PROB. II.

To find the side of a square, whose content is given. Extract the square root of the given content in perches, and you have the side in perches, and consequently in chains.

EXAMPLE.

It is required to lay out a square piece of ground which shall contain 12A. 3R. 16P. Required the number of chains in each side of the square; and to lay down a map of it, by a scale of 40 perches to an inch.

12. 3. 16, 2056; the square root of which is

From a scale where 4 of the large, or 40 of the small divisions are an inch, take 45. 34. the perches of the side, of which make a square.

PROB. III.

To find the content of an oblong piece of ground.

Multiply the length by the breadth, for the content.

Let ABCD be an oblong piece of ground, whose length AB is 14C. 25L. and breadth 8C. 37L. 1 demand the content in acres, and also to lay down a map of it, by a scale of 20 perches to an inch.

Make an oblong (by schol. to prob. 9. sect. 1.) whose length, from a scale of 20 to an inch, may

be

PROB. IV.

The content of an oblong piece of ground, and one side given, to find the other.

Divide the content in perches, by the given side in perches, the quotient is the required side in perches ; and thence it may be easily reduced to chains.

EXAMPLE.

Ch. L.

There is a fence 14.25 long, by the side of which it is required, to lay out an oblong piece of ground, which shall contain 3A. OR. 27P. what breadth must be laid off at each end of the fence, to enclose the 3A. OR. 27P.?

A. R. P.

P.

P

3. 0.27,=507. and 507 divided by 29,=17.28=

Ch. L.

8. 37.

The map is done as the last.

PROB. V.

To find the content of a piece of ground, in form of an oblique angular parallelogram ; or of a rhombus, or rhomboides.

Multiply the base into the perpendicular height. The reason is plain from theo. 13, sect. 1.

Let ABCD be a piece of ground in form of a rhombus, whose base AB is 22 chains, and perpendicular DE, or FC, 20 chains. Required the content.

A.

The content will be found=11.

The converse of this is done by prob. 4. and the map is drawn, by laying off the perpendicular on that part of the base from whence it was taken: joining the extremity thereof to that of the base, by a right line; and thence complete the parallelogram.

PROB. VI.

To find the content of a triangular piece of ground.

Multiply the base by half the perpendicular, or the perpendicular by half the base; or take half the product of the base into the perpendicular.

The reason hereof is plain, from cor. 2. theo, 12.

EXAMPLE.

fig. 16.

A

B

D

Let ABC be a triangular piece of ground, whose longest side or base BC, is 24C. 38L. and perpendicular AD, let fall from the opposite angle, is 13C. 28L. Required the content.

The map may be readily drawn, having the distance from either end of the base, to the perpendicular given; as may be evident from the figure.

PROB. VII.

The content of a triangular piece of ground, and the base given, to find the perpendicular.

Divide the content in perches, by half the base in perches and the quotient will give you the perpendicular in perches, and so in chains.

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