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144.16-length of the arch 5207-76=area. [ABC, by Art. 23.

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ART. 26. To find the Area of a Segment of a Circle.

Definition. A segment of a circle is any part of a circle cut off by a right line drawn across the circle, which does not pass through the centre, and is always greater or less than a semicircle.

EXAMP. 1. To find the area of the segment ABC, whose chord AC is 172, the chord of half the arch ABC, viz. BC=102, and the versed sine BD=58.48.

RULE. By Art. 23, find the length of the arch line ABC, and by Art. 24, the diameter FB; then multiply half the chord of the arch ABC by half the diameter, and the product will be the area of the sector ABCE: then find the area of the triangle AEC, whose base AC is 172, and perpendicular height 34, found by subtracting the versed sine BD from half the diameter; and the area of the triangle AEC, being subtracted from the area

D

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F

of the sector ABCE,' will leave the area of the segment ABC.

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EXAMP. 2. In the segment ABCD greater than a semicircle, given the chord of the whole segment AD=136, the chord AC of

half the arch ACD=146, the chord AB or BC one fourth of the arch ACD-86, and the radius AE or ED =80, to find the area of the segment ABCD.

B

First find the area of the sector ABCDE, by Art. 25, at the second Example; then find the area of the triangle AED, by Art. 6, and, adding the area of the triangle to the A area of the sector, you will have the area of the segment.

86-chord AB.

2

98

146

80

C

E

136

8.666

172

=double of AB, add.

172

146 chord AC, subtract.

3)26

180.666 arch line ABC. 80-radius.

14453.280 area of the sector Carried over.

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C

D

Note 1. The area of a Lune or Crescent, is calculated by the preceding rule. A Lune is a figure made by two circular arcs which intersect each other, as ACBD. The area of the Lune is the difference of the two segments, which are contained by the arcs and the chord. Thus the A difference of the segments ACBE and ADBE is the area of the crescent ACBD.

Note 2. A Circular Zone is a figure contained between two parallel chords. If the chords be equal, it is called a middle zone, as A ABCD. The area of a zone is evidently the difference between the area of the circle and L the areas of the two segments.

E

B

B

C

ART. 27. To find the Area of an Ellipsis. Definition. An ellipsis, or oval, is a curve which returns into itself like a circle, but has two diameters, one longer than the other, the longest of which is called the transverse, and the shortest the conjugate diameter.

RULE. Multiply the two diameters of the ellipsis together; then multiplying the product by 7854, this last product will be the area of the ellipsis.

EXAMP. In the ellipsis ABCD, the transverse diameter AC is 88, and the conjugate diameter BD is 72, to find

the area.

B

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176

616

6336

•7854

25344

31680

50688

44352

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4976.2944-area

Mensuration of Superficies is easily applied to Surveying thus, take the angles of the plot with a good compass then measure the

sides with Gunter s chain, which note down in links (or chains and links, which is done by separating the two right hand figures of your links by a comma, your chain being 100 links) then cast up the contents, according to the rule of the figure, cutting off the five right hand figures of the product, and those at the left hand, if any, are acres; then multiply the five figures cut off, by 4, by 40, and by 2724, cutting off as before, and those at the left hand, will be roods, poles, and feet, respectively.

SECTION II. OF SOLIDS.

Solids are measured by the solid inch, foot, or yard, &c. 1728 of these inches, that is 12×12×12, make one cubick or solid foot. The solid content of every body is found by rules adapted to their particular figures.

ART. 28. To measure a Cube.*

Definition. A cube is a solid of six equal sides, each of which is an exact square.

* Here follows a Table of the Proportions, which the following Solids have to the Cube and Cylinder, having the same Base and Altitude.

1. A Cube whose side s 12 inch 8, contains

Solid Inches.

1728

2. A Prism, having an equilateral triangle, whose side is 12 inches from its Base, and its Altitude 12 inches, contains

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3. A Square Pyramid, whose height and the side of its base are each 12 inches, is of the above cube, and therefore contains

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249'413

44 Triangular Pyramid, whose height and side of its triangu-}

lar base are each 12 inches, is near of the cube, and contains
5. A Cylinder, whose diameter and height are each 12 inches,
114 of the above cube, and contains

is 1.

