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It is required to divide 400 acres of land among ABC, whose claims on it are respectively in proportion to the numbers 2, 3, 5.

2: 80

2+3+5=10: 400:3: 120

5: 200

It is required to divide 800 acres of land among A, B, C, whose estates are worth £1000, £2000, £3000 a year; the ground in their shares being worth 5, 10, and 20 shillings an acre, respectively.

Here the claims are as 1, 2, 3; and the qualities of their land are as 1, 2, 4: therefore their quantities must be as,,; or as 1, 1, &: then

1+1+3=23:800::

1: 29010 acres

A's share.

1: 29010 acres

B's share.

3 218 acres C's share.

These respective quantities may be cut off by the method described in one of the preceding problems.

PROP. 7.- To measure any inaccessible distance by the chain-as the breadth of a river.

Set up a mark at A, and from B measure any distance BC, leaving marks at B and C. Retire back to E and D, keeping EBA, and DCA, in a direct line, respectively; measure the lines BE and ED; then say, as ED-BC: BE :: BC: BA.

B

A

E

D

Let us suppose BC to measure 4 perches, BE 3, and ED 6; required the distance AB.

6

4

2:54: 10 perches, the distance AB.

This problem is very useful in conducting a chain survey. Let us suppose, in the progress of the survey, that one of your lines runs in the direction of EB; you then place a mark, in a line with EB, at A; and proceed as above. But if you wish to shew, on your plan, the method employed in crossing the river, EC or BD, which, with the measurements already made, will enable you to plot the work to a scale; then applying the scale to BA, you at once see the distance.

PROP. 8.- To reduce a Field, in the form of a trapezium, to a triangle of equal area.

Let it be required to construct a triangle CEB, which shall be equal in area to the trapezium ABCD.

A

D

B

C

E

Produce the line BA, towards E; lay a parallel ruler on AC, and slide it back to DE; then join EC, and

the triangle ECB will be equal to the trapezium ABCD.

This problem sometimes facilitates the finding of areas, as it easily transforms the trapezium into a triangle, the area of which is found from its base and perpendicular.

PROP. 9.-To reduce a five-sided figure to a triangle, having the same area.

Let ABCDE be the figure to be transformed into a triangle of equal area. Produce AB both ways,

[blocks in formation]

towards F and G. Apply the parallel ruler to AD, move it back to EF, and join FD. Again, apply it to BD, move it back to GC, and join GD; then the triangle FDG is equal in area to the figure ABCDE.

PROP. 10.—In a triangular Field, to draw from one of its angles, right lines to the opposite sides, dividing it into any number of parts, which shall be to each other in any assigned proportion.

Divide the base AB into the same number of parts,

[blocks in formation]

=

25

A

y

C

2 w D

E

5A. 3R. 18p. 586250 square links, and 586250÷3= 195416 square links each man's share. Draw EO, and set off the guess line ox; measure the trapezium AEor suppose you find its area to be⇒200216, or too much by 4800 square links, and the length of ox to be 400 links. Then 4800 200 24 links, the perpendicular from or to y. The trapezium OEAy is the first share.

From O set off a guess line to w; measure the trapezium OEDw; suppose you find its area to be = 190916 square links, or too little by 4500, and the length of Ox to be 450 links. Then 4500÷225= 20 links, the perpendicular from w to Oz. The trapezium Oy BCz is another share.

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