36 x 7 с S0 X 8-48 240-48 3 192 Example. Required the area of an octagon whose side is 20 feet Here 202=400, and the tabular area 4-8284271 ; Therefore 4-8284271 x 400=1931-37084 feet, area required. 1920. PROBLEM VII. To find the diameter and circumference of any circle, either from the ker. Rule 1. As 7 is to 22, or as 1 is to 3•1416, so is the diameter to the circumference. Or as 22 is to 7, so is the circumference to the diameter. Example. Required the circumference of a circle whose diameter is 9. Here 7 : 22::9: 283; or, *-*°=289, the circumference required. Required the diameter of a circle whose circumference is 36. Here 22 : 7::36 : 114%; or, o=11the diameter required. 1991. PROBLEM VIII. To find the length of any arc of a circle. Rule 1. Multiply the decimal 01745 by the number of degrees in the given arc, and that by the radius of the circle ; then the last product will be the length of the arc. This rule is founded on the circumference of a circle being 6.2831854 when the diameter is 2, or 3:1415927 when the diameter is 1. The length of the whole circumference then being divided into 360 degrees, we have 360° : 6.2831854 ::1° : 01745. Example. Required the length of an arc of 30 degrees, the radius being 9 feet. Here 01745 x 30 x 9=4-7115, the length of the arc. Rule 2. From 8 times the chord of half the arc subtract the chord of the whole arc, and one third of the remainder will be the length of the arc nearly. Example. Required the length of an arc DCE (fig. 516.) whose chord DE is 48, and versed sine 18. and 1900=30= DC. =64, the length of the arc required. 1299. PROBLEM IX. To find the area of a circle. Rule 1. Multiply half the circumference by half the diame Or multiply the whole circumference by the whole diameter, and take of the product. Rule 2. Square the diameter, and multiply such square by ·7854. Rule 3. Square the circumference, and multiply that square by the decimal •07958. Example. Required the area of a circle whose diameter is io, and its circumference 31-416. 31.416x 10 =78-54. By rule 3., 31-416 * 31416 * •07958 = 78.54. 1223. PROBLEM X. To find the area of a circular ring, or of the space included between he circumferences of two circles, the one being contained within the other. Rule. The difference between the areas of the two circles will be the area of the ring. Or, multiply the sum of the diameters by their difference, and this product again by -7854, and it will give the arca required. Example. The diameters of two concentric circles being 10 and 6, required the area of the ring contained between their circumferences. Here 10+6=16, the sum, and 10-6=4, the difference. Therefore •7854 x 16 x 4= -7854 x 64 = 50-2656, the area required. 1224. PROBLEM XI. To find the area of the sector of a circle. Rule 1. Multiply the radius, or half the diameter, by half the arc of the sector for the Or multiply the whole diameter by the whole arc of the sector, and take of the product. This rule is founded on the same basis as that to Problem IX. Rule 2. As 360 is to the degrees in the arc of the sector, so is the area of the whole circle to the area of the sector. This is manifest, because it is proportional to the length of the arc. Example. Required the area of a circular sector whose arc contains 18 degrees, the diameter being 3 feet. 360 : 18::9.4248 : 47124, the length of the arc. ter. Fig. 516. area. By the second rule, -7854 x 39=7-0686, area of the whole circle. S60: 18::7-0686 : 35343, the area of the sector. 1225. Problem XII. To find the area of a segment of a circle . Rule 1. Find the area of the sector having the same are with the segment by th problem. Then find the area of the triangle formed by the chord of the seg semicircle. (fig. 517.), its chord AB being 12, and the radius AE or CE 10. in the arc AC. ACBE. AEB. =16:9504, area of segment ACBDA. column of heights in the following table. Take out the corresponding area in next column on the right hand, and multiply it by the square of the circle's diam for the area of the segment. This rule is founded on the principle of similar p figures being to one another as the squares of their like lineal dimensions, segments in the table are those of a circle whose diameter is 1. In the first colo is contained the versed sines divided by the diameter. Hence the area of similar segment taken from the table and multiplied by the square of the diam gives the area of the segment to such diameter. When the quotient is not fo exactly in the table, a proportion