5. If the diameter of a cylindrical wine-vat be 78.6 inches, what is the area of its base, in wine gallons?

Ans. By Division, the area is 21.0048; and by the table of wine areas, 21.005 wine gallons.

6. The diameter of a cylindrical vessel is 53.2 inches, required its area and content in ale gallons, when the depth of the liquor is 35.7 inches.

Ans. By Multiplication, the area is 7.8822, and the content 281.39454 ale gallons.

GENERAL RULES.

The following Rules will solve most of the useful Problems relating to the circle and its equal or inscribed square, &c.

RULE 1. The diameter of a circle multiplied by .8862269, will give the side of a square equal in area.

2. The circumference of a circle multiplied by .2820948, will give the side of a square equal in area.

3. The diameter of a circle multiplied by .7071068, will give the side of the inscribed square.

4. The circumference of a circle multiplied by .2250791, will give the side of the inscribed square.

5. The area of a circle multiplied by .6366197, and the square root of the product extracted, will give the side of the inscribed square.

6. The side of a square multiplied by 1.414214, will give the diameter of its circumscribing circle.

7. The side of a square multiplied by 4.442883, will give the circumference of its circumscribing circle.

8. The side of a square multiplied by 1.128379, will give the diameter of a circle equal in area.

9. The side of a square multiplied by 3.544908, will give the circumference of a circle equal in area.

PROBLEM XIV.

To find the area of the sector of a circle.

RULE.

By the Pen.

Multiply the length of the arc by the radius of the

sector, and half the product will be the area in square inches; which being divided by 282, 231, and 2150.42, the respective quotients will be the area in ale and wine gallons, and malt bushels.

Note 1. The length of the arc may be found by Problem XII.

2. The area of a quadrant may be obtained by taking, and the area of a semi-circle by taking of the area of the whole circle.

EXAMPLES.

2

1. What is the area of the sector ABCD, in ale gallons, when the radius A B or AD measures 45, the chord BD of the whole arc 72, and the versed sine CE 18 inches?

Here √36+18=√1296+324v 1620 40.2492

=DC, the chord of half the arc; and

40.2492 x 8-72

3

As the square divisor on A, is to the radius of the sector on B; so is half the length of the arc on A, to the

area on B.

2. The radius of a sector is 31.5, and the length of the arc 42.6 inches; what is its area in wine gallons? Ans. 2.9045 wine gallons. 3. The radius of a sector is 60, and the chord of the whole arc 72 inches; what is its area in malt bushels ? Ans. 1.0768 malt bushels.

4. The radius of a cooler, placed in an oblique angle of a brew-house, measures 80, the chord of the whole are 124.8, and the chord of half the arc 69.3 inches; required the area and content in ale gallons, when the depth of the liquor is 8.7 inches.

Ans. The area is 20.312, and the content 176.7144 ale gallons.

PROBLEM XV.

To find the area of the segment of a circle.

RULE I.

By the Pen.

To two thirds of the product of the chord and height of the segment, add the cube of the height divided by twice the chord, and the sum will be the area in square inches. Divide this area by 282, 231, and 2150.42, and the respective quotients will be the area in ale and wine gallons, and malt bushels.

Note 1. If the segment be greater than a semi-circle, find the area of the remaining segment, which subtract from the area of the whole circle; and the remainder will be the area of the segment required.

2. When the chord and versed sine are given, divide the square of half the chord by the versed sine; to the quotient add the versed sine, and the sum will be the diameter of the circle; hence its area may be found by Prob. 13. Or, multiply the square of the diameter by .7854, and the product will be the area in square inches; from this take the area of the remaining segment, and the difference will be the area of the segment required, in square inches.

EXAMPLES.

1. The chord A B measures 120, and the versed sine

CE 30 inches; what is the area of the segment AC B, in ale and wine gallons, and malt bushels ?

C

B

E

7200
3

Here 120 × 30x =3600 × }= =2400, two-thirds

of the product of the chord and versed sine, or height of the

segment; and

303 120 x 2

27000

240

-=112,5, the cube of the

height divided by twice the chord; then 2400+112.5= 2512.5, the area in square inches; whence, 2512.5÷282= 8.9095, the area in ale gallons; 2512.5÷231=10.8766, the area in wine gallons; and 2512.5÷2150.42=1.1683, the area in malt bushels.

2. The chord A B is 128, and the versed sine CE 32 inches; what is the area of the segment in ale and wine gallons?

Ans. The area is 10.137 ale, and 12.375 wine gallons.

3. Required the area of the segment A D B, greater than a semi-circle, in ale and wine gallons, and malt bushels; the chord A B measuring 60, and the versed sine ED 40 inches.

Ans. 7.351 ale gallons, 8.974 wine gallons, and .964 malt bushels.

RULE II.

Divide the versed sine or height of the segment by the diameter of the circle of which the segment is a part; and find the quotient in the column of heights or versed sines, in the Table at the end of Part IV.

Take out the corresponding Area Seg. which multiply

by the square of the diameter, and the product will be the area of the segment in square inches. Divide the area thus obtained by 282, 231, and 2150.42, and the respective quotients will be the area in ale and wine gallons, and malt bushels.

Note 1. If the quotient of the height by the diameter do not terminate in three places of figures, without a fractional remainder, find the Area Seg. answering to the first three decimals of the quotient; subtract it from the next greater Area Seg.; multiply the remainder by the fractional part of the quotient, and the product will be the corresponding proportional part to be added to the first Area Seg.

EXAMPLE. If the versed sine be 25, and the diameter 55, we have =.454, the tabular height. The Area Seg. answering to .454, is .346764; the next greater Area Seg. is .347759; their difference is 000995; then .000995 × 1 = .000542, which being added to .346764, gives .347306, the Area Seg. corresponding to .454.

Note 2. When the area of a segment greater than a semi-circle is required, subtract the quotient of the height by the diameter, from 1; find the Area Seg. corresponding to the remainder, which take from .785398, and the difference will be the Area Seg. answering to the quotient.

EXAMPLE. If the versed sine be 66, and the diameter 80, we have = .825, the tabular height; then 1.825.175; and the Area Seg. answering to.175, is .092313, which being taken from .785398, leaves .693085, the Area Seg. corresponding to .825.

EXAMPLES.

1. The chord AB measures 120, and the versed sine CE 30 inches; what is the area of the segment ACB, in ale and wine gallons, and malt bushels ?

By Note 2, under Rule 1, we have

602

30

3600

+30=- +30

30.0

30

120+30=150, the diameter CD; then =.2, the ta

150

bular height, or quotient of the versed sine divided by the diameter. The Area Seg. corresponding to this quotient, is .111823; then .111823 x 150.111823 x 22500= 2516.0175, the area in square inches; and 2516.0175÷ 282 8.922, the area in ale gallons; 2516.0175-23110.8918, the area in wine gallons; and 2516.0175÷2150.42 1.17, the area in malt bushels.