square of the distance of the focus from the centre, is equal to the difference of the squares of the semi-diameters: hence, we have √2252-141.52 = √30602.75 175 links, equal F o, or fo: and, 225-175 = 50 links, equal A F, or B f.

m

F

C

D

Again, by another property of the ellipse, the sum of the two lines drawn from the foci, and meeting in any B point in the circumference, is equal to the transverse diameter; that is, Fm + f m = A B.

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 A B; putting down a stake at each focus, and one at the centre o. At o, erect the perpendiculars o C and o D, making each = 141.5 links.

Put the two loops over the stakes at F, f, and stretch the cord, so that the two parts F m, ƒ m, 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. NOTE. - As surveyors are frequently requested to lay out, in various figures, small quantities of land for plantations, &c. it is presumed that the foregoing problems will be found not without their use.

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 A B C D containing 6A. 1R. 26P., part off 3A. by a line parallel to A B.

Then,

300000
800

640000 (800 links, the side of the square.
64

..0000

= 375 links, the side A E, or B F required.

2. From the rectangle A B C D containing 8A. IR. 24P., part off 2A. 1R. 32P. by a line parallel to A D = 700 links. Then, from the remainder of the rectangle, part off 2A. 3R. 25P. by a line parallel to A B.

850)290625(342 links, the side E G, or B H.

2550

.3562

3400

. 1625

1700

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 A B C D 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 A D.

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

=

2

9,00)4950,00

550 links, the perpendicular A E.

Hence A B E is the triangle required.

2. From A B C D 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 C D.

758)337740(445.5 links, the perpendicular C E.

3032 Hence A B ED is the trapezoid required.

.3454

3032

.4220

3790

.4300

3790

.510

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

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 A B C representing a triangle, whose base A B is 1200, and sides A C and B C, 1000 and 800 links respectively, part off 2a. 2R. 24P. by a line drawn from the angle B to the side A C.

441.6 links, the perpendicular D E.

At A, erect the perpendicular A F, which make 441.6 links; then draw F E parallel to A B, and it will intersect the side A C, in the point to which the division-fence B E must be made.

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

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

2. From A B C representing a triangle, whose base A B is 1300, and sides B C and A C, 1100 and 900 links respectively, part off la. 3R. 36P. by a line drawn from the angle A to the side B C, so that the triangle A E C may contain the proposed quantity.