is used between the next less and greater area. Example. As before, let the chord AB be 12, and the radius 10 or diameter 20. Having found as above DE=8: then CE-DE=CD=10-8=2. He by the rule CD+CF=2+20='1, the tabular height; this being found the first column of the table, the corresponding tabular area is 0408 then *040875 x 20= *040875 × 400=16.340, the area nearly the same before. Areas of the SEGMENTS OF A Circle WHOSE DIAMETER, UNITY, IS LUPPOSED TO E DIVIDED INTO 1000 EQUAL Parts. *001 000042022 +004322 043 011734064 021168 -085 032186106 04459 ·002 000119|023 +004618|044 012142-065 -021659086 032745107 04519 003 000219 024 004921 | .045 012554066 -022154 087 033307 108 | 04575 004 000337025 005230046 -012971067022652 088 033872-109 04638 ·005 000470026 005546 047 013392068 -029154|089 034441-110-04700 *006 000618027 005867048 013818069 023659 090 035011 11104763 -007 -000779 028 006194049 01 4247 070 •024168091 035585 112 04826 *008 000951 | 029 006527 | 050 014681 071 024680.092 03616211304889 009001135030 006865) 051 015119-072 025195 -093 036741 |'114 04952 *010 001329 031 007209.052 01 5561 073 025714 094 037323.115 -05016. 011 001533032 -007558 053 016007074 026236 095 037909.116.05080 012 ·001 746 033 007913.054 016457 075 026761 096 038496 117 051 44 019 .003471 040 010537)-061 .019716.082 030526 -103 .042687|124056003 Hght. Area Seg. Hght. Area Seg. Hght. Area Seg Hght. Area Seg. Hght. Area Seg. Hght. Area Seg. -127 057991 190 -103900] -253 -156149/315 212011377 270951 1439 391850 -128, 058658-191 104685 -254 157019-316 2129401.378 .271 920 440 3328431 *129 059327 192 -105472 255 .157890] -317-213871 379 .272890 441 333836 130, 059999·193 106261-256 •158762 -318 -214802 +380 273861 442 +934829 •131 060672 •194 •107051 | .257 •159636 319.215733.381 / -2748321 443 / -335822 •132 061 348 -195 •107842.258 •160510 320 -216666|.382 -275803) -444 336816 *133 062026 -196 •108636 -259-161386) 321-217599 -383.276775) -445 -337810 *194 062707-197 109430-260 -162263 -322 -218533384 -277748 446 338804 135 063389) 198 110226 -261 -1631 40-323 -21 9468-385 -278721447 -339798 -136, 064074|199 111024 -262 •164019-324 220404 -386 -279694 448 -340793 -137 064760-200 -111823.263 •164899)-325 -221340-387-280668) 449 .341787 -138 065449|.201-112624 .264 165780) -326 -222277388 -281642.450 -342782 -139 0661 40202 •113426].265 166663 -327 .223215389 -282617):451 / 943777 -140, 066833 -203 -114230-266 -167546-328 -224154 .390 -283592-452 -344772 -141 067528 -204 -115095 .267 168430•329 -225093 391 -284568/453 -345768 •142 068225 -205 .115842 -268 169315-330 -226033 392 -285544-454 •346764 -143 068924 206 116650-269•170202) 931 / -226974393 -286521 455 •347759 -1 44 069625 -207 •117460270 -171089:332 -227915 394 •287498 .456 348755 145 0703281-208 •118271 271 -171971 / 333 .228858 395 .288476):457 349752 -146 071033 -209 •119083/ -272 •172867) 334 .229801 | 1996 | .289453:458 350748 •1 47 071741 -210.119897) -273 •173758 -335 -230745 .397 290432459 351745 •148 072450-211 •1 20712 -274 •174649 -336 -231689-398 291411-460 •952742 *149 073161-2121.121529) -275 :175542):337 232634:399 292390] -461 353739 •150 073874 -213 •1223471-276 •176435 .398 -233580|400 / -293369462 354736 •151 / 074589214 •123167 277|.177330 339 -234526401 | .294949463355782 *152 075306 .215 •123988.278 •178225/340 -235473.402 -295330-464-356730 -153 076026 -216 :124810.279 .179122):341 | .236421-403 .296311 465 | .357727 -154 076747 -217 •125634 -280.180019.342 -237369| 404 -297292466 / 358725 -155 077469 -218 +1 26459-281.180918/3431.238318|405) .298273/ .467 .359729 -156 078194 -219 127285) -282 -181817:344 -239268-406 -299255):408 •360721 -1571 078921-220.128113-283 -182718-345 .240218.407•300238.469.361719 -158 079649 -221 -128942_-284.183619.346 241169/ 408 301220:470:362717 •159 080380|-222 :129773 285 184521 | .347-242121-409 -302203) 471-363715 -160 081112 .223 +130605 286 185425):348-243074 410 .303187.472.364713 -161 081 846 224 131438.287) •1 86329-349 244026:411 304171 473-365712 *162' 082582 .225 •1 32272288 -187234 :350-244980.412 .3051 551.474.366710 -163 083320-226-133108 289/-188140351 245934 :113.306140| 475 367709 -164 084059] 227 •193945 -290 189047 -352 246889414.3071 25.476.368708 -165 084801 -228 134784) -291 189955 -353 -247845):415 .308110.477.369707 •166 085544 -229.135624.292)190864|354 -248801 .416-309095478 370700 -167 086289 -230 •136465 293 •191775355 249757-417310081.479 371704 -168 087036 231•137307 -294 •192684):356 -250715 418 .3110681-480:372704 -169 087785 -232 •1381501-295 :193596):357 -251673/ 419:312054 481 | :373703 1.170 088535)-233 138995 -296 •194509):358-252631-420 313041 482 •374702 -171 089287 234 139841297|195422 359 253590 -421 3140291.483-375702 -172 090041 -235 140688)-298-196337-360 -254550:422-315016].484 .376702 -173 090797|236 •141537299-197252-361-255510423 316004 485/377701 •174 091554] -237 •142387-300 .198168.362 -256471-424 316992486 378701 -175 092313 -238 •1432381-301 •199085 363257433425 .317981 487 .379700 -176 093074) -239/144091 302 -200003 364 258395 426 :318970 488.380700 .177 -093836) -240 -144944 .303 .200922:365 .259357|:427:319959489381699 -1781 094601 -241 .145799:304 -201841-366 -260320-428-320948.490.382699 -179 095366) -242 -146655)-305 -202761 -367 -2612841.429 321938-491-383699 -180 096134-243 •147512-306 203683.368 262248430 .322928 492 •384699 •181 *096903-244 -148371 307 -204605 .369.263213.431 | 323918].493.385699 .182 097674 +245 +1 49230|.308.205527) 370 .264178-432 324909.494986699 183 098447-246 -150091309 -206451.371 •2651441.433 3259001:495 .387699 .184 099221/247 .150953/ 310 -207376 372 -266111.494 .326892 496 388699 -185 099997)-248 151816-311 -208301):373 .267078 435 -3278824971-389699 .186 100774 -249 •1526801:312 -209227-374 -268045-436.328874):498 390699 •187 -101 553 -25):153546):913 210154)-375 -269013.437 .329866 499 -391699 '188 102334 -251 154412:914 211082-376 -269982-438-390858-500-392699 -189 1031161-252|155280 1226. PROBLEM XIII. To find the area of an ellipsis. which will give the area required. This rule is founded on Theorem 3. Cor. 2. Conic Sections. (1098, 1100.) Here 70 x 50 x 7854=2748-9. vertical axis or diameter ; then, as the said vertical axis is to the other axis (para to the base of the segment), so is the area of the circular segment first found to area of the elliptic segment sought. This rule is founded on the theorem allud to in the previous problem. Or, divide the height of the segment by the vert: axis of the ellipse; and find in the table of circular segments appended to Prob. (1224.) the circular segment which has the above quotient for its versed sine ; th multiply together this segment and the two axes of the ellipse for the area. Example. Required the area of an elliptic segment whose height is 20, the verti axis being 70, and the parallel axis 50. Here 20=70= -2857142, the quotient or versed sine to which in table answers the segment .285714. Then .285714 x 70 x 50 = 648-342, the area required. 1228. PROBLEM XV. To find the area of a parabola or its segment. Rule. Multiply the base by the perpendicular height, and take two thirds of the pi duct for the area. This rule is founded on the properties of the curve alrea described in conic sections, by which it is known that every parabola is of circumscribing parallelogram. (See 1097.) Example. Required the area of a parabola whose height is 2 and its base 12. Here 2 x 12 = 24, and 3 of 24=16 is the area required. MENSURATION OF SOLIDS. 1229. The measure of every solid body is the capacity or content of that body, co sidered under the threefold dimensions of length, breadth, and thickness, and the measu of a solid is called its solidity, capacity, or content. Solids are measured by units whi are cubes, whose sides are inches, feet, yards, &c. Whence the solidity of a body said to be of so many cubic inches, feet, yards, &c. as will occupy its capacity or spa or another of equal magnitude. 1230. The smallest solid measure in use with the architect is the cubic inch, from whi other cubes are taken by cubing the linear proportions, thus, 1728 cubic inches=1 cubic foot ; 27 cubic feet =1 cubic yard. 1231. Problem I. To find the superficies of a prism. Multiply the perimeter of one end of the prism by its height, and the product will be t surface of its sides. To this, if wanted, add the area of the two entds of the pris Or, compute the areas of the sides and ends separately, and add them together. Example 1. Required the surface of a cube whose sides are 20 feet. Here we have six sides ; therefore 20 x 20 x 6=2400 feet, the area required. each side of its end or base 18 inches. base; two bases; then, 3 x 20 x 1.5 +1.948 = 91 ·948 is the area required. Example 3. Required the convex surface of a round prism or cylinder whose leng is 20 feet and the diameter of whose base is 2 feet. Here, 2 x 3.1416 = 6•2832, and 3:1416 * 20 = 125-664, the convex surface required. 