Page images
PDF
EPUB

105. An acre is a surface equal in extent to 10 square chains; that is, equal to a rectangle of which one side is ten chains, and the other side one chain.

One-quarter of an acre, is called a rood.

Since the chain is 4 rods in length, 1 square chain contains 16 square rods; and therefore, an acre, which is 10 square chains, contains 160 square rods, and a rood contains 40 square rods. The square rods are called perches.

106. Land is generally computed in acres, roods, and perches, which are respectively designated by the letters A. R. P.

When the linear dimensions of a survey are chains or links, the area will be expressed in square chains or square links, and it is necessary to form a rule for reducing this area to acres, roods, and perches. For this purpose, let us form the following

TABLE.

1 square chain=10000 square links.
1 acre=10 square chains=100000 square links.
1 acre=4 roods=160 perches.

1 square mile=6400 square chains=640 acres. Now, when the linear dimensions are links, the area will be expressed in square links, and may be reduced to acres by dividing by 100000, the number of square links in an acre: that is, by pointing off five decimal places from the right hand.

If the decimal part be then multiplied by 4, and five places of decimals pointed off from the right hand, the figures to the left will express the roods.

If the decimal part of this result be now multiplied by 40, and five places for decimals pointed off, as before, the figures to the left will express the perches.

If one of the dimensions be in links, and the other in chains, the chains may be reduced to links by annexing two ciphers: or, the multiplication may be made without annexing the ciphers, and the product reduced to acres and decimals of an acre, by pointing off three decimal places at the right hand.

When both the dimensions are in chains, the product is reduced to acres by dividing by 10, or pointing off one deci

From which we conclude; that,

1st. If links be multiplied by links, the product is reduced to acres by pointing off five decimal places from the right hand. 2d. If chains be multiplied by links, the product is reduced to acres by pointing off three decimal places from the right hand. 3d. If chains be multiplied by chains, the product is reduced to acres by pointing off one decimal place from the right hand. 107. Since there are 16.5 feet in a rod, a square rod is equal to 16.5 × 16.5=272.25 square feet.

If the last number be multiplied by 160, we shall have 272.25 × 160=43560=the square feet in an acre. Since there are 9 square feet in a square yard, if the last number be divided by 9, we obtain

4840=the number of square yards in an acre.

PROBLEM I.

108. To find the area of a square or rectangular piece of ground.

Multiply the two sides together, and the product will express the area (Geom. Bk. IV, Prop. IV).

1. To find the area of the rectangular field ABCD.

Measure the two sides AB, BC: let us suppose that we have found AB=14 chains 27 links, and BC=9 chains 75 links. Then,

AB=1427 links,
BC= 975 links,

AB×BC=1391325 square links,

13.91325 acres.

4

3.65300 rods,

40

26.12000 perches.

D

A

Ans. 13A 3R 26P.

C

B

2. What is the area of a square field, of which the sides are each 33 ch 81?

3. What is the content of a rectangular field, of which the longest side is 49 ch 271, and the shorter 38 ch 71?

Ans. 187A 2R 11P.

PROBLEM II.

109. To find the content of a piece of land in the form of a triangle.

FIRST METHOD.

A

Measure either side of the triangle as BC, and from the opposite angle A let fall a perpendicular AD, and measure this perpendicular; then, multiply the base and perpendicular together, and divide the product by 2, the result will express the area of the triangle. Or, the area is equal to the base multiplied by half the perpendicular, or to the perpendicular multiplied by half the base (Geom. Bk. IV, Prop. II).

B

D

C

1. What is the content of a triangle whose base is 25 ch 11, and perpendicular 18 ch 141?

Ans. 22A 2R 29P.

2. What is the content of a triangle whose base is 15.48 chains, and altitude 9.67 chains?

Ans. 7A 1R 38 P

SECOND METHOD.

Measure two sides and their included angle. Then, add together the logarithms of the two sides and the logarithmic sine of their included angle; from this sum subtract the logarithm of the radius, which is 10, and the remainder will be the logarithm of double the area of the triangle. Find, from the table, the number answering to this logarithm, and divide it by 2; the quotient will be the required area (Geom. Mens. Prob. II).

1. In a triangle ABC, suppose that we have found AB= 57.65 ch, AC=125.81ch, and the included angle CAB=

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

REMARK. In this example, the links are treated as decimal parts of the chain; the result, therefore, is in square chains and decimal parts of a square chain.

2. What is the area of a triangle whose sides are 30 and 40 chains, and their included angle 28° 57′?

THIRD METHOD.

Ans. 29A OR 7P.

Measure the three sides of the triangle. Then, add them together and take half their sum. From this half sum subtract each side separately. Then, multiply the half sum and the three remainders together, and extract the square root of the product: the result will be the area (Gcom. Mens. Prob. II).

Or, after having obtained the three remainders, add together the logarithm of the half sum and the logarithms of the respective remainders, and divide their sum by 2: the quotient will be the logarithm of the area.

1. Find the area of a triangular piece of ground whose sides are 20, 30, and 40 chains.

[blocks in formation]

45×25×15×5=84375: and √84375=290.4737=the area.

Ans. 29A OR 8P.

2. What is the area of a triangle whose sides are 2569, 4900, and 5035 links?

[blocks in formation]

110. To find the area of a piece of land in the form of a trapezoid.

Measure the two parallel sides, and also the perpendicular distance between them. Add the two parallel sides together, and take half the sum; then multiply the half sum by the perpendicular, and the product will be the area (Geom. Bk. IV. Prop. VII).

1. What is the area of a trapezoid, of which the parallel sides are 30 and 49 chains, and the perpendicular distance between them 16 ch 601, or 16.60 chains? 30+49=79; dividing by 2, gives multiply by

gives for the area in square chains,

30

16.60

49

39.5

16.60

655.700

Ans. 65A 2R 11P.

2. Required the content, when the parallel sides are 20 and 32 ch, and the perpendicular distance between them 26 ch. Ans. 67A 2R 16P.

PROBLEM IV.

111. To find the area of a piece of land in the form of a quadrilateral.

Measure the four sides of the quadrilateral, and also one of the diagonals: the quadrilateral will thus be divided inta

« PreviousContinue »