Example 1. Required the area of a parallelogram whose length is 12-25 feet, and height 8.5 feet. 12.25 x 8.5=104·125 feet, or 104 feet 11 inches. Example 2. Required the content of a piece of land in the form of a rhombus whose length is 6-20 chains, and perpendicular height 5·45. Recollecting that 10 square chains are equal to a square acre, we have, 6.20 x 5'45=33.79 and =3.379 acres, which are equal to 3 acres, 1 rood, 20.6 perches. 33-79 Example 3. Required the number of square yards in a rhomboid whose length is 37 feet, and breadth 5 feet 3 inches (= 5·25 feet). Recollecting that 9 square feet are equal to 1 square yard, then we have 1215. PROBLEM II. To find the area of a triangle. area. =21584 yards. Rule 1. Multiply the base by the perpendicular height, and take half the product for the Or multiply either of these dimensions by half the other. The truth of this rule is evident, because all triangles are equal to one half of a parallelogram of equal base and altitude. (See Geometry, 904.) Example 1. To find the area of a triangle whose base is 625 feet, and its perpendicular height 520 feet. Here, 625 x 260=162500 feet, the area of the triangle. Rule 2. When two sides and their contained angle are given: multiply the two given sides together, and take half their product; then say, as radius is to the sine of the given angle, so is half that product to the area of the triangle. Or multiply that half product by the natural sine of the said angle for the area. This rule is founded on proofs which will be found in Chap. IV., on which it is unnecessary here to say more. Example. Required the area of a triangle whose sides are 30 and 40 feet respectively, and their contained angles 28° 57'. Rule 3. When the three sides are given, take half the sum of the three sides added together. Then subtract each side severally from such half sum, by which three remainders will be obtained. Multiply such half sum and the three remainders together, and extract the square root of the last product, which is the area of the triangle. This rule is founded on one of the theorems in Trigonometry to be found in the section relating to that branch of the subject. Example. Required the area of a triangle whose three sides are 20, 30, and 40. 20+30 +40=90, whose half sum is 45. 45-20=25, first remainder; 45-30-15, second remainder; 45-40=5, third remainder. Then, 45 x 25 x 15 x 5 =84375, whose root is 290·4737, area required. 1216. PROBLEM III. To find the area of a trapezoid. Add together the parallel sides, multiply their sum by the perpendicular breadth or distance between them, and half the product is the area. (See Geometry, 932.) Example 1. Required the area of a trapezoid whose parallel sides are 750 and 1225, and their vertical distance from each other 1540. 1217. PROBLEM IV. To find the area of any trapezium. Divide the trapezium into two triangles by a diagonal; then find the areas of the two triangles, and their sum is the area. Observation. If two perpendiculars be let fall on the diagonal from the other two opposite angles, then add these two perpendiculars together, and multiply that sum by the diagonal. Half the product is the area of the trapezium. Example. Required the area of a trapezium whose diagonal is 42, and the two perpendiculars on it 16 and 18. Here, 16+18=34, whose half = 17; Then, 42 x 17=714, the area. 1218. PROBLEM V. To find the area of an irregular polygon. Draw diagonals dividing the proposed polygon into trapezia and triangles. Then, having found the areas of all these separately, their sum will be the content required of the whole polygon. Example. Required the content of the irregular AC=55, GC=44, Bn=18, Ep=8, 55 x 6.5 =357.5 44 x 11.5=506 A G B 1878 5, area required. To find the area of a regular polygon. E Fig. 514. Rule 1. Multiply the perimeter of the polygon, or sum of its sides, by the perpendicu- Example. Required the area of the regular pentagon ABCDE 25 × 5 2 Here =625-half the perimeter, and 62.5 x 17.2=1075 square feet area required. D P Fig. 515. Rule 2. Square the side of the polygon, and multiply the square by the tabular area or multiplier set against its name in the following table, and the product will be the area. This rule is founded on the property, that like polygons, being similar figures, are to one another as the squares of their like sides. Now the multipliers of the table are the respective areas of the respective polygons to a side =1; whence the rule is evident. In the table is added the angle OBP in the several polygons. Example. Required the area of an octagon whose side is 20 feet. Here 202-400, and the tabular area 4.8284271; 1220. 