Example. Required the area of an octagon whose side is 20 feet. Here 202-400, and the tabular area 4-8284271; Therefore 4-8284271 × 400=1931-37084 feet, area required. 1920. PROBLEM VII. To find the diameter and circumference of any circle, either from the other. 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. 22 9 Here 7: 22::9: 284; or, x=284, the circumference required. Required the diameter of a circle whose circumference is 36. 36x7 Here 22:7::36:11; or, 2211, the diameter required. and 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 are, 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. Rule 2. Required the length of an arc of 30 degrees, the radius being 9 feet. Here 01745 x 30 x 94-7115, the length of the arc. 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, 1222. PROBLEM IX. To find the area of a circle. Rule 1. Multiply half the circumference by half the diame ter. Or multiply the whole circumference by the whole Rule 2. Square the diameter, and multiply such square by 7854. Fig. 516. Rule 3. Square the circumference, and multiply that square by the decimal 07958. Example. Required the area of a circle whose diameter is 10, and its circumference By rule 2., 10 x ·7854=100 × 7854=78.54. So that by the three rules the area is 78.54. 1223. PROBLEM X. To find the area of a circular ring, or of the space included between the 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. 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 area. 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. By the first rule, 3·1416 x 3=9.4248, the circumference. 360 18:9-4248 47124, the length of the arc. •47124 × 3+4=1·41372÷ 4 = 95343, the area of the sector. By the second rule, 7854 × 32=7·0686, area of the whole circle. 1225. PROBLEM XII. To find the area of a segment of a circle. the segment is greater than a semicircle, and their difference when less tha semicircle. Example. Required the area of the segment ACBDA (fig. 517.), its chord AB being 12, and the radius AE As AE sin. 4 D 90°:: AD: sin. 36° : 52}=36·87 degrees in the arc AC. Their double 73-74 degrees in are ACB. Now, 7854 x 400=314·16, the area of the whole circle. Again, AE2-AD2= √100-36= √648.= DE. Therefore, AD DE=6 x 8=48, the area of the triangle Hence the sector ACBE (64·350), less triangle AEB (48) =16·3504, area of segment ACBDA. Fig. 517. Rule 2. Divide the height of the segment by the diameter, and find the quotient in t column of heights in the following table. Take out the corresponding area in t next column on the right hand, and multiply it by the square of the circle's diamet for the area of the segment. This rule is founded on the principle of similar pla figures being to one another as the squares of their like lineal dimensions. T segments in the table are those of a circle whose diameter is 1. In the first colun is contained the versed sines divided by the diameter. Hence the area of t similar segment taken from the table and multiplied by the square of the diamet gives the area of the segment to such diameter. When the quotient is not four 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. Hend Having found as above DE 8: then CE-DE=CD=10−8=2. AREAS OF THE SEGMENTS OF A CIRCLE WHOSE DIAMETER, UNITY, IS EUPPOSED TO EX Hght. Area Seg. Hght. Area Seg. Hght. Area Seg Hght. Area Seg. Hght. Area Seg. Hght. Area Seg. 011734| 064 021168 085 032186 106 044522 032745 107 045189 066 021659 086 033307·108 045759 022652 088 033872109 046381 068 023154 089 034441110 047005 035011111 047632 035585112 048262| 096162 113 048894 013818 069 023659 090 037323·115 001 000042 022 004322 043 002 000119 023 004618 044 003 000219 024 004921 045 004 000337 025 005230 046 005 000470 026 005546 047 006 000618027 005867 048 007 000779 028 006194049 008 000951029 006527 050 009 001135 030 006865 051 010001 329 031 007209 052 011 001533032 007558 053 012 001746 033 007913 054 |013 001968 034 008273 055 016911 076 027289 097 014 002199035 008638 056 017369 077 027821 098 039680119 052736. 015002438 036 009008 057 017831 078 028356 099 |016 002685 037 009383 058 01 8296 079 028894 100 017002940038 009763 059 018766 080 029435101 018 003202 039 010148 060 01 9239 081 029979102 019 003471 040 010537 061 019716 082 030526103 020 003748 041 010931062 020196 083 031076104 021 004031 042 011330063 020680! 