15)63274 Square Links

4

2)53096

40

21)23840

Answer. 15 Acres 2 Roods and 21 Rods.

PROBLEMS for finding the Area of Right Lined Figures, and also of Circles.

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 half 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 half 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+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; 2387X13=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 and the Logarithms of the three Remainders be added together, the Number corresponding to one half the Sum of those Logarithms will be the Area of the Triangle.

The half Sum, 31

1.49136

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 28° 5'. Demanded the Area.

One Side, 105

The other Side, 85

Sine Angle, 28° 5'

2.02119

1.92942

9.67280

Note. Radius may be subtracted by cancelling the Left hand figure of the Index, or subtracting 10, without the trouble of setting down the Cyphers.

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 28° 5' 0.47076

105X85-8925, and 8925x0.47076-4201 the double Area of the Triangle.

PROBLEM X. To find the Area of a Trapezoid. 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.

PRÓBLEM 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

Note. A little practice will suggest the most convenient way of drawing the Diagonals; but whichever 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 Circumfer

ence.

SECTION II.

The following CASES teach the most usual methods of taking the Survey of Fields; also 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

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.

FIELD BOOK. See PLATE II. Fig. 46.

Chains.

Note. When there are cyphers 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