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Problem 10.

To find the Area of any Space of Archimedes' Spiral. Let the Space given be ADBCA, to find its Area.

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Make the Distance AC the Radius of a Circle circumscribing the Spiral; then, find the Area of the whole Circle, and divide it by 3; the Quotient will give the Spiral Area A DBCA required. *

* The Area of the Spiral Space of Archimedes is always equal to (one-third) Part of the circumfcribing Circle.

Problem 11.

To find the Area of a Cycloid.

Let the Cycloidical Space given be ABE, to find its Area.

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If a Circle or a Wheel roll along upon the right Line A B, till it perform one Revolution, i. e. till it measure out a right Line equal in Length to its Circumference, then that Point in the Circle, as D, which first touched the right Line A B, will describe, when it arrives at B, the Curve A EB, called a Cycloid.

Rule to find the Area.

Find the Area of the generating Circle HDCK, and multiply it by 3, the Product will give the Cycloidical Space required.

To find the Length of the Arch A E B.

Multiply the Diameter of the generating Circle HDCK by 4, and the Product will give the Length of the Arch required.

The Area of a Cycloid is equal to three Times the Area of the generating Circle; and the Length of the Arch is equal to 4 Times the Diameter of the fame Circle.

Problem 12.

To find the Area of a Segment, or Part of a Sector of a

Circle.

Rule.

Multiply half the Sum of the two Arches by the Diftance between them, (or by one of the Ends) and the Product will give the Area.

Example.

Suppose the Length of the Arch AB be 84 Inches, the Arch CD 72.5 Inches, and the Distance between them EF 3.5 Inches; what is the Area of the Segmental Space ABDC?

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The Sum of the two Arches AB and CD is 156.5; the half is 78.25; which multiplied by EF or AC 3.5, gives 273.875 Inches, the Area required.

Problem 13.

To describe a Parabola, by having only the Bafe or greatest Ordinate AB, and the Height or Axis VD given.

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Draw a Line from the Vertex V to the End of the Ordinate or Base at B. Divide that Line into two equal Parts, as at b; upon b erect a Perpendicular, and where it cuts the Axis V D, as at C, set one Foot of the Compaffes, and with the other opened to the Vertex V, describe the Circle as in the Figure. So will the Distance between the Base or Ordinate A B, and that Point where the Axis continued cuts the Circle, as DR, be the Latus Rectum fought. One Quarter of which is always the true Distance of the Focus from the Vertex or Top of the Parabola. This being obtained, we may proceed to delineate the Parabola as follows.

To delineate the Parabola.

Take of the Latus Rectum DR in the foregoing Scheme, and fet it from the Vertex V in the next Figure both Ways, upwards and downwards, to fand F, (equal to the Distance of the Focus from V.)

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Next, let a Number of Points, as a, a, a, &c. be taken in the Axis, and through each draw perpendicular Lines, as ee, ee, ee, &c. and all parallel to one another. Then with a Pair of Compasses take the Distance af, and with one Foot in the Focus F strike Dashes across each Parallel respectively, in e and e, &c. then with an even Hand draw a Curve through those Points, and it will form the Parabola required.

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