the centre is equal to the difference of the squares of the semi-dia

Procure, therefore, a cord, and upon it make two loops, so that the distance between them may be equal to the transverse diameter; then measure in the field the diameter AB, putting down a stake at each focus, and one at the centre o. At o erect the perpendiculars oc and oD, making each = 141.5 links.

Put the two loops over the stakes at F, f, and stretch the cord so that the two parts Fm, fm, may be equally tight; at m put down a stake as one point in the circumference of the ellipse, and in the same manner determine as many others as you please.

But if the ellipse be very large, so that you cannot conveniently procure a cord as long as the transverse diameter, you must then erect perpendiculars, called ordinates, at every 50 links, or at every chain's length, &c., upon that diameter, and measure the lengths of these perpendiculars by the scale.

Then measure in the field the transverse and conjugate diameters, and erect the perpendiculars in their proper places, always remembering to put down a stake at the end of each perpendicular. 2. Lay out an ellipse which shall contain 8A. 3R. 8P., one of the diameters being given equal to 800 links.

Ans. The other diameter is = 1400 links.

PROBLEM XI.

To part from a square or rectangle any proposed quantity of land by a line parallel to one of its sides.

RULE. Divide the proposed area by the side upon which it is to be parted off, and the quotient will be the length of the other side of the figure required.

Examples.

1. From the square ABCD, containing 6A. 1R. 26P., part off 3A. by

2. From the rectangle ABCD, containing 8A. 1R. 24P., part off 2A. 1R. 32P. by a line parallel to AD 700 links. Then, from

=

the remainder of the rectangle, part off 2A. 3R. 25P. by a line parallel to AB.

3. Part off 6A. 3R. 12P. from a rectangle containing 15A. by a line parallel to the longer side, the shorter being 1000 links.

Ans. The longer side of the given rectangle is 1500, and the shorter side of the rectangle required is 455 links.

PROBLEM XII.

To part from a square or rectangle any proposed quantity of land, either in a right-angled triangle or trapezoid, by a line drawn from any of the angles to either of the opposite sides.

RULE.-When the proposed area is to be parted off in a triangle, divide double this area by the base or side upon which it is to be parted off, and the quotient will be the perpendicular.

When the proposed area is to be parted off in a trapezoid, subtract it from the area of the square or rectangle, and part off the remainder in a triangle, as above directed.

Examples.

1. From ABCD, representing a square whose side is 900 links, part off a triangle which shall contain 2A. 1R. 36P. by a line drawn from the angle B to the side AD.

2A. 1R. 36P. 247500 square links

=

2

9,00)4950,00

550 links, the perpendicular AE.

Hence ACE is the triangle required.

D

B

2. From ABCD, representing a rectangle whose length is 1265 and breadth 758 links, part off a trapezoid which shall contain 7A. 3R. 24P. by a line drawn from the angle B to the side CD.

758) 337740 (445.5 links, the perpendicular CE.

3032

.3454

3032

.4220

3790

.4300

3790

.510

B

Hence ABED is the trapezoid required

3. From a rectangular field, whose length is 1560 and breadth 1000 links, it is required to part off a trapezoid which shall contain

P

12A. 3R. 12P. by a line drawn from any of the angles to the longer opposite side.

Ans. The area of the rectangle is 15A. 2R. 16P.; consequently the area of the triangle is 2A. 3R. 4P., and its perpendicular 555 links.

PROBLEM XIII.

To part from a triangle, upon the base or longest side, any proposed quantity of land by a line drawn from either of the angles at the base to the opposite side.

RULE.-Divide twice the proposed area by the base upon which it is to be parted off, and the quotient will be the perpendicular. Or, if the proposed area be divided by half the base, the quotient will be the perpendicular.

NOTE-A parallel ruler may be used with advantage in this and several of the following problems.

Examples.

1. From ABC, representing a triangle whose base AB is 1200, and sides AC and BC 1000 and 800 links respectively, part off 2a. 2R. 24P. by a line drawn from the angle B to the side AC.

4416 links, the perpendicular DE.

At A erect the perpendicular AF, which make = 4416 links; then draw FE parallel to AB, and it will intersect the side AC in the point to which the division-fence BE must be made.

Or, by the plotting scale, erect the perpendicular DE = 441.6 links, which will determine the point E.

By the scale you will find AE = 664 links; measure, therefore, in the field, 664 links from A to E; stake out the line BE, and ABE will be the triangle required.

2. From ABC, representing a triangle whose base AB is 1300, and sides BC and AC 1100 and 900 links respectively, part off 1A. 3R. 36P.