1357'17

6. A Sphere or Globe, whose axis or diameter is 12 inches, equal 904'78 to the side of the cube, is 11 of it, and contains

7. A Cone, whose base and altitude are each 12 inches, equal to the side of the cube, is -5, of it, and contains

9

8. A Parabolick Conoid, whose diameter at the base and height, are each 12 inches, being its circumscribing cylinder, contains

9 A Hyperbolick Conoid, whose height, and diameter at the base, are each 12 inches, is 5 of its circumscribing cylinder, and 7/2 contains

10. A Parabolick Spindle, whose height and middle diameter are each 12 inches, is of its circumscribing cylinder, and contains Hence arises a different method of finding their contents.

15

452*38829

678'583

665'49

723-824

General Rule. If the base of the solid, whose contents you would find, be rectilinear, consider it as Parallelopipedon; if curved, as a Cylinder, and find the content accordingly: then take such a part of the content, thus found, as is specified in the preceding Table, which if the parts be taken in inches, will be the solid content of the given figure, in inches, which, divided by 1728, will give the cubick feet.

EXAMP. 1. There is a triangular prism, the side of whose base is 48 inches, and whose perpendicular height is 108 inches: what is its solid content?

The base being right lined, I consider it as a parallelopipedon, the side of whose base is 48 inches, and whose length is 108 inches, and as 784-24 is contained 2.20340712 times in a cubick foot; 2.20340712 is a divisor, to divide the

The solid foot is composed of 1728 inches; for a solid, that is 1 foot, or 12 inches every way, that is 12x12x12, contains 1728 inches.

RULE. Multiply the side by itself and that product by the same side, and this last product will be the solid content of the cube.*

EXAMP. The side of a cube AB, being 18

inches, or 1 foot and 6 inches, to find the con

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content of the parallelopipedon by; therefore 48X48×108÷220340712=112930-56 solid inches 65 353 solid feet.

Had the dimensions been given in feet, it would have been 4×4X9+ 2 20340712-65 353 feet.

EXAMP. 2. There is a square pyramid whose height is 12 feet, and the side of whose base is 3-5 feet; what is its content?

3.5X35X12+3=49 feet, Ans. EXAMP. 3. There is a triangular pyramid, whose height is 15 feet, and the side of whose base is 5 feet: what is its content?

5X5X15+7=53'57 feet, Ans. EXAMP. 4. There is a cylinder whose diameter is 25 feet, and whose length is 24 feet; what is its content?

Here, the diameter is to be considered as the side of the base of a parallelopipedon. Therefore, 25X25X24X11÷14=117'857 feet, Ans. EXAMP. 5. There is a spherical balloon, whose diameter is 50 feet; how many cubick feet of air does it contain?

Here, the diameter is to be considered as the side of a cube. Therefore,

50X50X50X11÷21=65476·19 feet, Ans. EXAMP. 6. There is a cone, whose height is 15 feet, and the diameter of whose base is 5 feet; what is its content?

Here, the diameter of the base is to be considered as the side of the base of a parallelopipedon, and its height, as the length. Therefore,

5X5X15X5+19-98'684 feet, Ans. EXAMP. 7. There's a parabolick conoid, whose diameter at the base is 2·9 feet, and whose height is 6 et; what is the content?

This solid being of a cylinder; we must first find the content as of that of a cylinder, and then halve ǹ Therefore,

29X2X6X1÷14-39'647, and 39'647÷2=19'823, Ans. EXAMP. 8. There is a hyperbolick conoid, whose diameter at the base is 2'9 feet, and whose height is 6 feet; what is the content?

First find the content of a cylinder.

2*9X2′9X6X11÷11=39′ 647, and 39 647X-5=16′519 feet, Ans. EXAMP. 9. There is a parabolick spindle, whose middle diameter is 2'9 feet, and whose length is 6 feet; required the content?

First, find the content of a cylinder.

2*9X29X6X11÷14=39'647, and 39‍647X-3=21145 feet, Ans.

15

* Multiplying a side by itself, or squaring a side, gives the area of the base, or the number of square inches, feet, &c. in the base; whence one inch, foot, &c. in height would give as many solid inches, feet, &c. as there are squares in the base; two inches, &c. in height, twice as many, and so on, and is the rule, when the sides are equal to each other. In the same way, the rule for the content of the Parallelopipedon is proved.

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