To inscribe a Circle in a Regular Polygon.

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

Demonstration. The distances OM, ON, OP, &c. are all equal, by art. 227, and therefore the circumference passes through the points M, N, P, &c.; and the sides AB, BC, CD, &c. are all, by art. 120, tangents to the circle; and the circle is, by art. 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).

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

Demonstration. Join OA, OB, OC, &c., OA', OB', OC, &c. The triangles OAB', OB'B are equal, by art. 63, for the hypothenuse OB' is common, and the legs OA, OB are equal, being radii of the same circle. The angle AOB is, therefore, bisected by OB', and, in the same way, it may be shown that the angles BOC, COD, &c. are respectively bisected by OC', OD' &c.

In the right triangles OAA', OAB', OBB', &c. the sides OA, OB, OC, &c. are equal, the right angles OAA', OAB', &c. are equal, and also the angles AOA', AOB', BOB', &c. are equal, because they are the halves of the equal angles AOB, BOC, &c. The hypothenuses OA', OB', OC', &c. are therefore all equal, by art. 53; and the circle drawn with OA' as a radius, passes through the points A', B', C", &c., and is circumscribed about the polygon A'B'C'D', &c.

The Perimeters of similar regular Polygons are as their Radii.

The angles A'OB', B'OC', &c. are equal, being the doubles of the equal angles A'OA, AOB', &c.; the arcs A'B', B'C', &c. are equal, by art. 95; and the chords A'B', B'C' are equal, by art. 111. The polygon A'B'C'D', &c. is, therefore, equilateral, and, being inscribed in a circle, it is regular, by art. 202.

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 arts. 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, and 228.

料

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.

Demonstration. 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 art. 173, for the angle 0=0', being, by art. 224, the same part of four right angles, and the angle OBA= O'B'A' for each is half the angle ABC= A'B'C'. Hence, by art. 198,

=

233. Corollary. Hence, the perimeters of regular polygons of the same number of sides are, by art. 207,

The Ratio of a Circumference to its Diameter.

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.

Demonstration. For circles are similar regular polygons, by art. 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 diam

eters.

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 π, we have

238. Corollary.

Area of Rectangle.

π is a 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,

ABCD: AEFG = AB × AC: AE × AF.

Demonstration. 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',

Area of the Rectangle.

&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', nn', &c. to AC are drawn through m, n, &c., each of the partial rectangles of ABCD is divided into 5 equal rectangles; and each of the partial rectangles of AEFG is divided into 3 equal rectangles; and all these small rectangles are, evidently, equal.

Hence ABCD contains 4 X 5 of them, and AEFG contains 3X7 of them; that is,

ABCD AEFG 4 X 5:7 X 3,

=

which is equal to the product of the ratio 4: 7 by 5:3, or of AB: AE by AC: AF, so that

ABCD: AEFG = AB × AC: AE × AF.

242. Corollary. The rectangle ABCD is, consequently, by art 240, to the unit of surface, as AB × AC to unity, or as the product of its base multiplied by its altitude to unity.

Hence the area of a rectangle ABCD is the product of its base by its altitude.

243. Corollary. The area of a square is the square of one of its sides.

244. Corollary. Rectangles of the same altitude. are to each other as their bases, and rectangles of the same base are to each other as their altitudes.

245. Theorem. Any two parallelograms ABCD,