BOOK VI.

REGULAR POLYGONS, AND THE AREA OF THE CIRCLE.

DEFINITIONS.

344. A REGULAR POLYGON is one which is both equilateral and equiangular.

345. Regular polygons may have any number of sides: the equilateral triangle is one of three sides the square is one of four.

PROPOSITION I.-THEOREM.

346. Regular polygons of the same number of sides are similar figures.

For, since the two polygons have the same number of sides, they have the same number of angles; and the sum of all the angles is the same in the one as in the other (Prop. XXIX. Bk. I.). Also, since the polygons are equiangular, each of the angles A, B, C, &c. is equal to each of the angles G, H, I, &c.; hence the two polygons are mutually equiangular.

Again; the polygons being regular, the sides A B, BC, CD, &c. are equal to each other; so likewise are the sides GH, HI, IK, &c. Hence,

AB: GH:: BC: HI:: CD: IK, &c.

Therefore the two polygons have their angles equal, and their homologous sides proportional; hence they are similar (Art. 210).

347. Cor. The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop. XXXI. Bk. IV.).

348. Scholium. The angle of a regular polygon is determined by the number of its sides (Prop. XXIX. Bk. I.).

PROPOSITION II.THEOREM.

349. A circle may be circumscribed about, and another inscribed in, any regular polygon.

Let ABCDEFGH be any regular polygon; then a circle may be circumscribed about, and another inscribed in it.

Describe a circle whose circumference shall pass through the three points A, B, C, the centre being 0; let fall the perpendicular OP from

H

G

F

A

B

C

E

D

O to the middle point of the side BC; and draw the straight lines O A, OB, OC, OD.

Now, if the quadrilateral OP CD be placed upon the quadrilateral OP BA, they will coincide; for the side OP is common, and the angle OPC is equal to the angle OP B, each being a right angle; consequently the side PC will fall upon its equal, PB, and the point C on B. Moreover, from the nature of the polygon, the angle PCD is equal to the angle PBA; therefore CD will take the

G F

H

E

A

D

direction B A, and CD being equal to B A, the point D will fall upon A, and the two quadrilaterals will coincide throughout. Therefore OD is equal to AO, and the circumference which passes through the three points A, B, C, will also pass through the point D. By the same mode of reasoning, it may be shown that the circle which passes through the three vertices B, C, D, will also pass through the vertex E, and so on. Hence, the circumference which passes through the three points A, B, C, passes through the vertices of all the angles of the polygon, and is circumscribed about the polygon (Art. 166).

B C

Again, with respect to this circumference, all the sides, AB, BC, CD, &c., of the polygon are equal chords; consequently they are equally distant from the centre (Prop. VIII. Bk. III.). Hence, if from the point O, as a centre, and with the radius OP, a circle be described, the circumference will touch the side BC, and all the other sides of the polygon, each at its middle point, and the circle will be inscribed in the polygon (Art. 168).

350. Scholium 1. The point O, the common centre of the circumscribed and inscribed circles, may also be regarded as the centre of the polygon. The angle formed at the centre by two radii drawn to the extremities of the same side is called the angle at the centre; and the perpendicular from the centre to a side is called the apothegm of the polygon.

Since all the chords A B, BC, CD, &c. are equal, all the angles at the centre must likewise be equal; therefore the value of each may be found by dividing four right angles by the number of sides of the polygon.

351. Scholium 2. To inscribe a regular polygon of any number of sides in a given circle, it is only necessary to

divide the circumference into as many equal parts as the polygon has sides; for the arcs being equal, the chords AB, BC, CD, &c. are also equal (Prop. III. Bk. III.); hence likewise the triangles A OB,

E

C

BOC, COD, &c. must be equal, since their sides are equal each to each (Prop. XVIII. Bk. I.); therefore all the angles ABC, BCD, CD E, &c. are equal; hence the figure ABCDEF is a regular polygon.

PROPOSITION III.-THEOREM.

352. If from a common centre a circle can be circumscribed about, and another circle inscribed within, a polygon, that polygon is regular.

Suppose that from the point 0, as a centre, circles can be circumscribed about, and inscribed in, the polygon ABCDEF; then that polygon is regular.

F

E

D

A

B

C

For, supposing it to be described, the inner one will touch all the sides of the polygon; therefore these sides are equally distant from its centre; and consequently, being chords of the outer circle, they are equal; therefore they include. equal angles (Prop. XVIII. Cor. 1, Bk. III.). Hence the polygon is at once equilateral and equiangular; consequently it is regular (Art. 344).

PROPOSITION IV.-PROBLEM.

353. To inscribe a square in a given circle.

Draw two diameters, A C, BD, intersecting each other at right angles; join their extremities, A, B, C, D, and the figure A B C D will be a square.

D

For, the angles AOB, BOC, &c. being equal, the chords AB, BC, &c. are also equal (Prop. III. Bk. III.); and the angles A B C, BCD, &c., being inscribed in semicircles, are right angles (Prop. XVIII. Cor. 2, Bk. III.). Hence ABCD is a square, and it is inscribed in the circle ABCD.

A

354. Cor. Since the triangle AO B is right-angled and isosceles, we have (Prop. XI. Cor. 5, Bk. IV.),

AB: A0::2:1;

hence, the side of the inscribed square is to the radius as the square root of 2 is to unity.

355. The side of a regular hexagon inscribed in a circle is equal to the radius of the circle.

Let ABCDEF be a regular

hexagon inscribed in a circle, the

centre of which is O; then any side, as B C, will be equal to the radius O A.

E

0

F

Join BO; and the angle at the centre, AO B, is one sixth of four

A

B

right angles (Prop. II. Sch. 1), or one third of two right angles; therefore the two other angles, OA B, OBA, of the same triangle, are together equal to two thirds of two right angles (Prop. XXVIII. Bk. I.). But A O and BO being equal, the angles OA B, OBA are also equal (Prop. VII. Bk. I.); consequently, each is one third of two right angles. Hence the triangle AOB is equiangular; therefore A B, the side of the regular hexagon, is equal to A0, the radius of the circle (Prop. VIII. Cor. Bk. I.).