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PROB.

PART II. To find the area of an offset.

Pl. III.
Fig. 9.

VIII.

Offsets are parts of a field which lie between a crooked hedge and the ftraight line which is measured for the fide of the field, and these spaces are measured, by taking the perpendicular diftances of the hedge from the measured line, at a great number of places, fo that the hedge between thefe places may be nearly a ftraight 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 fum 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 bafe, half their fum should be added to the other perpendiculars before divifion. 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 bafe. This rule, however, gives the area too great, when there are not perpendiculars at the ends of the bafe, 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 fum of 16 and 8, gives 59, which, divided by 4 the number of parts of the base, gives 14 for the mean perpendicular; and this, multiplied by the base 44, gives the area 649.

But the fum of all the perpendiculars 71, divided 5 their number, gives 14% for the mean perpendicular, and this, mul tiplied by 44, makes the area 633.6 only.

PROB. IX.

To find the fuperficies of any prism.

The bafes, if they are regular figures, are measured by Prob. 5. of this Part; or if they are irregular, they are measured by Prob. 4. and the fides being parallelograms, are measured by Prob. 1. of this Part: and the fum of thefe is the fuperficies of the prifm.

PROB

PROB. X.

To meafure the fuperficies of a pyramid.

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The base is measured by Prob. 5. if it be regular, or by Prob. 4. if it be irregular; and the fides being triangles, are measured by Prob. 2. or 3. of this Part.

PROB. XI.

To measure the fuperficies of any regular body.

Regular bodies are thofe bounded by equilateral and equiangular figures. The fuperficies of the tetrahedron confifts of four equal and equilateral triangles; the hexahedron, or cube, of fix equal fquares; the octahedron, of eight equal equilateral triangles; the dodecahedron, of twelve equal regular pentagons; and the fuperficies of the icofahedron, of twenty equal equilateral triangles. Therefore it will be eafy to measure them from what has been fhown. And in the fame manner may the fuper, ficies of any folid contained by planes, be measured.

PROB. XII.

To measure the fuperficies of a cylinder ABCD.

PART II.

Pl. III.

Fig. 10,

Let AE be equal to the circumference of the base, and complete the rectangle DE; it is equal to the fuperficies of the cylinder. Let AF be any line lefs than AE, and in the bafe, let a polygon be infcribed, greater than the rectangle contained by FA, and the radius AO, and at its angles, let perpendiculars to the bafe be erected, which will conftitute a prifm within the cylinder, of which the fuperficies is lefs than that of the cylinder: but by the 9th of this part, the fuperficies of the prism is equal to the rectangle contained by DA, and the perimeter of the polygon, which is greater than AF, because the rectangle contained by AO, and that perimeter, is greater than the polygon, a 5.of this. and therefore greater than the rectangle FA, AO; much more, therefore, is the fuperficies of the cylinder greater than the rectangle DA, AF, that is, than any rectangle lefs than AE. And in the fame 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 bafe by the altiYtude, and the product is the fuperficies of the cylinder, excluding the bafes, which, being circles, are meafured by Prob. 6. Note, It is evident, that if the fuperficies of the cylinder be fpread out, it will coincide with DE.

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Pl. III. To meafure the fuperficies of a cone ABC.

Fig. 11.

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Let BE be at right angles to AB, and equal to the circumfe rence of the base, and join AE; and the triangle ABE is equal to the fuperficies of the cone: let BF be any line less than BE, and join AF, and in the base, let a polygon be infcribed greater than the rectangle FB, BO, and draw lines from A, to all its angles, which will conftitute a pyramid within the cone: let HK be a fide of the polygon, bifected by BC in G, and join 212. 6. E. AG; and as OG to GA, fo make BG to GL; therefore GL BA. E. is 5. b greater than GB, and AL greater than AB; alfo OG is to 12. 5. E. GA, as OB to AL; and therefore the rectangle contained by OG, and the perimeter of the polygon, is to the rectangle cond1. 6. E. tained by AG, and the fame perimeter, as the rectangle OB, es.of this. BF to the rectangle AL, BF; that is, the polygon is to twice fro. of the fuperficies of the pyramid, as OB, BF to AL, BF: but the polygon is greater than the rectangle OB, BF; therefore 14. 5. E. twice the fuperficies of the pyramid is greater & than the rectangle AL, BF, and therefore much greater than the rectangle AB, BF: Wherefore, the fuperficies of the pyramid is greater than the triangle ABF, and much more is the fuperficies of the cone greater than any triangle ABF, that is, lefs than ABE. In the fame manner, it may be proved, that it is less than any triangle greater than ABE. Therefore, the fuperficies of the cone is equal to the triangle ABE; and therefore its area is got by multiplying the half of the circumference of the bafe, by the flant fide AB. This is alfo proved, by fuppofing the fuperficies to be fpread 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.

