BOOK V.

ON THE PROPORTIONALITIES AND MEASUREMENT OF POLYGONS

AND CIRCLES.

THEOREM 1.

The area of any circle is equal to the product of its radius into half of its circumference.

Let CA be the radius of the circle, and AB a very small portion of its circumference, and CAB will be a sector; and we may conceive the whole circle made up of a great number of such sectors; and each sector may be as small as we please; and when very small, AB, BD, &c., each one taken separately, may be considered a right line; and the sectors CAB, CBD, &c., will be triangles. The triangle CAB, is measured by the base, CA, multiplied into half the altitude, (th. 30, b. 1) AB; and the triangle CBD is measured by CB, or its equal, CA, into half BD: then the area, or measure of the two triangles, or sectors, is CA, multiplied by the half of AB, plus the half of BD, and so on for all the sectors that compose the circle; therefore, the area of the circle is measured by the product of the radius into half the circumference. Q. E. D.

THEOREM 2.

Circumferences of circles are to one another as their radii, and their areas are to one another as the squares of their radii.

Let CA be the radius of a circle (see last figure), and Ca the radius of another circle. Conceive them to be placed upon each other so as to have the same center.

Let AB be a certain definite portion of the circumference of the larger circle, so that m times AB will represent that circumference.

But whatever part AB is of the greater circumference, the same part ab is of the smaller; for the two circles have the same number of degrees, and of course susceptible of division into the same number of sectors. But by proportional triangles we have,

CA: Ca AB : ab

Multiply the last couplet by m (th. 4, b. 2), and we have,
CA: CamAB : mab

That is, as the radius of one circle is to the radius of the other, so is the circumference of the one to the circumference of the other.

Q. E. D.

To prove the second part of the theorem, represent the larger circle by C, and the smaller by c; and whatever part the sector CAB is of the circle C, the sector Cab is the same part of the circle c.

Scholium. 1. Circles are to one another as the squares of their diameters; for if squares be described about any two circles, such squares will be squares on the diameters of the circles. But each circle is the same proportional part of its circumscribed square; and as like parts of things have the same proportion to each other as the wholes (th. 4, b. 2); therefore, circles are to one another as the squares of their diameters.

Scholium 2. As the circumference of every circle, great or small, is assumed to contain 360 degrees, if we conceive the circumference to be divided into 360 equal parts, and one such part represented by AB, on one circle, or ab on the other, AB and ab will be very near straight lines, and the length of such a line as AB will be greater or less according to the radius of the circle; but its absolute length cannot be determined until we know the absolute relation between the diameter of a circle and its circumference.

To measure the circumference of a circle, or, to discover exactly how many times, and part of a time, it is greater than its diameter, is a problem of some difficulty, and requires patience and care; and it can only be done approximately; for as far as investigations have extended, the circumference of a circle is incommensurable with its diameter.

To acquire a very clear and distinct idea of the ratio between the diameter and circumference of a circle, the pupil must commence with first approximations, and proceed with great deliberation.

Conceive a circle described on the radius CA, and in it describe a regular polygon of six sides (problem 26), and each side will be equal to the radius CA; hence the whole perimeter of this polygon must be six times the radius, or three times the diameter. Let CA bisect bd in a. Produce cb and cd, and through the point A, draw DB parallel to db; DB will then be a side of a regular polygon of six sides, described about the circle, and we can compute the length of this line, DB, as follows: The two triangles, Cod, and CBD, are equiangular, by construction; therefore,

Ca: db : CA: DB.

Now, let us assume CA, or cd, or the radius of the circle, equal unity; then db=1, and the preceding proportion becomes

But the whole perimeter of the circumscribing polygon is six

6.000000

6.9282032

Thus we have shown, that when the radius of a circle is 1, the perimeter of an inscribed polygon of six sides, is And of a similar circumscribed polygon, is But, if we call the diameter 1, the perimeter of the inscribed polygon of six equal sides will be,

And of the circumscribed, will be

3.0000000

. 3.4641016

As we would avoid all metaphysical verbiage in science, and come to the point at once, we lay it down as an axiom, that when the radius of a circle is 1, and of course the diameter 2, the circumference is greater than 6, and less than 6.9282032; and if the diameter is 1, the circumference must be greater than 3, and less than 3.4641016; and this we may call the first approximation to the ratio between the diameter and circumference of a circle.

Scholium 2. As the area of a circle is numerically equal to the radius multiplied by half the circumference (th. 2, b. 5), therefore, if we represent the radius by R, and half the circumference by л, and the area of the circle by a, then we shall have this equation:

If we now make R=1, this equation gives x=a; that is, when the radius of a circle is 1, the half circumference is numerically equal to the area. We will, therefore, seek the area of a circle whose radius is unity; and that area, if found, will be numerically the half circumference, and by inspecting the last figure, we perceive that it is perfectly axiomatic (the whole is greater than a part), that the area of the sector CbAd, is greater than the triangle Cbd, and less than the triangle CBD; and the area of the whole circle is greater than one polygon, and less than the other. Finding the AREA of a circle, or finding a square which shall be equal to a circle of given diameter, is known as the celebrated problem of squaring the circle.

THEOREM 3.

Given, the area of a regular inscribed polygon, and the area of a similar circumscribed polygon, to find the areas of a regular inscribed and circumscribed polygon of double the number of sides.

AB`a

Let C be the center of the circle; AB side of the given inscribed polygon; EF parallel to AB, a side of the circumscribed polygon.

If AM be joined, and AR and BQ be drawn as tangents, at A and B, AM will be a side of an inscribed polygon of double the number of sides; and AR=RM (scholium 2, th. 18, b. 3), BQ QM, and AR+RM=RQ= the side of the circumscribed polygon of double the number of sides.

The As ARC and RMC, are equal, for AC=CM. CR is common to both triangles, and AR=RM, tangents from the same point, R; therefore, CR bisects the angle ECM.

Now, as the same construction, and the same reasoning would take place at every one of the equal sectors of the circle, it is sufficient to consider one of them, and whatever is true of that arc, would be true of every one, and true for the whole circle, and its polygons.

To avoid confusion, let p represent the area of the given inscribed polygon, and P the area of the similar circumscribed polygon. Also let p' represent the area of an inscribed polygon of double the number of sides, and P' the circumscribed polygon of double the number of sides.

As the As ACD and ACM have the common vertex A, they are to each other as their bases, CD to CM; they are also to each other as the polygons of which they form a part.

But, because of the common vertex, M, the two As, CAM and CEM, are to each other as CA to CE. But the As are like parts of the polygons p' and P; we have,