084 031629105 040276·120 053385 040875 121 | 054036| 041476122 054689 042080123 055345 042687124 ·056003| 043296125 056663 043908 126 057326 Hight. Area Seg. Hght. Area Seg. Hght. Area Seg Hght. Area Seg. Hght. Area Seg. Hght. Area Seg 127 057991 190103900 253 156149 315 212011377 270951 439 331850 128 058658191104685 254157019 316 212940 378-271920 440 332843 129 059327 192 105472 255 157890 317 213871 379 272890441933836) 130 059999 -193106261 256 158762 318 214802 380 273861 442 334829| 131 060672-194107051 257 159636 319 215733 381 274832 443 335822 132 061348-195107842 258 160510 320 216666 382 275803444 336816 133 062026 -196 108636 259161386 321 217599 383 276775 445 337810 134 062707-197109430 260 162263 322 218533 384 277748 446 338804 135 063389-198 110226 261 1631 40 323 219468 385 278721 447 339798 136 064074-199111024 262 164019 324 220404 386 279694 448 340793 137 064760-200-111823 263 164899 325 221340387 280668 449 341787 138 065449-201112624 264 165780 326 222277 388 281642 450 342782 139 066140 202-113426 265 166663 327 223215 389 282617451 343777 40 066833 203114230 266 167546328 224154 390 283592452 344772 141 067528 204 115035 267 168430 329 225093 391 284568453 345768 142 068225205115842 268 169315 330 226033 392 285544 454 346764 143 068924 -206-116650 269 170202 331 226974 393 286521 455 347759 144069625 -207-117460 270 171089 332 227915 394 287498 456 348755 145 070328208-118271 271 171971333 228858 395 288476 457 349752 146 071033 -209-119083 272 172867 334 229801 396 289453 458 350748 147 071741 210-119897 273 173758 335 230745 397 290432 459 351745 148 072450 211-120712 274 174649 336 231689 398 291411 460 352742 149 073161 212-121529 275 175542 337 232634 399 292390461 353739 150 073874 213-122347 276 176435 338 233580 400 293369 462 354736 151 074589 -214-123167 -277-177330 339 234526 401 294349 463 355732 152 075306 215-123988 278 178225 340 235473 402 295330 464 356730 153 076026 -216-124810 279 179122 341 236421403 296311 465 357727 154 076747 -217-125634 280 180019342 237369404 297292466 358725 155 077469 218 126459 281 180918 343 238318 405 298273 467359723 156 078194 219-127285 282 181817344 239268 406 299255 408 360721 157 078921-220-128113 283 182718 345 240218407 300238 469 361719 158 079649 221 128942 -284-183619-346 241169 408 301220470 362717 159 080380-222-129773 285 184521 347 242121 409 302203 471 363715 160 081112 -223-130605 286 185425-348 243074 410 303187472364713 161 081846 -224-131438-287-186329 349 244026 411 304171 473 365712 162 082582 -225-132272288 187234 350 244980 412 305155 474 366710 163 083320-226 133108 -289 188140 351 245934 413 306140 475 367709 164 084059 227 133945 -290 189047 352 246889 414 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 309095 478 370706 167 086289 -230-136465 -293-191775 355 249757-417 310081 479 371704 168 087036 231 137307-294-192684 356 -250715 418 311068 480 372704 169 087785 -232 138150-295-193596 357 251673 419 312054 481 373703 170 088535-233-138995 296 194509 358 -252631 420 313041 482 374702 171 089287-234-139841 297-195422 359 253590 421 314029 483 375702 090041-235-140688-298-196337-360 254550 422 315016 484 376702 173 090797 -236 141537 299 197252 361 255510 423 916004 485 377701| 174 091554 237 142387 300 198168 362 256471424316992486 378701| 175 092313238143238-301199085 363 257433 425 317981 487 379700 176 0930742391 44091 302 200003 364 258395426 318970 488 380700 177 093836-240 144944 303 200922 365 259357427 31995 489 381699] 178 094601241145799304 201841 366260320428 320948490382699 179 095366 242 146655305 202761 367 261 284429 321938491383699 180 096134 243 147512 306 203683 368 262248 430 322928 492 384699| 181 096903 244148371 307 204605 369 263213431323918 493 385699 182 097674245 149230308 205527 370 264178 432 324909494 386699 184 099221247150953 310 207376 372 266111494326892496388699] 183 098447 246150091 309 206451|371 2651 44 433 325900 495 387699 185 099997248 151816311208301|373267078 435 327882497 389699 18710155325) 15 35463132101 54375 269013437 329866499 391699] 186 100774249152680 312 209227 374 268045 436 328874 498 390699| 188 102334 251154412 314 211082376 269982 438 350858 500 392699 172 189 103116 252·155280 1226. PROBLEM XIII. To find the area of an ellipsis. Rule. Multiply the longest and shortest diameter together, and their product by 785 which will give the area required. This rule is founded on Theorem 3. Cor. 2. i Conic Sections. (1098, 1100.) Example. Required the area of an ellipse whose two axes are 70 and 50. Here 70 x 50 x 7854=2748.9. 