To circumscribe a Circle about a Regular Polygon.

Solution Find, by § 149, the circumference of a circle which passes through the three vertices A, B, C ; and this circle is circumscribed about the given polygon.

Proof. Suppose the circumference divided into the same number of equal arcs AB', BC', &c. as that of the sides of the given polygon. The chords AB', B'C', &c. form, by 202, a regular polygon, which, by § 206, is similar to ABCD &c.

Hence,

the angle ABC= the angle AB'C' ;

and, consequently, by § 108, the arc ABC, which is twice the arc AB, is equal to the arc AB'C', which is twice the arc AB'. We have then,

and the chord AB is equal to the chord AB', and coincides with it. The polygons AB'C'D' &c., ABCD &c., must, therefore, by 195, coincide; and the circle is circumscribed about the given polygon.

223. Corollary. There is a point 0 in every regular polygon equally distant from all its vertices, and which is called the centre of the polygon.

224. Corollary. If we join AO, BO, CO, &c., the angles AOB, BOC, COD, &c. are all equal, and each has the same ratio to four right angles, which the arc AB has to the circumference.

225. Corollary. The isosceles triangles AOB, BOC, COD, &c. are all equal.

226. Corollary. The angles OAB, OBA, OBC, OCB, &c. are all equal, and each is half of the angle ABC.

227. Problem. To inscribe in a given circle a regu

To inscribe in a Circle any Regular Polygon.

lar polygon, similar to a given regular polygon ABCD &c. (fig. 123)

Solution. From the centre 0 of the given polygon draw the lines AO, BO; at the centre O' of the given circle make the angle A'O'B' equal to AOB, and the chord A'B', being applied round the circumference as many times as ABCD &c. has sides, gives the required polygon AB'C'D' &c., as is evident from § 224.

228. Theorem. The sides of a regular polygon are all equally distant from its centre.

Proof. Let fall the perpendiculars OM, ON, OP, &c. (fig. 122), from the centre 0, upon the sides AB, BC, &c. In the right triangles OAM, OBM, OBN, OCN, OCP, &c., the hypothenuses OA, OB, OC, &c. are all equal, by § 223, and the legs AM, MB, BN, NC, CP, &c. are equal, since each is, by § 116, half of AB, or of its equal BC, &c. The triangles OAM, OBM, OBN, &c. are, consequently, equal, by § 64; and the perpendiculars OM, ON, OP, &c. are equal.

229. Problem. To inscribe a circle in a given regular polygon ABCD &c. (fig. 124).

Solution. From the centre 0 of the polygon, with a radius equal to OM, the distance of AB from O, describe a circle, and it is the required circle.

Proof. The distances OM, ON, OP, &c. are all equal, by § 228, and therefore the circumference passes through the points M, N, P, &c.; and the sides AB, BC, CD, &c. are all, by § 120, tangents to the circle; and the circle is, by § 118, inscribed in the polygon.

230. Problem. To circumscribe about a given circle a polygon similar to a given inscribed polygon ABCD &c. (fig. 125). 6

Homologous Sides of Regular Polygons.

Solution. Through the points A, B, C, D, &c. draw the tangents A'B', B'C', C'D', &c. and A'B'C'D' is the required polygon.

Proof. The triangles AB'B, BC'C, &c. are by § 151, isosceles; they are also equal, for the sides AB, BC, &c. are equal, and the angles BAB', ABB', CBC', BCC', &c. are equal because they are measured by the halves of the equal arcs AB, BC, &c. Hence the angles A', B', &c. are equal, and the sides A'B', B'C', &c. are equal, and A'B'C' &c. is a regular polygon of the same number of sides with ABC &c.

231. Corollary. A regular polygon of 4, 8, 16, &c.; 3, 6, 12, &c. ; 5, 10, 20, &c.; 15, 30, 60, &c. sides; or, one similar to any given regular polygon may, therefore, be circumscribed about a circle by means of §§ 212-221, and 228.

232. Theorem. The homologous sides of regular polygons of the same number of sides are to each other as the radii of their circumscribed circles, and also as the radii of their inscribed circles.

Proof. Let ABCD, &c., A'B'C'D', &c. (fig. 126) be regular polygons of the same number of sides, and let O, O' be their centres; OA, O'A' are the radii of their circumscribed circles, and the perpendiculars OP, O'P' are the radii of their inscribed circles.

==

Join OB, O'B'. The triangles OAB, O'A'B' are similar, by 173, for the angle OAB = OBA O'B'A' = O'A'B' for each is half the angle ABC=A'B'C'. Hence, by § 198,

OP : O'P' = AB ; A'B' — OA : O'A'.

The Ratio of a Circumference to its Diameter.

233. Corollary. Hence, the perimeters of regular polygons of the same number of sides are, by § 207, to each other as the radii of their inscribed circles, and also as the radii of their circumscribed circles.

234. Theorem.

The circumferences of circles are

to each other as their radii.

Proof. For circles are similar regular polygons, by § 208, and the radii of their inscribed and circumscribed circles are their own radii.

235. Corollary. The circumferences of circles are to each other as twice their radii, or as their diameters. 236. Corollary. If we denote the circumference of a circle by C, its radius by R, and its diameter by D ; also the circumference of another circle by C', its radius by R', and its diameter by D', we have

Hence, the circumference of every circle has the same ratio to its radius; and also to its diameter.

237. Corollary. If we denote the ratio of the circumference, C, of a circle to its diameter, D, by л, we have

238. Corollary.

Unit of Surface.

is the circumference of a circle

whose diameter is unity, and the semicircumference of a circle whose radius is unity.

CHAPTER XIII.

AREAS.

239. Definitions. Equivalent figures are those which have the same surface.

The area of a figure is the measure of its surface.

240. Definition. The unit of surface is the square whose side is a linear unit; so that the area of a figure denotes its ratio to the unit of surface.

241. Theorem. Two rectangles, as ABCD, AEFG (fig. 127) are to each other as the products of their bases by their altitudes, that is,

Proof. a. Suppose the ratio of the bases AB to AE to be, for example, as 4 to 7, and that of the altitudes AC: AF to be, for example, as 5 to 3.

AE may be divided into 7 equal parts Aa, ab, bc, &c., of which AB contains 4; and, if perpendiculars aa', bb', &c. to AE are drawn through a, b, c, &c., the rectangle ABCD is divided into 4 equal rectangles Aaa' C, abb'a', &c., and the rectangle AEFG is divided into 7 equal rectangles Aaa"F, abb"a", &c.

Again, AC may be divided into 5 equal parts Am, mn, &c., of which AF contains 3; and, if perpendiculars mm',