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Ex. 10. Wanting to know the extent of a piece of water, or distance between two headlands, I measured from each of them to a certain point inland, and found the two distances to be 735 yards and 840 yards; also the horizontal angle subtended between these two lines was 55° 40′, What then was the distance required?

Ans. 741. 2 yards.

Ex. 11. Wanting to know the distance between a house and a mill which were seen at a distance on the other side of a river, I measured a base line along the side where I was of 600 yards, and at each end of it took the angles subtended by the other end and the house and mill, which were as follow, viz. at one end the angles were 58° 20′ and 95° 20′, and at the other end the like angles were 53° 30′ and 98° 45'. What then was the distance between the house and mill?

Ans. 962.5866 yards.

Ex. 12. Wanting to know my distance from an inaccessible object 0, on the other side of a river, and having no instrument for taking angles, but only a chain or cord for measuring distances; from each of two stations, A and B, which were taken at 500 yards asunder, I measured, in a direct line from the object 0, 100 yards, viz. AC and BD each equal 100 yards; also the diagonal AD measured 550 yards, and the diagonal BC 560. What then was the distance of the object 0 from each station A and B?

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* These practical examples are taken from Hutton's Mathematics. Vol. II.

Seventh London Edition.

SECTION III.

MENSURATION OF AREAS, OR THE VARIOUS ME

THODS OF CALCULATING THE SUPERFICIAL
CONTENT OF ANY FIELD.

DEFINITION.

THE area or content of any plane surface, in

perches, is the number of square perches which that surface contains.

PL. 7. fig, 1.

Let ABCD represent a rectangular parallelogram, or oblong: let the side AB, or DC, contain 8 equal parts; and the side AD, or BC, three of such parts; let the line AB be moved in the direction of AD, till it has come to EF; where AE, or BF (the distance of it from its first situation) may be equal to one of the equal parts. Here it is evident, that the generated oblong ABEF, will contain as many squares as the side AB contains equal parts, which are 8; each square having for its side one of the equal parts, into which AB, or AD, is divided. Again, let AB move on till its comes to GH, so as GE, or HF, may be equal to AE, or BF; then it is plain that the oblong AGHB, wili contain twice as many squares as the side AB contains equal parts. After the same manner it will appear,that the oblong ADCB will contain three times as many squares as the side AB contains equal parts; and in general, that every rectangular parallelogram, whether square or oblong, contains as many squares as the product of the number of equal parts in the base, multiplied into the number of the same equal parts in the height, contains units, each square having for its side one of the equal parts.

Hence arises the solution of the following prob

lems.

PROB. I.

To find the content of a square piece of ground.

1. Multiply the base in perches, into the perpendicular in perches, the product will be the content in perches; and because 160 perches make an acre, it must thence follow, that

Any area, or content in perches, being divided by 160, will give the content in acres; the remaining perches, if more than 40, being divided by 40, will give the roods, and the last remainder, if any, will be perches.

Or thus:

2. Square the side in four-pole chains and links, and the product will be square four-pole chains and links: divide this by 10, or cut off one more than the decimals, which are five in all, from the right towards the left: the figures on the left are acres; because 10 square four-pole chains make an acre, and the remaining figures on the right, are decimal parts of an acre. Multiply the five figures to the right by 4, cutting 5 figures from the product, and if any figure be to the left of them, it is a rood, or roods; multiply the last cut off figures by 40, cutting off five, or (which is the same thing) by 4, cutting off four; and the remaining figures to the left, if any, are perches.

1. The first part is plain, from considering that a piece of ground in a square form, whose side is a perch, must contain a perch of ground; and that 40 such perches make a rood, and four roods an

acre; or which is the same thing, that 160 square perches make an acre, as before.

2. A square four-pole chain (that is, a piece of ground four poles or perches every way) must contain 160 square perches; and 160 perches make an acre, therefore 10 times 16 perches, or 10 square four-pole chains, make an acre.

NOTE. The chains given, or required, in any of the following problems, are supposed to be twopole chains, that chain being most commonly used; but they must be reduced to four-pole chains or perches for calculation, because the links will not operate with them as decimals.

EXAMPLES.

PL. 1. fig. 17.

Ch. L.

Let ABCD be a square field, whose side is 14.29, required the content in acres.

Ch. L.

By problem 4. section 1. part 2. 14. 29 are equal to 29.16 perches

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14. 29 are equal to 7. 29 of four-pole chains, by prob. 1. sect. 1. pt. 2. 7. 29

6561

1458

5103

A. R.P.

Acres 531441 cont. as before 5. 1. 10

4

Rood 125764

40

Perches 10130560

It is required to lay down a map of this piece of ground, by a scale of twenty perches to an inch.

Tale 29. 16 the perches of the given side, from the small diagonal on the common surveying scale, where 20 small, or two of the large divisions, are an inch make a square whose side is that length (by prob. 9. geom.) and it is done.

PROB. II.

To find the side of a square, whose content is given.

Extract the square root of the given content in perches, and you have the side in perches, and consequently in chains.

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