Rule. As the sum total of the proportionate shares is to the whole quantity of the estate to be divided, so is each proportionate share to its respective claim; thus:

Example 4. The same estate is to be divided between A B C, the value of the whole being 7501.; A is to be allotted land equal in value to 350l., B equal to 250l., and C equal to 1507.; thus:

To set out the above quantities on the ground.

Rule. Divide each separate quantity by the length of the side DE equal to 675 links; thus:

And so on in like manner for any number of shares.

Fig. 4, Plate 27.

Problem 41.

Example 5. To divide a triangular piece of ground into a given number of parts by right lines drawn from an angle to its opposite side.

Rule. Divide the base of the triangle into the required number of parts, then draw a line from the vertex to the division point; it will be the proportion required.

K

A triangular piece of ground, containing 26 acres 2 roods, is to be divided into three parts, bearing the proportion to the numbers 4, 2, 1, equal to 7; the length A B equal to 28 chains. By construction :

Divide the base A B, equal 28 chains, into 7 equal parts; from the angle C draw a line to the fourth division as Ca, and from C to the sixth division as C b, dividing the triangle as required.

Arithmetically:

Rule. As the sum of the ratios is to the length of the base A B, so is each respective part to the length required. As:

To find the quantity contained in each. As:

Example 6. The triangular field A B C is to be divided between two persons in the proportion of 2 to 5, having an equal right to the pond, or an occupation road to that point.

Geometrically:

Rule. Divide the side of the triangle A B into 7 equal parts; from C draw the line CE; parallel to which from a, the fifth point, draw the line a D; then draw the line D E, the division fence required.

Problem 43.

To lay this out on the ground:

From the plan, scale the length from C to D equal to 300 links, and from D to B 312 links; if correct, fix a stake at the point D.

Fig. 6, Plate 27.

Problem 44.

Example 7. The triangular field A B C is to be divided into three equal parts, reserving to each the right of water at the pond, or as an entrance from the occupation road.

Rule. Divide the line A B into three equal parts, as A a, a b, bB; draw the line C D; then from a, draw the line a c parallel to CD; and from b, draw the line bd parallel to CD; draw the lines c D and d D, which will be the division fences required. To lay out the same on the ground, measure the lengths from the plan as before, and drive stakes at the points c and d.

Fig. 7, Plate 27.

Problem 45.

Example 8. To divide a triangular field A B C, containing 4A. 1R. 8P., or 4.300, into three equal parts, parallel to the side B C, the length of each side being 10 chains.

Rule. Divide the side A B into three equal parts as a b; then find the mean proportion between A B and A a; by multiplying the two together, the square root of their product will be the mean proportion required, as A c; then draw a line from c parallel to B C, and B c C d will be equal to 4 of the triangle.

Thus: A B = 1000 × A a = 666′ = √666000 816 A c the mean

=

In the same manner divide the remaining part, A c, into two equal parts as at b; find the mean proportion between A c and Ab as before; mark off the length from A to e; draw the line from e parallel to B c, then c d e f will be the second part, and the triangle A ef will be the third part.

Thus : A c = 816 × A b = 408 = √ 332928 = 576 = A e the mean

Another method:

Problem 46.

Example 9. Rule. Find the content of the whole triangle and subtract from it the quantity to be cut off. As similar triangles are in proportion to the squares of their like sides, the triangle

K 2

Acd will be to the triangle A B C, as the square of A c is to square of A B.

Then, as the whole quantity is to the square of the side A B, so is the remaining quantity to the square of Ac; extract the square root, and the product will give the division required.

Note. The same method must be repeated for any number of parts the triangle is required to be divided.

Fig. 8, Plate 27.

Problem 47.

Example 10. It is required to cut off a portion of land equal to two acres, in the direction shown by the dotted line A B.

Note. When a piece of land has to be cut off, and the boundary very crooked, and no fixed point to measure from, to fix the stakes of the true line of division, that portion must be accurately measured, and a chain line fixed as near to the division as can be guessed, as at A B, and at each end put in temporary stakes. The survey must then be plotted, and scaled to this assumed line, from which the correct line CD can be calculated, returning then to the field to make the amendment.

When calculated, it was found to be minus 1 rood 3 poles, or .268 decimals; therefore divide the decimals .26800 by the length of AB equal to 660 links, the quotient will be 40 links nearly.

Then set off 40 links at each end of the line A B, which will be the exact division required.

Fig. 9, Plate 27.

Problem 48.

Example 11. It is required to cut off a portion of a field from a given point, A, to its opposite side, equal to 1 acre 1 rood.

Note.-Let this also be surveyed and the quantity calculated, as in the last example.

From the plan draw an assumed line as A B equal to 640 links; find the contents of the piece to the line A B equal to 1.887, or 1A. 3R. 22P., being minus the quantity required by .362 decimals, or OA. 1R. 18p.

Then divide .36200 by 640, the length of A B, the quotient will be 56 links, the width to be added if the amended line was parallel. As the increased quantity would form a triangle, this quotient must be doubled, or 112 links for the perpendicular to a, then will the line A a be the exact division required.

To set this out on the ground:

From the plan, scale the distance B a equal to 130 links, and at the point a put in a stake. Scale, 4 chains to the inch.

Example 12. A field containing 3A. OR. 6P. is to be divided into four equal portions, in such manner that access to the pond of water be given to each. By the Table, No. 13, 3 acres 6 poles = 3.03525 square links.

Thus: 3.035254.75881 to each share

No. 1. Set off the line A F as a fence; draw B F an assumed line; calculate the quantity A B F = .59040, which is less than the quantity required by .16841; divide this quantity by the length of B F = 380 equal to 44 links, being half the perpendicular; therefore set off 88 links, which will complete the first trapezium A Ba F.

No. 2. Draw the guess line C F, and calculate the quantity of a CF equal to .49099, less than the quantity required by .26782; divide this by the length CF equal to 37871, the half; set off the perpendicular, 142 links, which completes the trapezium a Cb F.

No. 3. Draw the guess line EF; calculate the quantity AEF equal to .56121 less than the quantity required by .19760; which divide by E F = 350 links; the quotient 56 the half, or 112 the perpendicular to c, which completes the trapezium A E C F.

No. 4. The remaining trapezium is to be cast up according to the previous examples, and is found to be .75881, the quantity required.

Problem 50.

Example 13. To divide a common field amongst sundry claimants according to the value per acre of the various parts of the common, and in proportion to the sum of each proprietor's share.

Rule. Divide the yearly value of each person's estate by the value per acre; then, as the sum of all the quotients is to the quantity of the whole common, so is each particular quotient to the quantity of each particular share.