Fig. 9. PARTU. To find the area of an offset. Offsets are parts of a field which lie between a crooked hedge and the straight line which is measured for the side of the field, and these spaces are measured, by taking the perpendicular distances of the hedge from the measured line, at a great number of places, so that the hedge between these places may be nearly a ffraight line ; and then the spaces between the perpendiculars are found by the rule for finding the area of a quadrilateral, given in Prob. 4. of this part. Or if the measured fide be divided into equal parts by the perpendiculars, and meet the hedge at both its ends, add the perpendiculars together, and divide the sum by the number of parts into which they divide the base, and the quotient is a mean perpendicular, which, multiplied by the base, gives the area. If there be perpendiculars at the ends of the base, half their sum should be added to the other perpendiculars before division. But the common rule is to add all the perpendiculars together, and divide by the number of them, for a mean perpendicular to be multiplied by the base. This rule; however, gives the area too great, when there are not perpendiculars at the ends of the base, and too little when there are. Let the base be 44, and the perpendiculars at its ends 8 and 16, and the other perpendiculars 14, 16, and 17, the sum of these three is 47, which, added to 12 half the sum of 16 and 8, gives 59, which, divided by 4 the number of parts of the base, gives 143 for the mean perpendicular; and this, multiplied by the base 44, gives the area 649. But the sum of all the perpendiculars 71, divided 5 their number, gives 143 for the mean perpendicular, and this, muls tiplied by 44, makes the area 633.6 only. To find the superficies of any prism. The bases, if they are regular figures, are measured by Prob. s. of this part; or if they are irregular, they are measured by Prob. 4. and the sides being parallelograms, are measured by Prob. 1. of this Part: and the sum of these is the superficies of the prism. PROB PROB. X. To measure the fuperficies of a pyramid. Part II. The base is measured by Prob. 5. if it be regular, or by Prob. 4. if it be irregular; and the sides being triangles, are measured by Prob. 2. or 3. of this Part. PROB. XI. To measure the superficies of any regular body. Regular bodies are those bounded by equilateral and equiangular figures. The superficies of the tetrahedron consists of four equal and equilateral triangles; the hexahedron, or cube, of fix equal squares; the octahedron, of eight equal equilateral triangles; the dodecahedron, of twelve equal regular pentagons ; and the superficies of the icosahedron, of twenty equal equilateral triangles. Therefore it will be easy to measure them from what has been shown. And in the same manner may the super, ficies of any solid contained by planes, be measured. PRO B. XII. Fig. 10, a 5. of this. and therefore greater than the rectangle FA, AO; much more, therefore, is the superficies of the cylinder greater than the redangle DA, AF, that is, than any rectangle less than AE. And in the same manner, it may be proved, that it is less than any rectangle greater than AE; wherefore, it is equal to AE : Therefore, PART II. Therefore, multiply the circumference of the base by the alti. Cu tude, and the product is the superficies of the cylinder, ex cluding the bases, which, being circles, are measured by Prob. 6. Note, It is evident, that if the superficies of the cylinder be spread out, it will coincide with DE. PROB. XIII, FI. III. To measure the superficies of a cone ABC. . 11. Let BE be at right angles to AB, and equal to the circumference of the base, and join AE; and the triangle ABE is equal to the superficies of the cone: let BF be any line less than BE, and join AF, and in the base, let a polygon be inscribed greater than the rectangle FB, BO, and draw lines from A, to all its angies, which will constitute a pyramid within the cone: let HK be a side of the polygon, bisected by BC in G, and join 2 12. 6. E. AG; and as OG to GA, To make a BG to GL; therefore GL A. 5. E. is greater b than GB, and AL greater than AB; also OG is to Ciz. 5. E. GA, as · OB to AL; and therefore the rectangle contained by OG, and the perimeter of the polygon, is to the rectangle cond 1.6. E. tained by AG, and the same perimeter, as the rectangle OB, € yo of this. BF to the rectangle AL, BF; that is, the polygon is to twice f 1o. of the superficies of the pyramid, as OB, BF to AL, BF: but this. the polygon is greater than the rectangle OB, BF; therefore & 14. 