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Problem 15.

The Chord and versed Sine (or Depth) of a Segment of a Circle being given, to find the (whole) Diameter of the Circle.

Rule.

Divide the Square of half the Chord by the versed Sine (or Depth), and the Quotient will be the other Part of the Diameter, which added to the versed Sine, the Sum will be the whole Diameter fought. *

Example.

Suppose ACB be a Segment given, whose Chord A B is 36 Inches, and the versed Sine (or Depth) CD 6 Inches, what is the Diameter of the whole Circle CE?

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Operation. AD 18 × 18 = 324, which DC6=54, the Part DE. Then DE 54 + DC6=60 Inches; the whole Diameter E C required.

* Half the Chord of the Segment is always a mean Proportion between the two Parts of the Diameter; it will therefore ever hold,-as the versed Sine: is to half the Chord:: so is half the Chord: to the remaining Part of the Diameter.

Problem 16.

To find the Area of a Circular Ring.

Def. The Space included between two concentric Circles of different Diameters is called a Circular Ring.

Rule.

Multiply the Sum of the two Diameters by their Difference, and multiply this Product again by .7854; this last Product will be the Area required. *

Example.

Suppose the Diameter of the larger Circle AB be 30 Inches, and the Diameter of the smaller CD be 18 Inches, what is the Area of the Ring formed by these two Circles?

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30+18=48×12=576×7854452.3904

Operation.
Inches, the Area of the Ring required.

* Or, find the Area of each Circle, and fubtract the less from the bigger, the Remainder will be the Area of the Ring.

Problem 17.

To find the Area of a Crescent or Lune.

Def. A Lune, &c. is the Space included between the Arches of two eccentric Circles intersecting or cutting each other.

Rule.

Find the Areas of the two Segments from which the Crefcent or Lune is formed; and their Difference will be the Area required.

Example.

Let the Chord A B in the following Figure be 40 Inches, and the Height D C 10 Inches, and DE 4 Inches, what is the Area of the Lune or Crefcent ACBDE?

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Operation. By Problem 13th.

The Area of the bigger Segment is 279.57
Area of the leffer Segment is

107.50

The Area of the Lune or Crefcent required 172.07 In.

Problem 18.

To find the Area of an Oval.

Def. An Ellipsis or Oval is a Figure bounded by a regular Curve Line returning into itself; having one Diameter longer than the other. The longer Diameter is called the Transverse; the shorter, the Conjugate.

Rule

Multiply the Length by the Breadth, and that Product by .7854; this last Product will be the Content.

Example.

Suppose the longer Diameter A B of the following Oval be 61.6 Inches, and the shorter Diameter CD be 44.4 Inches, what is the Content?

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Operation. 61.6×44.4 2735.04×.7854 2148.100416 Inches, the Area in the Conic Sections.

Problem 19.

To find the Length of the Circumference of an Oval.

Rule.

Multiply half the Sum of the two Diameters by 3.1416, and the Product will be the Circumference near enough for Practice.

Example.

Suppose the longer Diameter AB be 50 Inches, and the shorter Diameter CD be 40 Inches, what is the Circumference of the Oval?

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Operation. Half 2 Diameters 45×3.1416 141.372 Inches, the Circumference required.

See another Method of finding the Circumference of an Oval or Ellipfis in the Conic Sections.

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