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will form the regular circumscribed polygon GHIK, &c., similar to the inscribed one.

It is evident, in the first place, that the three points O, B, H, lie in the same straight line; for the right-angled triangles OTH, OHN, having the common hypothenuse OH, and the side OT=ON, must be equal, and consequently the angle TOH=HON, wherefore the line OH passes through the middle point B of the arc TN. For a like reason, the point I is in the prolongation of OC; and so with the rest. But since GH is parallel to AB, and HI to BC, the angle GHI=ABC; (27.1;) in like manner, HIK=BCD; and so with all the rest: hence the angles of the circumscribed polygon are equal to those of the inscribed one. And farther, by reason of these same parallels, we have GH AB :: OH : OB, and HI BC :: OH: OB; therefore GH: AB :

: HI : BC. But AB=BC, therefore GH-HI. For the same reason, HI IK, &c.; hence the sides of the circumscribed polygon are all equal; hence this polygon is regular, and similar to the inscribed one. In like manner, if the arcs PD, DQ, QE, &c., are severally bisected, the chords ab, bc, cd, &c., will form a regular inscribed polygon abcd, &c,. of twice as many sides as the given polygon. (Prop. 5. 5. Sch.) By completing the construction as above we shall also have a regular circumscribed polygon fghk, &c., similar to the inscribed one.

Cor. 1. Reciprocally, if the circumscribed polygon GHIK, &c., were given, and the inscribed one ABC, &c., were required to be deduced from it, it would only be necessary to draw from the angles G, H, I, &c., of the given polygon, straight lines OG, OH, &c., meeting the circumference in the points A, B, C, &c.; then to join those points by the chords AB, BC, &c.; which would form the inscribed polygon.

But an easier solution of this problem would be simply to

join the points of contact T, N, P, &c., by the chords TN, NP, &c., which likewise would form an inscribed polygon similar to the circumscribed one.

Cor. 2. Hence we may circumscribe about a circle any regular polygon, which can be inscribed within it; and conversely.

Cor. 3. The circumscribed polygon of six sides is greater than the circumscribed one of twelve sides. For if the dodecagon fghk, &c., be taken from the hexagon IKLM, &c., there will remain the triangles fKg, hLk, &c.

In like manner it may be shown that any regular circumscribed polygon of any number of sides, is greater than one of twice that number of sides; and consequently, the greater the number of sides of a regular circumscribed polygon, the nearer its area approaches to the area of the circle.

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The area of a regular polygon GHIKLM, is equal to its perimeter multiplied by half the radius OT of the inscribed circle.

For, the triangle GOH will be measured by GHXOT; (6. 4;) the triangle OHI by HI XON: but ON=OT; hence the two triangles taken together will be measured by (GH+HI xOT. And, by continuing the same operation for the other triangles, it will appear that the

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sum of them all, or the whole polygon, is measured by the sum of the bases GH, HI, IK, &c., or the perimeter of the polygon, multiplied into OT, or half the radius of the inscribed circle. Hence,

The area of any regular polygon is equal to its perimeter multiplied by half the radius of the inscribed circle.

Scholium. The radius OT of the inscribed circle, is nothing else than the perpendicular let fall from the centre to one of the sides: it is sometimes named the apothem of the polygon.

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The perimeters of two regular polygons, having the same number of sides, are to each other as the radii of the circumscribed circles, and also as the radii of the inscribed circles; and their surfaces are to each other as the squares of those radii.

Let AB be a side of one of the

polygons, O the centre, and consequently OA the radius of the circumscribed circle, and OD, perpendicular to AB, the radius of the inscribed circle; let ab, in like manner, be a side of the other poly

A D B

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gon, o its centre, oa and od the radii of the circumscribed and the inscribed circles. The perimeters of the two polygons are to each other as the sides AB and ab; (27. 4;) but the angles A and a are equal, being each half of the angle of the polygon; so also are the angles B and 6; hence the triangles ABO, abo are similar, as are likewise the right-angled triangles ADO, ado; hence AB: ab:: AO : ao :: DO do; therefore the perimeters of the polygons are to each other as the radii AO, ao of the circumscribed circles, and also as the radii DO, do of the inscribed circles.

The surfaces of those polygons are to each other as the squares of the homologous sides AB, ab; (27. 4;) therefore they are likewise to each other as the squares of AO, ao the

radii of the circumscribed circles, or as the squares of OD, od the radii of the inscribed circles. Hence, The perimeters, &c.

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Two regular polygons similar to each other, can be formed, the one inscribed in a circle, and the other circumscribed about it, which shall not differ from each other by any assignable space.

If we inscribe a polygon ABCD, &c., of eight sides in a circle, bisecting AC a fourth part of the circumference, and again bisecting AB the half of this fourth, and continue to bisect the arcs produced by the last bisection, we shall at length have a polygon of an indefinite number

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of sides, and it will also be regular. (Prop. 5. 5. Sch.) And if a similar polygon abcd, &c. be described about the circle, (6. 5. Cor. 2,) these two polygons will not differ from each other by any assignable space.

For, since a regular inscribed polygon of any number of sides is less than one of twice that number of sides, (4. 5. Cor.,) and since a regular circumscribed polygon of any number of sides is greater than one of double that number, (Prop. 6. 5. Cor. 3,) it follows that an inscribed and circumscribed polygon continually approach to an equality with each other as the number of their sides is increased. Hence, if their number of sides be indefinitely increased, the difference of their areas must at length become indefinitely small. But when this difference is reduced to the smallest conceivable space, their number of sides may still be

doubled, quadrupled, or multiplied by any number, and their areas will approach nearer to an equality by each successive multiplication of their sides; consequently their difference will become less than any assignable space. Hence, Two regular polygons, &c.

Cor. 1. Since the inscribed and circumscribed polygons continually approach to an equality as the number of their sides is increased, it follows that their perimeters, by increasing their number of sides indefinitely, will at length differ insensibly from each other, and will ultimately unite on the circumference of the circle, and coincide with it.

Cor. 2. Hence, if the number of sides of the polygons is sufficiently increased, the polygons will ultimately become equal to the circle, and equal to each other.

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The circumferences of circles are to each other as their radii ; and their areas are as the squares of their radii.

Let the circumference whose radius is CA, be designated by circ. CA; and its area, by area CA: then is circ. CA: circ. OB:: CA: OB; also

area CA area OB :: CA2: OB2.

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For, if two regular polygons of the same number of sides are inscribed in the circles, their perimeters will be to each

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