Example.

Suppose the Diameter of a Circle A B to be 22.6 Inches, what is its Circumference, and also its Area?

First. As 7: 22 :: 22.6 : 71.028, according to Archimedes.

Or, as 113: 355 :: 22.6 : 71, according to Metius. Or, as 1: 3.141593 :: 22.6: 71.0000018, according to Van Geulen.

Then half the Circumference multiplied by half the Diameter gives the Area = 401.15 Inches.

Note. If the Circumference of a Circle be given, the Diameter may be found by reverting the above Proportion, thus,

As 22:7:: Circumference to the Diameter;

Or, as 355: 113 :: Circumference to the Diameter; Or, as 3.14159: 1 :: Circumference to the Diameter.

Other Proportions are as 106 : 333

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Problem 11.

To find the Area of a Circle; of a Semi-circle; or of a Quadrant.

Rule.

Square the Diameter, and multiply the Product by.7854 (a Decimal); this last Product will be the Area of the Whole Circle; which divided by 2 gives the Area of the Semi-circle; or by 4 gives the Area of the Quadrant.

Example.

Suppose the Diameter A B of the following Semi-circle ABC is 41.2 Inches, what is its Area?

Operation. 41.2×41.2=1697.44.7854-1333.169376

nches, Area of the Whole Circle.

Then, 1333.169376÷2=666.584688, Area of the Semiircle A B C.

And, 1333.169376-4-333.292344, Area of the Quadrant

LEC, or BEC.

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Problem 12.

To find the Area of the Sector of a Circle.

Def. The Sector of a Circle is a Figure bounded by two Semi-diameters, or Radius's, having fome Part of the Periphery or Circumference for its Bafe.

Rule. *

Multiply the Semi-diameter or Radius by half the Arch of the Sector, and the Product will be the Area.

Example.

Let ABC be the Sector of a Circle, whose Radius A C, or BC, is 20.2 Inches; and half the Arch A B, 15 Inches, what is its Area?

C

B

Operation. 20.2×15=303.0 Inches, the Area fought.

* A Sector of a Circle is equal to a Triangle whose Base is equal to the Arch, and Height equal to the Radius or Semi-diameter.

Problem 13.

To find the Area of a Segment of a Circle.

Def. A Segment of a Circle is a Part cut off by a right Line, less than the Diameter, drawn within the Circle, as ED, fo that EBD is a Segment less than a Semi-circle; and E AD is greater than a Semi-circle.

Rule.

To the Square of half the Chord add of the Square of the Depth; the Square Root of this Sum multiplied by of the Depth will give the Area of the Segment. *

Example.

Suppose the Chord ED of the Segment EDB be 40

Inches, and the Depth FB 10 Inches, what is its Area?

Operation. 20×20=400+40=440, whose Square Root is 20.973×13.33 279.57, the Area fought.

4

Problem 14.

To find the Length of an Arch of any Circle?

Rule.

Multiply the Chord of the Arch by 8; from the Product fubtract the whole Chord; divide the Remainder by 3; and the Quotient will be equal to the Length of the Arch required. *

Example.

Suppose the Chord AC of the Arch ABC be 50.8 Inches, and the Chord A B of half the Arch be 30.6, what is the Length of the Arch ABC?

B

C

Operation.
Length of the Arch required.

30.6×8=244.8-50.8=194.0÷3=64.6, the