PROBLEM VII. To find the Area of a Square or Parallelogram. Rule. Multiply the length into the breadth ; the Product will be the Area. PROBLEM VIII. To find the Area of a Rhombus or Rhomboides. RULE. Drop a Perpendicular from one of the Angles to its opposite Side, and multiply that Side into the Perpendicular; the Product will be the Area. PROBLEM IX. To find the Area of a Triangle. Rule 1. Drop a Perpendicular from one of the Angles to its opposite Side, which may be called the Base; then multiply the Base by half the Perpendicular, or the Perpendicular by half the Base; the Product will be the Area. Or, multiply the whole Base by the whole Perpendicular, and half the Product will be the Area. Rule 2. If it be a Right Angled Triangle, multiply one of the Legs into half the other; the Product will be the Area.. Or, multiply the two Legs into each other, and balf the Product will be the Area. RULE 3. When the three sides of a Triangle are known, the Area may be found Arithmetically, as follows: Add together the three Sides ; from halt their Sum subtract each side, noting down the Remainders ; multiply the half Sum by one of those Remainders, and that Product by another Remainder, and that Product by the other Remainder; the Square Root of the last Product will be the Area. Example. Suppose a Triangle whose three Sides are 24, 20, and 18 Chains. Demanded the Area. 24 +20 +18=62, the Sum of the three Sides, the half of which is 31. From 31 subtract 24, 20, and 18; the three Remainders will be 7, 11, and 13. 31x7=217; 217x11=2387 ; 2387 X 13=31031, the Square Root of which is 176.1, or 17 Acres 2 Roods and 17 Rods. By Logarithms. As the Addition of Logarithms is the same as the Multiplication of their corresponding Numbers; and as the Number answering to the one half of a Logarithm will be the Square Root of the Number corresponding to that Logarithm: it follows, That if the Logarithm of the half Sum of the three sides 40 SURVEYING. the Number corresponding to one half the Sum of those Logo arithms will be the Area of the Triangle. The half Sum, 31 1.49136 The first Remainder, 7 0.84510 The second Remainder, 11 1.04139 The third Remainder, 13 1.11394 The Square of the Area, 31000 4.49179 Area 176 Square Chains 2.24589 RULE 4. When two sides of a Triangle and their contained Angle, that is, the Angle made by those Sides, are given, the Area may be found as follows : Add together the Logarithms of the two Sides and the Logarithmic Sine of the Angle; from their sum subtract the Logarithm of Radius, the Remainder will be the Logarithm of double the Area. EXAMPLE. Suppose a Triangle one of whose Sides is 105 Rods and another 85, and the Angle contained between them 280 5'. Demanded the Area. One Side, 105 2.02119 The other Side, .85 1.92942 Sine Angle, 280 5 9.67280 13.62341 10.00000 Subtract Radius Double Area, 4200 Rods 3.62341 Answer. 2100 Rods. Note. Radius may be subtracted by cancelling the Left-hand figure of the Index, or subtracting 10, without the trouble of setting down the Ci: phers. By Natural Sines. Multiply the two given Sides into each other, and that Product by the Natural Sine of the given Angle; the last Product will be double the Area of the Triangle. Nat. Sine of the Angle 280 57 0.47076 105 X 85=8925, and 8925 X 0.47076—4201 the double Area of the Triangle. Rule. Multiply half the Sum of the two parallel Sides by the perpendicular distance between them, or the sum of the two parallel Sides by half the perpendicular distance, the product will be the Area, PROBLEM XI. To find the Area of a Trapezium, or irregular Four Sided Figure. Rule. Draw a Diagonal between two opposite Angles, which will divide the Trapezium into two Triangles. Find the Area of each Triangle and add them together. Or, multiply the Diagonal by half the Sum of the two perpendiculars let_fall upon it, or the Sum of the two perpendiculars by half the Diagonal, the product will be the Area. Note. Where the length of the four Sides and of the Diagonal is known, the Area of the two Triangles, into which the Trapezium is divided, may be calculated Arithmetically, according to PROB. IX. Rule 3. PROBLEM XII. To find the Area of a Figure containing more than Four Sides. Rule. Divide the Figure into Triangles, and Trapezia, by drawing as many Diagonals as are necessary, which Diagonals must be so drawn as not to intersect each other; then find the Area of each of the several Triangles or Trapezia, and add them together; the sum will be the Area of the whole Figure. Note. A little practice will suggest the most convenient way of drawing the Diagonals; but whicherer way they are drawn, provided they do not intersect each other, the whole Area will be found the same. PROBLEM XIII. Respecting Circles. Rule 1. If the Diameter be given the Circumference may be found by one of the following proportions: as 7 is to 22, or more exactly, as 113 is to 355, or in Decimals, as 1 is to 3.14159, so is the Diameter to the Circumference. RULE 2. If the Circumference be given the Diameter may be found by one of the following proportions: as 22 is to 7, or as 355 is to 113, or as 1 is to 0.31831, so is the Circumference to the Diameter. Rule 3. The Diameter and Circumference being known, multiply half the one into half the other, and the product will be the Area. RULE 4. From the Diameter only, to find the Area: multiply the Square of the Diameter by 0.7854, and the product will, be the Area. RULE 5. From the Circumference only to find the Area : multiply the Square of the Circumference by 0.07958, and the product will be the Area. Rule 6. The Area being given to find the Diameter : divide the Area by 0.7854, and the Quotient will be the Square of the Diameter; from this extract the Square Root, and you will have the Diameter. RULE 7. The Area being given to find the Circumference : divide the Area by 0.07958, and the quotient will be the Square of the Circumference; from this extract the Square Root, and you will have the Circumference. SECTION 11. The following Cases teach the most usual methods of taking the Survey of Fields; alsс, how to protract or draw a Plot of them, and to calculate their Area. Note. The Field Book is a Register containing the length of the Sides of a Field, as found by measuring them with a Chain ; also the Bearings or Courses of the Sides, or the Quantity of the several Angles, as found by a Compass or other instrument for that purpose; together with such Remarks as the Surveyor thinks proper to make in the Field. CASE I. To survey a Triangular Field. Measure the sides of the Field with a Chain, and enter their several lengths in a Field Book, protract the Field on Paper, and then find the Area by PROB. IX. Rule 1. Or, without plotting the Field, calculate the Area by PROB. IX. Rule 3. Fig. 46. C Chains. 20 24 CA 18 To find the Area. Ch. L. Base BC 24.00 Half Perp. AD - 17.34 A B 20 Ch. 9600 24.Ch. D 18 Ch. Gà L 14 68 O 9600 7200 16800 Acres 17)61600 4 Roods 2) 46400 40 Rods 18) 56000 Acres Roods Rods Area 17 2 18.56 Note. When there are ciphers at the Right Hand of the Links, they may be rejected; remembering to cut off a proper number of figures according to Decimal Rules. Observe, That in measuring with a Chain, slant or inclined Surfaces, as the Sides of Hills, should be measured horizontally, and not on the Plane or Surface of the Hill; otherwise, a survey cannot be accurately taken. To effect this, the lower end of the Chain must be raised from the ground, so as to have the whole in a horizontal Line ; and the end thus raised must be directly over the Point where the Chain begins or ends, according as you are ascending or descending a Hill; which Point may be ascertained by a Plummet and Line. CASE II. To survey a field in the form of a Trapezium. Measure the several sides, and a Diagonal between two opposite Angles ; protract the Field, and find the Area by ProbLEM XI. Or, without protracting the Field, calculate the Area according to the Note at the end of that PROBLEM. Fig. 51. 21 27.50 11.70 CD, 21.50 DA 14.70 50 a. 28 |