x 5 = 84375, Ex. 2. How many square yards of plastering are in a triangle, whose sides are 30, 40, 50, feet? Ans. 66. Ex. 3. How many acres, &c. contains the triangle, whose sides are 2569, 4900, 5025 links? Ans. 61 acres, 1 rood, 39 perches. PROBLEM III. To find the Area of a Trapezoid Add together the two parallel sides; then multiply their sum by the perpendicular breadth, or the distance between them; and take half the product for the area. By Geom. theor. 29. Ex. 1. In a trapezoid, the parallel sides are 750 and 1225, and the perpendicular distance between them 1540 links : to find the area. 1225 1975 X 770 = 152075 square links = 15 acr. 33 perc. Ex. 2 How many square feet are contained in the plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches ? Ans. 134 feet. Es. 3. In measuring along one side as of a quadrangular field, that side, and the two perpendiculars let fall on it from the two opposite corners, measured as below, required the content |