are infinitely fmall and coincide with the periphery of the circle, and whofe vertices meet in the center; and the laft rule for regular polygons will ferve for obtaining the area of the circle; viz. Multiply the radius of any circle (which is half the diameter) by half the periphery of the circle, and the product is the area.

Now by feveral operations and tedious proceffes, the peripherys are found in proportion to the diameters of circles very near, as 3.1416 to I. Hence it appears that the area of that circle whofe diameter is multiply'd by half its circumference is equal to .7854; and the area of every circle bears proportion to the fquare of its diameter as 7854 does to 1; fo that if you multiply the fquare of any circle's diameter by .7854, the product will give the area required.

Note. The fame rules ferve for a femicircle, or quadrant, by taking the parts accordingly.

Here follows feveral useful propofitions about a Circle.

The diameter of a circle being given, to find the circumference.

As I is to 3.1415927, fo is the diameter to the circumference.

The circumference being given to find the diameter. As 3.1416 is to I, fo is the circumference to the diameter.

The circumference being given to find the area. As I is to .0795775, fo is the fquare of the circumference to the area required.

By the area of a circle to find the diameter. As I is to 1.2732395, fo is the content to the fquare of the diameter.

By the area of a circle to find the fide of a fquare equal to it.

Extract the fquare root of the area.

By the area of a circle to find the circumference. As I is to 12.5663706, fo is the area to the fquare of the circumference.

Having the circumference of a circle to find the fide of the infcribed fquare.

As I is to .2250791, fo is the circumference to the fide infcribed.

Having the area of a circle to find the area of the infcribed fquare.

As is to .6366197, fo is the area of the circle drawn about the fquare, to the infcribed fquare.

PROBLEM IV. Of an Ellipfis.

An ellipfis or oval bears the fame proportion to its circumfcribing parallelogram as the circle does to its circumfcribing fquare. Therefore multiply the tranfverfe by the conjugate diameter, and that product again by .7854 for the

PROBLEM V. Of a Parabola.

A parabola is of the parallelogram circumferibing it. Therefore multiply the greatest ordinate by of the correfponding abscissa, and the product is the area required.

PROBLEM VI. Of a Sector.

A fector is equal to a triangle whofe bafe is equal to the fectoral arch, and perpendicular altitude equal to the radius which defcribes that arch. Therefore multiply the radius by half the arch-line for the area of the sector.

But first to find the length of the arch-line, multiply the chord of half the fegment A B by 8, and from the product fubtract the chord of the whole fegment A C, divide the remainder by 3, and the quotient is the arch-line A B C fought.