EXAMPLE. PL. 1. fig. 3. Let ABCD be an oblong piece of ground, whose length AB is 14C. 25L. and breadth 8C. 37L. required the content in acres, and also to lay down a map of it, by a scale of 20 perches to an inch. Ch. L. Perches. 14.25=29.00 8.37 27.48} By prob. 4. sect. 1. pt. 2. 15732 3496 A. R. P. 160)506.9200(3. 0. 27. content. 26 perches, or near 27. Or thus: 4 pole ch. Ch. L. Ch. L. 14.25=7.25} By prob. 1. sect. 1. pt. 2. 5075 2175 2900 Acres 3/16825 4 Rood 167300 4 Perches 2619200 To draw the map. Make an oblong (by schol. to prob. 9. geom.) whose length, from a scale of 20 to an inch, may be 29 perches, and breadth, 17.48 perches. 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 side required in perches; and thence it may be easily reduced to chains. EXAMPLE. There is a ditch 14 Ch. 25 L. long, by the side of which it is required to lay out an oblong piece of ground, which shall contain 3A. OR. 37P: what breadth must be laid off at each end of the ditch to enclose the 3A. 0R. 37P? The map is constructed like the last. PROB. V. To find the content of a piece of ground, in form of an oblique an÷ gular parallelogram; or of a rhombus, or rhomboides. Multiply the base into the perpendicular height, The reason is plain from theo. 13. geom. EXAMPLE. PL. 7. fig. 2. 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 con 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 completing 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 of this is plain, from cor. 2. theo. 12. geom. EXAMPLE. PL. 1. fig. 16. Let ABC be a triangular piece of ground, whose longest side or base BC, is 24C. 38 L. and perpendicular AD, let fall from the opposite angle, is 13 C. 28L. Required the content. Ch. L. Ch. L. 1. Base 24. 38 12. 38 1 = perp. 3. 39 11142 3714 3714 Acres 4/19682 4 Rood |78728 40 Perches 31 49120 A. R. P. Content 4. 0. 31. Ch. L. Perp. 13.28 perp. 6.39 Ch. L. 6.78 3.39 = four-pole chains by Or 2dly. Perp. 6.78 of four-pole chains. Or 3dly. Base 12.38 four-pole chains. Or the base and perpendicular may be reduced to perches; and the content may be thence obtained, thus: E e |