EXAMPLES. Ex. 1. To find the area of a triangle, whose base is 625, and perpendicular height 520 links? Here 625 × 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 feet? Ans. 663 square yards. Ex. 3. To find the number of square yards in a triangle, whose base is 49 feet, and height 25 feet? Ans. 685, or 68.7361. Ex. 4. To find the area of a triangle, whose base is 18 feet 4 inches, and height 11 feet 10 inches? Ans. 108 feet, 5 inches. RULE II. 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, multiply 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' ? 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 trigonometry, as sin. Z P, or radius: AC:: sin. A: CP, which is therefore AC X sin. ▲ A, taking radius 1. Therefore the area AB X CP is AB X AC X sin. ZA, to radius 1; er, as radius: sin. 4 A :: C I AB X AC: the area. 2 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. 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, RD, meeting the two sides AC, CB, produced, in D and E, and making CD = CB, and CECA. Also draw CFg bisecting DB and AE perpendicularly in F and G; and FHI parallel to the side AB, meeting AC in H, and AE produced in I. 1 K B Lastly, with centre H, and radius HF, describe a circle meeting AC produced in K; which will pass through G, because G is a right angle, and through I, because, by means of the parallels, AI FB = DF, therefore HD = HA, and HF = HI=LAB. HC = = IF Hence HA or HD is half the difference of the sides AC, CB, and half their sum or = ACCB; also HK = HI or AB; conseq. CK = AC + CB + AB half the sum of all the three sides of the triangle ABC, or CK of those three sides. Again HK = HI = is, calling s the sum IF AB, or KL AB; CAS theref. CL CK KLS AB, and AK = CK - ▲ ACE, and AG. FG = Also by the parallels, (AACB =) CG. IA = CG. DF, conseq. AG. CF. CG. DF A2ACB. But CG.CFCK. CL=1s. (1s — AB), and AG . DF = AK. AL = (s — AC). (SBC); theref. AG. CF. CG. DF= A2 ACB S. (S-AB).(S-AC). S-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: DF, drawing the first and second terms into CF, and the third and fourth into AG, the propor. becomes CG. CF AG.CF:: AG.CF:AG.DF, of CG. CF: A ABC:: AABC: BG. DF, that is, the AABC is a mean proportional between CG.CF and AG. DF, or between is. (s-AB) and (IS-AC). (‡S—BC). Q. E. D. Ex. 1. To find the area of the triangle whose three sides Then 45 x 25 x 15 x 5 = 84375, The root of which is 290-4737, the area. Ex. 2. How many square yards of plastering are in a triangle, whose sides are 30, 40, 50 feet? Ans. 663. Ex. 3. How many acres, &c. contains the triangle, whose sides are 2569, 4900, 5025 links? Ans. 61 acres, I 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 750 1975 × 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. 13 feet. Ex. 3. In measuring along one side AB of a quadrangular field, that side, and the two perpendiculars let fall on it from the two opposite corners, measured as follow: required the content. AP AP = 110 links 4Q= 745 AB 1110 CP = 352 DQ= 595 C Ans. 4 acres, 1 rood, 5'792 perches. AP PROBLEM IV. To find the Area of any Trapezium. DIVIDE the trapezium into two triangles by a diagonal; then find the areas of these triangles, and add them together. Or thus, let fall two perpendiculars on the diagonal from the other two opposite angles; then add these two perpendiculars together, and multiply that sum by the diagonal, taking half the product for the area of the trapezium. Ex. 1. To find the area of the trapezium, whose diagonal is 42, and the two perpendiculars on it 16 and 18. Ex. 2. How many square yards of paving are in the trapezium, whose diagonal is 65 feet, and the two perpendiculars let fall on it 28 and 33 feet? Ans. 222 yards. Ex. 3. In the quadrangular field ABCD, on account of obstructions there could only be taken the following measures, viz. the two sides BC 265 and AD 220 yards, the diagonal AC 378, and the two distances of the perpendiculars from the ends of the diagonal, namely, AE 100, and CF 70 yards. Required the construction of the figure, and the area in acres, when 4840 square yards make an acre ? Ans. 17 acres, 2 roods, 21 perches. PROBLEM V. To find the Area of an Irregular Polygon. DRAW diagonals dividing the proposed polygon into trapeziums and triangles. Then find the areas of all these separately, and add them together for the content of the whole polygon. EXAMPLE. EXAMPLE. To find the content of the irregular figure ABCDEFGA, in which are given the following diagonals and perpendiculars: namely, To find the Area of a Regular Polygon. RULE I. MULTIPLY the perimeter of the polygon, or sum of its sides, by the perpendicular drawn from its centre on one of its sides, and take half the product for the area *. Ex. 1. To find the area of a regular pentagon, each side being 25 feet, and the perpendicular from the centre on each side 17.2047737. Here 25 x 5 125 is the perimeter. RULE II. Square the side of the polygon; then multiply that square by the tabular area, or multiplier set against its name in the following table, and the product will be the area t. No. *This is only in effect resolving the polygon into as many equal triangles as it has sides, by drawing lines from the centre to all the angles; then finding their areas, and adding them all together. This rule is founded on the property, that like polygons, being similar figures, are to one another as the squares of their like sides; which is proved in the Geom. theor. 89. Now, the multipliers in the table, are the areas of the respective polygons to the side 1. Whence the rule is manifest, Note. |