он

H

T

G

A

and OHI equal, from what has just been shown; the angles GOH and HOI equal, because they are measured by the equal arcs AB and BC, and the side OH common; they are, therefore, equal in all their parts: hence, GH is equal to HI. In like manner, it may be shown

that HII is equal to IK, IK to KL, and so on:

hence, the circumscribed polygon is equilateral.

The circumscribed poly

N

I

P

M

F

R

E

K

gon being both equiangular and equilateral, is regular; and since it has the same number of sides as the inscribed polygon, it is similar to it.

Cor. 1. If lines be drawn from the centre of a regular circumscribed polygon to its vertices, and the consecutive points in which they intersect the circumference be joined by chords, the resulting figure will be a regular inscribed polygon similar to the given polygon.

Cor. 2. The sum of the lines HT and HN is equal to the sum of HT and TG, or to HG; that is, to o~of the sides of the circumscribed polygon.

Cor. 3. If at the vertices A, B, C, &c., of the inscribed polygon, tangents be drawn to the circle and prolonged till they meet the sides of the circumscribed polygon, the resulting figure will be a circumscribed polygon of double the number of sides.

Cor. 4. The area of any regular circumscribed polygon

is greater than that of a regular circumscribed polygon of double the number of sides, because the whole is greater than any of its parts.

Scholium. By means of a circumscribed and inscribed square, we may construct, in succession, regular circumscribed and inscribed polygons of 8, 16, 32, &c., sides. By means of the regular hexagon, we may, in like manner, construct regular polygons of 12, 24, 48, &c., sides. By means of the decagon, we may construct regular polygons of 20, 40, 80, &c., sides.

PROPOSITION VIII. THEOREM.

The area of a regular polygon is equal to half the product of its perimeter and apothem.

Let GHIK be a regular polygon, its centre, and OT its apothem, or the radius of the inscribed circle: then will the area of the polygon be equal to half the product of the perimeter and the apothem.

For, draw lines from the centre to the vertices of the polygon. These lines will divide the polygon into triangles whose bases will be the sides of the the polygon, and whose altitudes will be equal to the apothem. Now, the area of any triangle, as OHG, is equal to half the product of the side HG

and the apothem: hence, the area

N

H T G

K

of the polygon is equal to half the product of the perimeter and the apothem; which was to be proved.

PROPOSITION IX. THEOREM.

The perimeters of similar regular polygons are to each other as the radii of their circumscribed or inscribed circles; and their areas are to each other as the squares of those radii.

1o. Let ABC and KLM be similar regular polygons. Let OA and QK be the radii of their circumscribed, OD and QR be the radii of their inscribed circles: then will the perimeters of the polygons be to each other as OA is to QK, or as OD is to QR.

is to the perimeter of KLM, as OA is to QK, or as OD is to QR (B. IV., P. XXVII., C. 2); which was to be proved.

2o. The areas of the polygons will be to each other as OA2 is to QK2, or as OD2 is to QR.

2

2

For, OA being homologous with QK, and OD with QR, we have, the area of ABC is to the area of KLM, as ŌÃ2 is to QK2, or as OD2 is to QR (B. IV., P. XXVII., C. 2); which was to be proved.

2

PROPOSITION X. THEOREM.

Two regular polygons of the same number of sides can be constructed, the one circumscribed about a circle and the other inscribed in it, which shall differ from each other by less than any given surface.

Let ABCE be a circle, O its centre, and the side of a square which is less than the given surface; then can two similar regular polygons be constructed, the one circumscribed about, and the other inscribed within the given circle, which shall differ from each other by less than the square of Q, and consequently, by less than the given surface. Inscribe a square in the

given circle (P. III.), and by means of it, inscribe, in succes

sion, regular polygons of 8, 16, 32, &c., sides (P. VII., S.), until one is found whose side is less than Q; let AB be the side of such a polygon.

Construct a similar circum

scribed polygon abcde: then

B

D

K

A

a

E

Q

will these polygons differ from each other by less than trọ square of Q.

For, from a and b, draw the lines a0 and 60; they will pass through the points A and B. Draw also OK to the point of contact K; it will bisect AB at I and Prolong 40 to E

be perpendicular to it.

Let P denote the

circumscribed, and p the inscribed

polygon; then, because they are regular and similar, we

shall have (P. IX.),

2

P: p :: Õa2 : ÕÃ3 :: OK2 or OA2 : 012;

hence, by division (B. II., P. VI.), we have,

But P is less than the square of AE (P. VII., C. 4); hence, P p is less than the square of AB, and conse quently, less than the square of Q, or than the given surface; which was to be proved.

Cor. 1. If the number of sides of the polygons be made greater than any assignable number; that is, infinite, the difference between their areas will be less than any assignable surface ; that is, it will be zero*.

Cor. 2. When the number of sides of the polygons is infinite, either polygon differs from the circle by less than any assignable quantity; for, the circumference of the circle lies between the perimeters of the polygons: hence, the circle differs from either polygon by less than they differ from each other.

* Univ. Algebra, Arts. 72, 73. Bourdon, Art. 71.