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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 fuperficies of the fruftum is equal to the quadrilateral BENM, of which BE and NM are parallel fides, and BM their distance.

PROB.

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PART II

Pl. III.

Fig. 12. 12.1.1.

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Let ABC be a femicircle, of which the diameter is AC and the centre D; and let GHKL be any regular polygon defcribed about it; and draw MO, HQ, NP, KR, perpendiculars to AC, a and HS parallel to it; and join DM, DN, and complete the 31. I. parallelogram CE. And because the triangles GMD, GQH, right angled at M, Q, are equiangular, GM: MD:: QG: c 32. 1. F. OH; and, alternately, GM: GQ: MD: QH, that is, as d 4. 6. F. the circumference of which MD is the radius to the circumfe- e 16. 5. E. rence of which QH is the radius f; therefore the rectangle f Cor. 2. contained by GM and the latter circumference is equal to the rectangle contained by GQ and the former : But the fuperfi- g 16.6. E. cies of the cone, defcribed by the revolution of the triangle CQH about GQ, is equal to the rectangle contained by GM, h 13. of and the circumference of which QH is radius; therefore, alfo, this. it is equal to the rectangle contained by GQ and the circumference ABC of which DM is radius. Again, because the angles

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12. E.

SKH, KHS are equal to the right angle HND, and SKH is c 32. 1. E. equal to HNP, the angle KHS is equal PND; therefore the k 29. 1. E. right angled triangles KHS, DPN are equiangular, and DN

is to NP, as d KH to HS, or QR; and therefore, it may be d 4. 6. E. proved, as before, that the fuperficies defcribed by the revolution of HKRQ about QR, which is equal to the rectangle contained by HK, and the circumference of NP, half the fum of QH and RK, is alfo equal to the rectangle contained by QR and the circumference of the infcribed circle ABC; and fo on : Wherefore, the whole fuperficies defcribed by the revolution of the polygon, is equal to the rectangle contained by the axis GL, and the circumference of the infcribed circle ABC: but the circle ABC is equal to the bafe of the cylinder, defcribed by the revolution of the rectangle EC, and the axis GL is greater than the height AC; therefore the fuperficies defcribed by the polygon GKL is greater than the superficies of the cylinder defcribed by CE. Let, in the fame manner, any polygon be infcribed in the femicircle ABC, and it may be proved, as before, that the fuperficies defcribed by its revolution is equal to the rectangle contained by its axis AC, and the circumference of the circle infcribed in it: but this circle is lefs than the bafe of the cylinder, and AC is the height of the cylinder; therefore the fuperficies defcribed by the polygon is lefs than that of the cylinder. Wherefore, the fuperficies of the cylinder is lefs than any fuper

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PART III. ficies greater than that of the sphere, and greater than any lefs than that of the sphere; it is therefore equal to the fuperficies of the sphere. And likewife any fegment of the cylindric fuperficies is equal to the fuperficies of the correfponding segment of the sphere, cut off by the fame plane.

m C. 3. E.

COR. 1. The furface 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.

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COR. 2. Because the rectangle CA, AP is equal to the square of the chord AN, if each of them be multiplied by n4,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 fuperficies of the fegment 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 fmalleft fuperficial measure, fo the cube defcribed from an inch is the smallest folid measure; and 1728 cubical inches make a folid 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 287 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.

4.

The Scots pint contains 103 cubical inches, but is fuppofed to contain 105 fuch inches; 2 pints make a quart, and 8 pints a gallon; and the pint is fubdivided into 2 chopins, or 4 mutchkins, and the mutchkin into 4 gills.

In dry measure, the Wincheiter gallon contains 2724 cubical inches; and 8 gallons make a bufhel, which fhould therefore contain 2178 cubical inches: but in levying the malt-tax, the bufhel is appointed to contain 2150 cubical inches; and the gal

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