5. E. twice the superficies of the pyramid is greater 8 than the rect angle AL, BF, and therefore much greater than the rectangle AĎ, BF: Wherefore, the superficies of the pyramid is greater than the triangle ABF, and much more is the superficies of the cone greater than any triangle ABF, that is, less than ABE. In the same manner, it may be proved, that it is less than any triangle greater than ABE. Therefore, the superficies of the cone is equal to the triangle ABE; and therefore its area is got by multiplying the half of the circumference of the base, by the flant fide AB. This is also proved, by supposing the superficies to be spread out on a plane, for it will then be a sector, of which AB is the radius, and the circumference of the base is the arch. Cor. It is evident, that if the cone be cut by a plane at M parallel to the base, and MN is parallel to BE, the superficies of the frustum is equal to the quadrilateral BENM, of which BE and NM are parallel fides, and BM their distance. PROB. PROB. XIV. To measure the superficies of a sphere. Part !!. I 2. E. Let ABC be a semicircie, of which the diameter is AG and PI. MUT. the centre D; and let GHKL be any regular polygon described Fig. 12. about it; and draw a MO, HQ, NP, KR, perpendiculars to AC, 12. ). and HS parallel to it t; and join DM, DN, and complete the b31.1.}.. parallelogram CE. And because the triangles GMD, GQH, right angled at M, Q, are equiangular, GM: MD::QG:C 32. 3. F. QHd; and, alternately", GM: GQ :: MD: QH, that is, as d 4.6. . the circumference of which MD is the radius to the circumfe. e 16.5.1, rence of which CH is the radius *; therefore the rectangle f Cor.,?. contained by GM and the latter circumference is equal to the rectangle contained by GQ and the formers: But the superfi- g 16.6.L. cies of the cone, described by the revolution of the triangle GQH about GQ, is equal to the rectangle contained by GM, h 13. of and the circumference of which OH is radius; therefore, also, this. it is equal to the rectangle contained by GQ and the circumfe. rence ABC of which DM is radius. Again, because the angles SKH, KHS are equal to the right angie HND , and SKH is C 32.1.1. equal to HNP, the angle KHS is equal PND; therefore the k 29. 1. I. right angled triangles KHS, DPN are equiangular, and DN is to NP, as KH to HS, or QR; and therefore, it may be d 4.6.1. proved, as before, that the superficies described by the revolution of HKRQ about OR, which is equal to the rectangle contained by HK, and the circumference of NP, half the sum of QH and RK, is also equal to the rectangle contained by QR and the circumference of the inscribed circle ABC; and so on : Wherefore, the whole superficies described by the revolution of the polygon, is equal to the rectangle contained by the axis GL, and the circumference of the inscribed circle ABC: but the circle ABC is equal to the base of the cylinder, described by the revolution of the rectangle EC, and the axis GL is greater than the height AC; therefore the superficies described by the polygon GKL is greater than the superficies of the cylinder defcri. hed by CE. Let, in the same manner, any polygon be inscribed in the semicircle ABC, and it may be proved, as before, that the fuperficies described by its revolution is equal to the rectangle contained by its axis AC, and the circumference of the circle inscribed in it: but this circle is less than the base of the cylinder, and AC is the height of the cylinder ; therefore the superficies described by the polygon is less than that of the cylinder. Wherefore, the superficies of the cylinder is less than any fuperUu ficies PART III. ficies greater than that of the sphere, and greater than any less mthan that of the sphere; it is therefore equal to the superficies of the sphere. And likewise any fegment of the cylindric superficies is equal to the superficies of the corresponding segment of the sphere, cut off by the same plane. OR. I. The surface of the sphere is four times a great circle of it, for a great circle is equal to the rectangle contained by half the circumference and half the diameter. m C, 3. E. Cor. 2. Because the rectangle CA, AP is equal to the square of the chord AN, if each of them be multiplied by n 4, of this. 3.1416, the circle of which AN is the radius is equal to the rectangle contained by AP, and the circumference of the circle ABC; that is, it is equal to the superficies of the segment made by the revolution of AMN about AP. PART III. OF SOLID FIGURES. As an inch is the smallest measure in length, and the square upon it the smallest superficial measure, so the cube described from an inch is the smallest folid measure ; and 1728 cubical inches make a solid foot. The English ale-gallon contains 282 cubical inches ; a pint is the eighth part of a gallon, and contains 354 cubical inches; in the country, 34 gallons make a barrel; but in London, 32 gallons make a barrel of ale, and 36 gallons a barrel of beer; and a firkin is the fourth part of a barrel. The English wine-gallon contains 231 cubical inches, and therefore the pint contains 283 cubical inches; there are 63 gallons in a hogshead, 84 gallons in a puncheon, 2 hofheads in a pipe, and 4 hofheads in a ton ; and a tierce is the half of a puncheon. The Scots pint contains 10314. cubical inches, but is supposed to contain 105 such inches ; 2 pints make a quart, and 8 pints a gallon; and the pint is subdivided into 2 chopins, or 4 mutchkins, and the mutchkin into 4 gills. In dry measure, the Wincheiter gallon contains 272 cubical inches; and 8 gallons make a buibel, which should therefore contain 2178 cubical inches : but in levying the malt-tax, the huhel is appointed to contain 2150 cubical inches ; and the gal. lon |