0000000::1.6434527: 9.8770399=48° 53′ 16′′ the arc, and 97° 46' 32" the arc. To find the area of the sector, (first reducing the terms in the first couplet to seconds,) 6.11 26050 5.5465300::4.0299765: 3.4639015 2910.175 = the area of the sector. The area of the triangle=38.40X 44 1689.60, which subtracted from 2910.175 the area of the sector, leaves 1220.575-the area of the segment ACB. The difference of the segments ADB, ACB=1477.469 = the area of the lune.

PROBLEM IX. To find the area of a ring, included between the peripheries of the two concentric circles.

Example 1. (2212 ×.7854)—(1062.7854)=(221+106)

×(221-106)×.7854=327×115.7854=37605 × .7854 =29534.9670=the area of the ring.

Example 2. 205000+190000=395000, and 205000 190000 15000. 395000×15000=5925000000, and 5925 000000.7854=4,653,495,000= the number of miles on one side of Saturn's ring.

Promiscuous examples of areas.

Example 1. 20×2=40=rods which multiplied by 2724=1 0890 the number of square feet in the street. 10890×5= $544,50.

Example 2. Let x = the number of square feet in the area; then the number of linear inches in one side, and x÷12=the number of linear feet in one side. By Art. 11, × √3=. × √3 the area=

12

2

X

x2 144

x2

576

by the supposition, x; therefore x2 √3=576x, dividing by x, x3=576, and x=576÷√3. Log. 576-2.7604225, and log. √3=0.2385606 which subtracted from 2.7604225, leaves 2.5218619, whose natural number=332.5530=the area of the triangle.

Example 3. Log. 4522-5.3102768, which added to 1.8 950909 the logarithm of .7854, 5.2053677, which divided by 2 2.6026838 whose natural number=400.575 a side of the square.

=

Example 4. Log. 362-3.1126050, which added to 1.89 50909 the logarithm of .7854,=3.2175141 which divided by 2=1.6087570 whose natural number 40.626011236 = the diameter of the circle.

=

Example 5. Log. 132=2.1208739, which multiplied by 2=4.2417478=log. 1322. Log. 0.7854=1.8950909, which added to 4.2417478=4.1368387 the area of the circle. To find the area of the inscribed square (Art. 33.) .7854: area of the circle area of the inscribed square; and, by the tables, 1.8950909 1.6989700: :4.1368387; 3.940 7178 whose natural number=8724=the required area as obtained in working by logarithms. Simply by arithmetical operations 1322=17424, and 17424X.7854-13684.8096= the area of the circle. To find the area of the inscribed square,.7854 : 0.5::13684.8096: 8712.00509=the answer.

Example 6. To find the perpendicular distance of one of the sides of the octagon from the center (Art. 16.) R: the length of one of the sides::cot 360° divided by twice the number of sides the required perpendicular; that is, R: 4

22° 30′ the perpendicular; and, by the tables, 10.0000 000 0.6020600::10.3827757;0.9848357, whose natural number=9.657 the required perpendicular. 9.657×4= 38.628 which multiplied by 8=309.024 the number of square feet in the octagon. As 9 square feet the number of which make one square yard, 309.024÷9=34.336=the

answer.

Example 7. See Figure 10. It is obvious from the figure that the diagonals cut each other at right angles. But, by Art. 14, sin N=R, and the area-half the product of the diagonals; that is 16×16=256 which divided by 2=128= the area.

Example 8. 3.14159×4=12.56636 the circumference of the wheel, and 12.56636 × 300=3769.908=the perimeter of the green. To find the diameter of the green, (Arts. 24, 25.) log. 3769.908-3.5763296, and log. 0.31831=1.5 028366, and their sum=3.0791662, whose natural number is 1200=the diameter of the green. By Art. 30, 12002 × .7854=the area. Log. 12002=6.1583624, which added to

1.8950909, the logarithm of .7854 becomes 6.0534533= 1131000 the number of square feet in the area of the green. By allowing 160 square rods and 272.25 square feet to a square rod, 1131000 square feet 4154.27 square rods 25 acres, 36 rods.

Example 9. 21x12x2=504=the number of square feet in the larger two sides of the room. 18×12×2=432=the

number of square feet in the smaller two sides, and their sum =936. 5×3×3=45=the deduction to be made for windows, 4×8×2=72=the deduction to be made for doors, and 4×6=27= the deduction to be made for fire place; and, their sum=144. 936-144-792 which divided by 9= 88 square yards which at 10 cents per yard=$8,80,

Example 10. As the width of the gravel walk=81 feet, twice this added to the diameter of the pond the diameter of the circular walk 10 rods and 16 feet. By Art. 4, 16 feet one rod, therefore the diameter of the walk 11 rods. 10+11=21, and 11-10=1, and 21×1=21=(Alg, 341.) the difference of the squares of the diameter of the pond and the walk. 21 x.7854-16.4934 = the area of the , ring.

Example 11. See Figure 17. 1200 the area of VCD, 30= the base, 1200÷30=40=the perpendicular. CN= ND=30. CV=√302 +402=√900+1600 = √2500 = 50 one side of VCD, and CV=VD=50. . By Euc. VI. 19, 1200 432: 2500 900, and 900-30=VG= VL. Again, by Euc. VI. 19, 1200; 432::3600; 1296, and √12 96-36-LG=the base.