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. 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, *=284, the circumference required. Required the diameter of a circle whose circumference is 36. 36x7 Here 22 7:36:1119; or, 2211%, the diameter required. 1221. PROBLEM VIII. To find the length of any arc of a circle. Or 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. Rule 2. 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. 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 diameter. 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. C Fig. 516. Rule 3. Square the circumference, and multiply that square by the decimal 07958. Required the area of a circle whose diameter is 10, and its circumference Example. By rule 2., 102 x 7854-100 × 785478·54. By rule 3., 31 416 × 31·416 x '07958=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 area 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 10-64, the difference. Therefore 7854 × 16 × 4= ·7854 × 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 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 39.4248, the circumference. 360 18:9-4248 47124, the length of the arc. By the second rule, 7854 x 32=7′0686, area of the whole circle. 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 the last problem. Then find the area of the triangle formed by the chord of the segment and the two radii of the sector. Take the sum of these two for the answer when the segment is greater than a semicircle, and their difference when less than a 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° : 521=36·87 degrees in the arc AC. Their double 73-74 degrees in arc ACB. Now, 7854 x 400-314-16, the area of the whole circle. Again, AE2- AD2 100-36=√64=8=DE. = Therefore, AD × DE=6 × 8=48, the area of the triangle Hence the sector ACBE (64·350), less triangle AEB (48) F Fig. 517. B Rule 2. Divide the height of the segment by the diameter, and find the quotient in the column of heights in the following table. Take out the corresponding area in the next column on the right hand, and multiply it by the square of the circle's diameter for the area of the segment. This rule is founded on the principle of similar plane figures being to one another as the squares of their like lineal dimensions. The segments in the table are those of a circle whose diameter is 1. In the first column is contained the versed sines divided by the diameter. Hence the area of the similar segment taken from the table and multiplied by the square of the diameter gives the area of the segment to such diameter. When the quotient is not found 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. by the rule CD÷CF=2÷20=1, the tabular height; this being found in the first column of the table, the corresponding tabular area is 040875; then 040875 × 20o = '040875 × 400=16·340, the area nearly the same as before. Hence AREAS OF THE SEGMENTS OF A CIRCLE WHOSE DIAMETER, UNITY, IS SUPPOSED TO BE DIVIDED INTO 1000 EQUAL PARTS. Hght. Area Seg. Hght. Area Seg. Hght. Area Seg. Hght. Area Seg. Hght. Area Seg. Hght. Area Seg 032745107 045139 045759 046381 011734 064 021168 085 032186 106 044522 033307·108 001 000042 022 004322 043 002 000119 023 004618 044 003 000219 024 004921045 004 000337 025 005230 046 005 000470 026 005546 | 047 006 000618 027 005867 048 007 000779 028 006194049 01 4247 070 024168091035585|·112 048262| 008 000951 029 006527 050 014681 071 024680 092 036162 113 048894 009 001135 030 006865|·051 015119 072 025195 093 036741114 049528 010001329 031 007209 052 015561073 025714 094 037323·115 050165 011 001 533 032 007558 053 016007 074 026236 095 | 037909 116 050804 012 001746 033 007913 054 016457 075 026761 096 038496·117 051446| 013 001968 034 008273 055 016911 076 027289 097 | 039087118 014 002199 035 008638 056 017369 077 027821 098 039680119 015002438 036 009008-057 017831 078 028356 099 040276·120 053385 016 002685 037 009383 058 01 8296 079 028894 