2. Required the number of square links in 96 acres, 2 roods, and 36 perches.

PROBLEM II.

Ans. 9672500.

To lay out, in a Square, any Quantity of Land proposed. RULE. Extract the square root of the proposed area, and it will be the side of the square required.

EXAMPLES.

1. Lay out, in a square, 7 acres, 1 rood, and 24 perches.

sides, which make AD, which make which make also

In laying out the square, in the field, let A B represent one of its 860.2 links. At A, erect the perpendicular AB; and at B, erect the perpendicular B C, А В. Then measure the line C D, and if you find it 860.2 links, the work is right.

=

=

2. Required the side of a square, which shall contain 15 acres, 2 roods, and 32 perches. Ans. 1253 links.

PROBLEM III.

Upon a given Line, to make a Rectangle that shall contain any proposed Quantity of Land.

RULE.-Divide the proposed area by the given side, and the quotient will be the other side of the rectangle.

EXAMPLES.

1. Lay out 3A. 3R. 26P. in the form of a rectangle, one side of which must be 850 links.

In laying out the rectangle in the field, let A B represent the given side. At A, erect the perpendicular A D, which make = 460.3 links ; and at B, erect the perpendicular B C, which make = AD. Then measure the line CD, and if you find it = A B, the work is right.

2. If one side of a rectangle be 525 links; required the other side, so that the figure may contain 6A. 2R. 23P. Ans. 1265.5 links.

PROBLEM IV.

To lay out any given Quantity of Land in a Rectangle, so that one of its sides shall be two, three, four, or any number of times as long as the other.

RULE.— Divide the given area by the given number, and the square root of the quotient will be the shorter side, which multiply by the given number, and the product will be the longer side.

EXAMPLES.

1. Lay out 3A. OR. 32P. in the form of a rectangle, one of the sides of which shall be twice as long as the other.

Let ABCD represent the rectangle, which you must lay out according to the directions in the last problem; AD being 800, and A B 400 links.

2. A rectangle contains 7A. 2R. OP.; what are its sides, one of them being three times the length of the other?

Ans. 1500 and 500 links.

PROBLEM V.

Upon a given Base, to lay out a Triangle that shall contain any given number of Acres, &c.

RULE. Divide the area by half the base, or twice the area by the whole base, and the quotient will be the perpendicular of the triangle.

EXAMPLES.

1. Lay out 3A. 2 R. 16P. in the form of a triangle, the base of which must be 1200 links.

Upon any part of the given base AB, suppose at D, erect the perpendicular DC, which make = 600 links; then stake out the line AC and BC; so will ABC be the required triangle. But if the perpendicular be erected at either end of the base, as at B, then the line AE must be staked out; and ABE will be the triangle required.

2. Required the perpendicular of a triangle, which contains 6a. 2R. 37P., its base being 1556 links. Ans. 865.2 links.

PROBLEM VI.

To lay out a Trapezium, that shall contain any Number of Acres, &c.; having one of its Sides or a base Line given.

RULE 1. Divide the given area into two parts, either equal or unequal; and then, by the last problem, find the perpendicular, that will lay out one of these parts in a right-angled triangle, upon the given base.

You must then consider this perpendicular as one of the diagonals of the trapezium, and also the base upon which you must lay out the other triangle.

RULE 2. Divide the given area into any two parts, as before and then, find the perpendicular that will lay out one of these parts in a right-angled triangle, upon the given base.

Add the square of the perpendicular thus found to the square of the given base, and the square root of the sum will be the hypothenuse. Consider this hypothenuse as one of the diagonals of the trapezium, and also the base upon which the other triangle must be laid out.

EXAMPLES.

1. Lay out 8A. in a trapezium, upon a given side of 800 links.

BY THE FIRST RULE.

Divide the given area into 5 and 3 acres, and let the triangle upon the given side contain the greater part.

1250 links, the perpendicular of the first triangle, and also the base of the second.

3A. 300000 square links.

2

125,0)60000,0(480 links, the perpendicular of the second triangle.

In laying out the trapezium, in the field, let AB represent the given side. At B, erect the perpendicular B C, which make = 1250 links. Then, upon any part of the line B C, as at D, erect the perpendicular DE, which make = 480 links. The four outlines being properly staked out, the work will be completed.

BY THE SECOND RULE.

5A. = 500000 square links.

2

8,00)10000,00

1250 links, the perpendicular of the first triangle.

Then, √12502 + 8002 = √1562500 + 640000

=

√2202500

1484 links, the hypothenuse of the first, and also the base of the

second triangle.