-

remainder; and 197 112 85, the third remainder; whence197 x 29 x 83 x 85√40305215 = 6348.638 inches, the area of the second triangle.

sum of the sides; then 146.5

114 32.5, the first remainder; 146.5-87 = 59.5, the second remainder; 146.5 92 = 54.5, the third remainder; whence 146.5 x 32.5.x59.5 x 54.5=1543954343.75 =3929.318 inches, the area of the third triangle; hence we have 6312.541 + 6348.638 + 3929.318 16590.497 square inches, the area of the whole polygon.

By Prob. IX., Part IV., we have 52.6 × 8 × 63.7

26804.96

=26804.96, and

= 13402.48,

the area in

2

58.019, the

4891.0017,

square inches; and 13402.48231

area in wine gallons; then, 58.019 × 84.3

the content in wine gallons.

EXAM. 22.

By Prob. X., Part IV., we have 87.6 × 87.6 × .021921 7673.76 x .021921 = 168.21649296, the area in ale gallons.

EXAM. 23.

By Prob. XIII., Part IV., we have 226.4 × 2264 → 359.0551256.96 359.05 142.757, the area of the base in ale gallons; and 142.757 x 258.8 36945.5116, the content in ale gallons.

EXAM. 24.

(See the Figure in Problem XIV., Part IV., of the Gauging.)

By the question, AD = 52, and DE = 43; then by Prob. V., Part I., we have 522 432 = √2704 1849 855 = 29.24 = AE; and AC AE 52 22.76, the versed sine

EC.

29.24

Again, by Prob. IV., Part I., we have √43% +22.76= √1849 +518.0176 = √2367.0176 = 48.653 = DC, the chord of half the arc.

IV., we have

Again, by Prob. XII., Part 48.653 × 8 86 389.224 86

3

=

3

= 101.0746, the length of the arc DCB.

=

303.224

3

Lastly, by Problem XIV., Part IV., we have 101.0746 x 52 5255.8792

= 2627.9396,

the

area of the sector ABCD, in square inches; and 2627.9396 ÷ 231 = 11.3763, the area in wine gallons.

EXAM. 25.

By Prob. XV., Part IV., we have 92 × 38 ×

3496 × % =

6992
3

=

=2330.6, two-thirds of the preduct of the chord and versed sine, or height of the

segment; and

383
92 × 2

54872

184

=298.217, the cube

of the height divided by twice the chord, then 2330.6 + 298.217 = 2628.817, the area in square inches ; and 2628.817282 = 9.322, the area in ale gal

lons; whence 9.322 x 6.4 = 59.6608, the content in ale gallons.

EXAM. 26.

By Prob. XVI., Part IV., we have 125.4 x 82.8 273810383.12 2738 3.792, the area in

malt bushels.

EXAM. 27.

By Prob. XVII., Part IV., we have

21.8

= 41611,

52.4

the tabular height; and the corresponding Area Seg. is .309125; then .309125 X 65.6 × 52.4 = 1062.59864, the area of the 1062.59864

and

282

segment in square inches ;

= 3.76808, the area in ale gallons.

EXAM. 28.

By Prob. XVIII., Part IV., we have 98.6 × 75.4 × 98.6 x 75.4 × 2 7434.44 x 2

==

14868.88

3

4956.293, the area in square inches; and 4956.293 231 = 21.455, the area in wine gallons.

EXAM. 29.

By Prob. IV., Part I., we have √80o + 602 6400+ 3600 = 10000 = 100 inches, the hypothenuse or chord of the segment.

Then, by Prob. V., Part IV., we have

4800

2

inches.

80 x 60

2

=2400, the area of the triangle in square

Again, by Prob. XV., Rule I., Part IV., we have

5000

100 x 25 x = 2500 x =

= 1666.600.

3

two-thirds of the product of the chord and versed sine

or height of the segment; and

253

=

100 x 2

1562

200

=78.125, the cube of the height divided by twic the chord; then 1666.666 + 78.125 = 1744.791, the area of the segment in square inches.

Now, 2240 + 1744.791 = 3984.791, the area a the cooler in square inches, and 3984.791 282 = 14.13, the area in ale gallons; whence 14.13 × 9.4 = 132.822, the content in ale gallons. (See Proo XIX., Part IV.)

EXAM. 30.

=

By Prob. XX., Part IV., we have 49.5 + 49.6 = 99.1, the sum of the first and last ordinates; (51.4 + 52.6 52.7 + 51.5) × 4 = 208.2 × 4 =832.8, four times the sum of the even ordinates; (52.2 + 53.2 + 52.3) × 2 = 157.7 x 2 = 315.4, twice the sum of the odd ordinates; and 97.6 ÷ 8 = 122, the common distance of the ordinates; then, (99.1 + 12.2

832.8+ 315.4) ×

= 5072.353, the 5072.353231

12.2

3

=1247.3 X

3

=

15217.06

3

area in square inches; hence = 21.958, the area in wine

By Prob. I., Part V., we have 64.3 × 64.3 × 643 = 4134.49 × 64.3 265847.707, the content in cubic inches; hence 265847.707 ÷ 282 = 942.722. the content in ale gallons; and 265847.707 ÷ 231 = 1150.855, the content in wine gallons.

EXAM. 32.

By Prob. II., Part V., we have 145 × 96 × 84 = | 13920 × 84 = 1169280, the content in cubic inches:

and 1169280282 4146.382, the content in ale =-gallons.

EXAM. 33.

By Prob. X., Part IV., we have 86.4 × 86.4 × .001689 = 7464.96 × .001689 = 12.60831744, the area of the base in malt bushels; and by Prob. III., Part V., 12.60831744 x 73.6 927.972163584, the content in malt bushels.

ЕХАМ. 34.

9025

5776 =3249, the square of the diameter of the base; then by Prob. IV., Part V., 3249 × 76 ÷ 359.05 =246924 ÷ 359.05 = 687.714, the content in ale gallons.

EXAM. 35.

By Prob. X., Part IV., we have 53.8 × 53.8 X .001875 = 2894.44 × .001875 = 5.42707500, the area of the top in wine gallons; and by Note 4, Prob. V., Part V., 5.42707500 × 20.8112.883, the content in wine gallons.

EXAM. 36.

By Rule II., Prob. VI., Part V., we have 45.7 x 45.7 2088.49, the square of a side of the greater end; 34.3 × 34.3 = 1176.49, the square of a side of the less end; and 45.7 × 34.3 =1567.51, the product of the sides; then (2088.49 + 1176.49 + 1567.51) × 22.8 = 4832.49 x 22.8 = 110180.772, the content in cubic inches. Hence 110180.772 x .0008 = 88.1446176, the content in malt bushels.

EXAM. 37.

By Prob. V., Part I., we have 2 852

752 =

27225
the diameter of the top; then by Rule I., Prob. VII.,

5625 = 2/1600 = 40 X 2 = 80,