1227. PROBLEM XIV. To find the area of any elliptic segment. Rule. Find the area of a circular segment having the same height and the sam vertical axis or diameter; then, as the said vertical axis is to the other axis (parall to the base of the segment), so is the area of the circular segment first found to th area of the elliptic segment sought. This rule is founded on the theorem allude to in the previous problem. Or, divide the height of the segment by the vertic axis of the ellipse; and find in the table of circular segments appended to Prob. 15 (1224.) the circular segment which has the above quotient for its versed sine; the 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 vertica axis being 70, and the parallel axis 50. Here 20÷702857142, the quotient or versed sine to which in th 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 pro duct for the area. This rule is founded on the properties of the curve alread described in conic sections, by which it is known that every parabola is of it circumscribing parallelogram. (See 1097.) Example. Required the area of a parabola whose height is 2 and its base 12. MENSURATION OF SOLIDS. 1229. The measure of every solid body is the capacity or content of that body, con sidered under the threefold dimensions of length, breadth, and thickness, and the measur of a solid is called its solidity, capacity, or content. are cubes, whose sides are inches, feet, yards, &c. said to be of so many cubic inches, feet, yards, &c. or another of equal magnitude. Solids are measured by units which Whence the solidity of a body i as will occupy its capacity or space 1230. The smallest solid measure in use with the architect is the cubic inch, from whic other cubes are taken by cubing the linear proportions, thus, 1728 cubic inches=1 cubic foot; 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 th surface of its sides. To this, if wanted, add the area of the two ends of the prism 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 × 20 - 6=2400 feet, the area required. Example 2. Required the surface of a triangular prism whose length is 20 feet and each side of its end or base 18 inches. Here we have, for the area of the base, 1 ·52 — ·752 = (2·25 — •5625 = ) 1·6875% for the perpendicular of triangle o base; and √1.6875=1299, which multiplied by 1·5=1·948 gives the area of th 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 lengt 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. PROBLEM 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 th product will be the surface of the sides or the sum of the areas of all the sides, o of the areas of the triangles whereof it consists. To this sum add the area of th end or base. Example 1. Required the surface of the slant sides of a triangular pyramid whos 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. 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, 85 × 3·1416 × 50÷2=667·5, the convex surface required. 1933. PROBLEM III. To find the surface of the frustum of a pyramid or come, being the lower part where the top is cut off by a plane parallel to the base. Rule. 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 x 4 = 13 feet 4 inches, and 2 feet 2 inches x 48 feet 8 inches; and 13 feet 4 inches + 8 feet 8 inches=22. Then 22 ÷ 2 x 10=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 × 12·5÷2=180÷2=90, the convex surface required. 1234. PROBLEM IV. To find the solid content of any prism or cylinder. Rule. 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, D 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 base 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 × 24 × 24=13824 cubic inches. B Fig. 518. 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 3x4 tained by the sides 3 and 4 is a right angle. Therefore x 1060 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 × 5·5 × ·7854 23·75835, area of base; and 23.75835 × 20=47·5167, content of cylinder required. 1235. PROBLEM V. To find the content of any pyramid or cone. Rule. 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 whose base are 30, and its perpendicular height 25. Example 2. Required the content of a triangular pyramid whose perpendicular height is 30 and each side of the base 3. and 4.5-3=1.5, one of the three equal remainders. (See Trigonometry, 1052.) but 45 x 1.5 × 1·5 x 15 x 30 ÷ 3 = 3.897117 × 10, or 38.97117, the solidity required. Example 3. Required the content of a pentagonal pyramid whose height is 12 feet and each side of its base 2 feet. Here, 1.7204774 (tabular area, Prob. 6. 1218.) × 4 (square of side) = 6·8819096 Then 3 275276384, content required. |