Cor. The perimeters of two regular polygons having 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 (B. IV, Prop. xvi).

(84.) It has been shown (B. I, Prop. xxiv), that the sum of all the interior angles of a polygon is found by multiplying two right angles, or 180°, by a number which is two less than the number of sides of the polygon; or, if n denote the number of sides, then the sum of all the angles will be (n-2)180°. But, since all the angles of a regular polygon are equal, the magnitude of one of these angles may be found by dividing the sum of all by the number of angles, or, which is the same, by the number of sides in the polygon; therefore each 360° angle is equal to n

n 2

n

.180°, or 180°.

From this formula, the angles of regular polygons, from the equilateral triangle upwards, have been calculated as in the following table:

No. of sides. 3 4 5

7

8 9 10 11

12

Mag. of angle. 60° 90° 1080 1200 12840 135° 1400 1440 147,30 150°

(85.) The expression for the angle of a regular polygon, which is 360° 180° n

shows that no polygon can have angles consisting of a whole number of degrees, unless the number of sides is an exact divisor of 360. Now the divisors of 360 are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360 (see Higher Arithmetic). Excluding 1 and 2, which evidently cannot give polygons, since a polygon cannot have less than 3 sides, we have twentytwo divisors left; therefore there are only twenty-two regular polygons, whose angles are expressed by a whole number of degrees. Had the French method of dividing the circumference into 400 degrees been adopted, the number of regular polygons, having an integral number of degrees in each angle, would have been only thirteen.

(86.) In ornamental architecture, polygons are used in the formation of surfaces produced by the juxtaposition of solid blocks, as in flooring, paving, or by their superposition as in masonry. The polygons used in such cases must always be such as will admit of being put together without leaving open spaces between them. If they be laid together, as is sometimes the case, leaving the vertices of their angles coincident, then no regular figures can be used, except those whose angles are of such a magnitude as will exactly fill the space surrounding a point. It is evident that the equilateral triangle and square will fulfil this condition, since six angles of an equilateral triangle, and four of a square, make up exactly 360°; thus the point O, in the first figure, is surrounded by six equilateral triangles, and in the second figure it is surrounded by four squares.

⊗田

In general, the condition necessary to be fulfilled is that.....

180°

360°

n

should be a divisor of 360° without a remainder; or, dividing both terms of this expression by 180°, the condition will be

2

that 1 shall be a divisor of 2. The only whole numbers for n

n

which will fulfil this condition, are 3, 4 and 6; hence it follows that a surface cannot be completely covered by any regular figures except the equilateral triangle, the square, and the hexagon.

The angles of the hexagon being 120°, three of them will fill the space round a point, as here represented.

In the formation of pavement, it is an object to avoid the combination of a great number of angles at the same point; the strength of the surface being weakened thereby, and the liability to fracture increased. The combination of equilateral triangles is objectionable on these grounds; and even the combination of squares is usually avoided, by causing the angles at which each pair of adjacent sides are united to coincide with the middle of the sides of a succeeding series, as here represented.

Where the angles of the component figures are intended to be invariably combined, the hexagonal arrangement will therefore have greater strength and stability for pavement than the others; but for upright masonry, the square or rectangular division is preferable, since

the surfaces of contact take the position best adapted to sustain the incumbent weight of the structure.

PROPOSITION II.

THEOREM. If a line be drawn bisecting an angle formed by two given lines, every point in this bisecting line will be equally distant from the two lines forming the angle.

Let AB, AC form the angle BAC, which is bisected by the line AD; then will every point in AD be equally distant from AB, AC.

For, in the line AD take any point as F, and draw FG, FH perpendicular respectively to

AB, AC. Then, comparing the two triangles AFG, AFH, we see that the angles AGF and AHF are equal, each being a right angle; also the angles FAG, FAH are equal, since AF bisects the angle GAH: hence the angle AFG is equal to AFH (B. I, Prop. xxiv, Cor. 1). Therefore in these two triangles we have the side AF common, and the two adjacent angles equal; consequently they are identical (B. I, Prop. Iv), and FG is equal to FH.

Cor. The centres of, all the different circles which can be described, touching the two lines AB, AC, must be situated in the line AD, which bisects the angle BAC.

PROPOSITION III.

PROBLEM. To inscribe a circle in a given triangle.

bisecting lines intersect, as a centre, a circumference be described touching AB, it will also touch AC and BC (B. V, Prop. 11, Cor.), and consequently be the inscribed circle required.

Cor. The three lines bisecting the three angles of a triangle meet at the same point, which point is the centre of the inscribed circle.