Example 12. See Figure 9. By Art 28, R: BC:: sin BCO BP substituting numbers, 1:26::sin 60°: BP; and by the tables, 10.0000000: 1.4147393;:9.9375306: 1.3525039=22.5115 the chord, and 45.0230=the chord. Again, by Art. 15, RBP::cot BCP: CP the perpendicular. Substituting numbers, 1:45.0230: :sin 60° the perpendicular; and, by the tables, 10.0000000; 1,6534055 ::9,9375306;1.5909361-38.9811834 the perpendicu lar, and the perpendicular=19.49. Now the area of a triangle the base the perpendicular hight=45.0230 ×

=

17

19.49. Log. 45. 0230=1.6534376, and log. 19.49=1.289 8118, and their sum=2.9432494 whose natural number= 877.52152174=the area of the triangle.

Example 13. Let x=the number of rods in the circumference, then 10 x=the number of rails and square rods; also, let a=the diameter of the field. By Art. 30, the area = ax=10x. ax=40x, and a=40=the diameter. 402 X .7854=1256.6400=the number of rods in the area. 1256. 6400-160=7.854-the number of acres, and 7.854×120

=$942,48.

Example 14. See Figure 7. Let x=one side, then x2÷ 4X√3 the area. √3=1.73219716; therefore, x2÷4× 1.732=219.5375. Clearing of fractions, 1.732x2 = 878. 1500, x2=507.015011547, and x=22.5167=a side. By Art. 28, CB R::BP sin PCB, by changing the order of the couplets BP: sin BCP::BC; R, by inversion, sin BCP BP::R: CB, substituting numbers, sin 60° 11. 25835::1: CB; and by the tables, 9.9375300; 1.0512213 ::10.0000000:0.1136907=12.9941006= R, and the di. ameter 26 nearly. 262 ×.7854-530.9304=the area.

Example 15. Let x=the diameter of the least, then 6x= the diameter of the greatest. By Art. 40, the ring between the 1st and the 6th (6x-x)x(6x+x)X.7854=x5×7xX .7854. By adding the several differences between the 1st and 6th, we have difference between 1st and 6th=247,4010, and making this equal to the former diff., we have 5×××× .7854-247.4010, 35x2 =315, x2=9, and x=3=the least diameter, and 6x=18=the greatest diameter. Again, let x= the 2d diameter, and as 3=the first, by Art. 40,(x+3)x(x-3) X.7854 the diff. between the 1st and 2d. By the conditions this diff=21.2058; therefore 21.2058=(x+3)x(x-3)X.78 54. By reduction x =9+27=36, and x=6=the 2d diameter. Again, let the 3d diameter, and as 6=the 2d, (x+6) ×(x-6)×.7854–35.343, and x2 −36=35.343÷.7854 = 45, and r2=45+36=81, and x=9=the 3d diameter.

Let x the 4th diameter, and as 9 the 3d, (x+9)× (x-9)X.7854-49.4802, and by reduction, x2-81-49.4802 ÷.785463, and x2=144, and x=12=the 4th diameter.

Let the 5th diameter, and as 12 the 4th, (x+12) X (x-12)×.7854=63.6174, and by reduction x2-144=81,

x2=144+81=225, and x-15=the 5th diameter.

There

fore the diameters of circles are severally 3, 6, 9, 12, 15, 18.

Note. Solutions to the 16th and 17th examples, may be found in the Appendix.

SOLIDS BOUNDED BY PLANE SURFACES.

PROBLEM I. To find the solidity of a prism.

Example 1. 1 foot 3'x9' 11' 3", and 31 feetX 11′ 3′′= 29 feet 0' 9"=29 feet 103 inches the solidity of the stick of timber.

Example 2. 2 feet 6' × 22 feet=55 feet which multiplied by 12 feet 660 cubic feet.

Example 3. By Art. 44, (2 ft. 3')3 the required area. 2 ft. 3'x2 ft. 3'5 ft. 0' 9" and this into 2 ft. 3'11 ft. 6' "3""=11 feet 675 inches the solidity.

Example 4. 36 ft.=432', and 432′ × 108′=46656, which divided by 1728=27 cubic feet.

Example 5. 10 ft. 4'x10 ft. 4' 106 ft. 9' 4", and this multiplied by 2 ft. 3'=240 ft. 3′-240 cubic feet.

Example 6. By Art. 16. R 1.18::cot 36° the perpendicular; and, by the tables, 10.0000000 : 0.2552725::10.1 387390 0.3940115. the perim. of the pentagon=9, whose logarithm=.9542425, which added to log. perpendicular= 1.3482540, and this added to 1.3617278, log. 23 becomes 2. 7099818 whose nat. number=512.77156118-the solidity.

Example 1. Art. 46. 4 ft. 2′=50 inches, and 11 ft. 9'= 141 inches. 502=2500=the base, and 141×2500=3525 00 the number of cubic inches. 352500282=1250=the

answer.

Example 2. 3 ft. 11'=47 inches, 3 ft. 36 inches, and 462 ft. 5544 inches. 47 × 36 × 5544-9380448, and this divided by 231, gives 40608 for the quotient.

Example 3. 43-64, and 64 × 1000=64000 ounces, which divided by 16=4000 pounds.