1232. Proslem II. To find the surface of a pyramid or cone. Rule. Multiply the perimeter of the base by the length of the slant side, and half t1 product will be the surface of the sides or the sum of the areas of all the sides, of the areas of the triangles whereof it consists. To this sum add the area of ti end or base. Example 1. Required the surface of the slant sides of a triangular pyramid who slant height is 20 feet and each side of the base 3 feet. Here. 20 x 3 (the perimeter) * 3+2 =90 feet, the surface required. A B Example 2. Required the convex surface of a cone or circular pyramid whose slant height is 50 feet and the diameter of its base 84 feet. Here, 8.5 x 3:1416 x 50-2=667-5, the convex surface required. 1233. PROBLEM III. To find the surface of the frustum of a pyramid or cme, being the lower pert where the top is cut off by a plane parallel to the base. Rale. Add together the perimeters of the two ends and multiply their sum by the slant height. One half of the product is the surface sought. This is manifest, because the sides of the solid are trapezoids, having the opposite sides parallel. Example 1. Required the surface of the frustum of a square pyramid whose slant height is 10 feet, each side of the base being 3 feet 4 inches and each side of the top 2 feet 2 inches. Here, 3 feet 4 inches *4=13 feet 4 inches, and 2 feet 2 inches x 4 = 8 feet 8 inches; and 13 feet 4 inches + 8 feet 8 inches =22. Then 22; 2x10=110 feet, the surface required. Example 2. Required the convex surface of the frustum of a cone whose slant height is 124 feet and the circumference of the two ends 6 and 8·4 feet. Here, 6 +8.4=14:4; and 14.4 x 12.5+2 = 180-2 =90, the convex surface required. 1234. PROBLEM IV. To find the solid content of any prism or cylinder. llule. Find the area of the base according to its figure, and multiply it by the length of the prism or cylinder for the solid content. This rule is founded on Prop. 99. (Geometry, 980.). Let the rectangular parallelopipedon be the solid to be measured, the small cube P(fig. 518.) being the measuring unit, its side being 1 inch, 1 foot, &c. Let also the length and breadth of the base ABCD, and also let the height AH, be divided into spaces equal to the side of the base of the cube P; for instance, H here, in the length 3 and in the breadth 2, making 3 times 2 or 6 squares in the base AC each equal to the base of the cube P. It is manifest that the parallelopipedon will contain the cube P as many times as the hase AC contains the base of the cube, repeated as often as the height AH contains the height of the cube. Or, in other words, the contents of a parallelopipedon is found by multiplying the area of the base by the altitude of the solid. And because all prisms and cylinders are equal to parallelopipedons of equal bases and altitudes, the rule is general for all such solids whatever the figure of their base. Example 1. Required the solid content of a cube whose side is 24 inches. Here, 24 x 24 x 24=13824 cubic inches. Example 2. Required the solidity of a triangular prism whose length is 10 feet and the three sides of its triangular end are 3, 4, and 5 feet. Here, because (Prop. 32. Geometry, 907.) 32 + 42=52, it follows that the angle con. tained by the sides 3 and 4 is a right angle. Therefore x 10=60 cubic feet, the content required. Example 3. Required the content of a cylinder whose length is 20 feet and its diameter 5 feet 6 inches. Here, 5.5 x 5.5 x 7854 =- 23.75835, area of base; and 23.75835 x 20=47:5167, content of cylinder required. 1335. PROELEM V. To find the content of any pyramid or cone. Kule. Find the area of the base and multiply that area by the perpendicular height. One third of the product is the content. This rule is founded on Prop. 110 (Geometry, 991.) Example 1. Required the solidity of the square pyramid, the sides of wiose base are 30, and its perpendicular height 25. Here, 30 * 25=7500, content required. Example 2. Required the content of a triangular pyramid whose perpendicular height is 50 and each side of the base 3. 3+3+3 Here, '=f=4.5, half sum of the sides : and 4:5—3=1.5, one of the three equal remainders. (See Trigonometry, 1052.) but v4.5 x 1.5 x 1.5 x 1.5 30:3 = 3.897117 ~ 10, or 38.97117, the solidity required. Esample 3. Required the content of a pentagonal pyramid whose height is 12 fect and each side of its baze 2 feet. Here, 1.7204774 (tabular area, Prob. 6. 1218.) 4 (square of side) = 6.8819096 area of base; and 6.8819096 x 12 = 82.5829152. Then =275276384, content required. z Fig. 518. 3x4 2 30 x 30 2 82.5829152 3 |