100 040875·121 | 054036 | 017 002940038 009763 059 018766 080 029435101 041476122 054689 018 003202|·039] ⋅010148 060 019239 081 029979102 042080·123·055345 019 003471 | 040 010537 061 019716082 030526·103 042687 124 056003 020 003748| 041 | 010931062 020196 083 031076 104 043296·125·056663 | 021004031042 011330' ·063' 020680084 031629 105 043908·126057326| 052090| 052736 Hght. Area Seg. Hght. Area Seg. Hght. Area Seg Hght. Area Seg. Hght. Area Seg. Hght. Area Seg. 273861 442 934829 274832 443 335822 275803 444 336816 276775 445 337810 277748 446 338804 339798 340793 341787 342782 343777 127 057991190 103900 253 156149 315 212011 377 270951 439 331850 128 058658191 104685 254·157019 316 212940 378 271920 440 332843 129 059327 192 105472 255 157890 317 213871 379 272890 441 333836 130 059999 193 106261 256·158762 318 214802 380 131 060672194 107051 257159636319 215733 381 132 061 348195 107842 258 160510 320 216666 382 •133 062026·196 108636259·161386 321 217599383 134 062707-197109430 260·162263 322 218533 384 135 063389-198 110226 261 1631 40 323 219468 385 278721 447 136 064074199 111024 262 164019 324 220404 386 279694448 137 064760 200 111823 263 164899 325 -221340|| 387 280668 449 138 065449201 112624 264165780 326 222277 388 281642 450 139 0661 40 202 -113426 265 166663 327 223215 389 282617451 140 066833 203 114230 266 167546 328 224154 390 283592 452 344772 141 067528 204-115035 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 331226974 393 286521 455 | 347759 •144 069625 207 117460 270 171089 332 227915 394 287498 456 348755 145 070328 208-118271 271 171971333 228858 395 288476 457 349752 146 071033 209 119083 272 172867 334 229801396 289453 458 350748 147 071741210 119897273 173758 335 230745 397 290432 459 351745 148 072450 211120712 274 174649 336 231689 398 291411 460 | ⚫352742 149 078161 212 121 529 275 175542 337 232634 399 292390 461 | ·353739| ·150 073874 213 122347 276 176435 338 233580 400 293369462 354736| •151 074589 214 123167 277 177330 339 234526401 294349 463 355732| 152 075306 215·123988 278 178225 340 235473 402 295330 464 356730 •153 076026 -216·124810 279 179122 841 236421403 296311 465 .357727 154 076747 217 125634 280 180019 342 237369404 297292 466 358725 •155 077469 218 126459 281 180918 343 238318405298273 467 359723 156 078194 219 127285 282 181817 344 239268406 299255 408360721 157 078921 220 128113 283 182718 345 240218 407 300238 469 361719 158 079649 221 128942 284 183619 346 241169408 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 081846 224 131438287 186329 349 244026 411 ·162 082582 -225·132272 288 187234 350 244980 412 ·163 083320 226 133108 289 188140 351 245934 113 164 084059 227 133945 290 189047 352 246889 414 165 084801 228 134784 291 189955353 247845 415 •166 085544 229 135624 292 190864354 248801 416 •167 086289 230 136465 293 191775 355 249757 417 168 087036 231 137307 294192684 356 250715 418 311068 480 372704| 169 087785 232 138150 295 193596 357 251673 419 312054481373703 170 088535 233138995 296·194509 358 252631 420 313041 482 374702 171 089287 234 139841 297 195422 359 253590 421 314029 483 375702 172 090041 235 140688 298196337 360 254550422 315016484376702| 173 090797 236 141537 299 197252 361 255510 423 316004 485 377701 174 091554 237 142387 300 198168 362 256471 424 316992 486 378701 175 092313 238 143238 301 199085 363 257433 425 317981 487379700 176 093074239144091 302 200003 364 258395 426 318970 488 380700 177 093836 240·144944 303 200922 365 259357 427 319959 489 381699 178 094601 241 145799304 201841 366 260320428 320948490 ·382699 179 095366 -242 ·146655 305 202761 367 261 284 429 321938491 | ·383699| 180 096134 243 147512 306 203683 368 262248 430 322928 492 384699 181 096903 244 148371 307 204605 369 263213 431323918 493 385699 182 097674 245 149230 308 205527 370 264178 432 324909 494 386699 183 098447 246 150091 309 206451 371 2651 44 433 325900495 387699 184 099221247150953 310 207376 372266111434 326892 496 388699| 185 099997 248 151816 311 208301 373 267078 435 327882 497 389699 186 100774 249 152680 312 209227 374 268045 436 328874 498 390699 187 101 553 250 153546 313 210154 375 269013437 329866 499 391699 188 102334 251154412 314 211082 376 269982 438 330858 500 392699 304171 473365712 309095478 367709 368708 369707 370706 310081 479 371704 |