A Treatise on Mensuration: Both in Theory and PracticeSaint, 1770 - 646 pages |
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Page 109
... zone or Space ABCD A in- cluded between two parallel chords AB , CD , and the two arcs AD , CB . RULE I. Find the area of each feg- ment ABEA , DCED , and their difference will be that of the zone required . EXAMPLE I. E D R A S OF PO B ...
... zone or Space ABCD A in- cluded between two parallel chords AB , CD , and the two arcs AD , CB . RULE I. Find the area of each feg- ment ABEA , DCED , and their difference will be that of the zone required . EXAMPLE I. E D R A S OF PO B ...
Page 110
... zone required . EXAMPLE II . Suppofe the greater chord AB to be 20 , the less DC 15 , D and their distance PQ 17 . Required the area of the zone ABCD . Then , as in the last exam- PAR PA - DO ple , TO = 2P2 35 × 5 = 2. Which being lefs ...
... zone required . EXAMPLE II . Suppofe the greater chord AB to be 20 , the less DC 15 , D and their distance PQ 17 . Required the area of the zone ABCD . Then , as in the last exam- PAR PA - DO ple , TO = 2P2 35 × 5 = 2. Which being lefs ...
Page 111
... zone ; and then it is evident that the whole circle diminished by the fum of the two fegments AFB , DEC , will give the zone . Now , OP = OTTO = 84-4 = 7. Hence 40 = AO 2 151 25 √OP1 + P¿1 = √ 13 ] 2 + 10 ? == the radius ; there- 2 2 ...
... zone ; and then it is evident that the whole circle diminished by the fum of the two fegments AFB , DEC , will give the zone . Now , OP = OTTO = 84-4 = 7. Hence 40 = AO 2 151 25 √OP1 + P¿1 = √ 13 ] 2 + 10 ? == the radius ; there- 2 2 ...
Page 112
... zone ABCD , the fame as before . = PROBLEM XI . To find the area of the ring included between the circum- ferences ABPA , DEQD of two concentric circles . RULE I. Multiply the fum of the diameters by the difference of the diameters ...
... zone ABCD , the fame as before . = PROBLEM XI . To find the area of the ring included between the circum- ferences ABPA , DEQD of two concentric circles . RULE I. Multiply the fum of the diameters by the difference of the diameters ...
Page 136
... zone LMNK into which the infcribed circle . is cut by thofe lines . First , √21 × 8 × 7 × 6 = 7 × 3 × 2 × 4 × 7 × 2 × 3 = 7 × 3 × 4 84 the area of the tri- angle ABC . = = Then 84 ÷ 7 ( AC ) = 12 = the perpendicular BQ ; and 84 ÷ 21 ...
... zone LMNK into which the infcribed circle . is cut by thofe lines . First , √21 × 8 × 7 × 6 = 7 × 3 × 2 × 4 × 7 × 2 × 3 = 7 × 3 × 4 84 the area of the tri- angle ABC . = = Then 84 ÷ 7 ( AC ) = 12 = the perpendicular BQ ; and 84 ÷ 21 ...
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abfciffa againſt alfo Alnwick alſo altitude angle area Verf bafe baſe becauſe breadth bung cafe cafk circle whofe circumference cofine cone confequently conftruction conjugate COROLLARY defcribe difference dimenfions diſtance ditto divided divifion ellipfe equal expreffed expreffion faid fame example fecond fection feet fegment feries fhall fide figure fimilar fince find the area firſt fixed axe fluxion folid fome fpheroid fpindle fquare fruftum fubtract fuch fuppofing furface gallons girt greateſt half head diameter hence hoof hyperbola inches interfecting laft laſt laſt problem lefs length meaſure method middle moſt multiply muſt Newcaſtle oppofite ordinate parabola paraboloid parallel perpendicular plane prob quotient radius rule Schoolmafter ſhall Sliding Rule ſphere ſquare ſtation tangent theſe thickneſs thofe thoſe tranfverfe trapezium triangle uſed vertex Wherefore whofe whofe height whole whoſe zone ΙΟ