shown in Prop. VI. Draw also OK to the point of contact K: it will bisect AB in I, and be perpendicular to it (Book III. Prop. VI. Sch.). Produce AO to E, and draw BE.

Let P represent the circumscribed polygon, and p the inscribed polygon: then, since the triangles aOb, AOB, are like parts of P and p, we shall have

aOb AOB: P p (Book II. Prop. XI.):

But the triangles being similar,

Hence,

aOb AOB Oa2 OA, or OK3.

:

Pp

Oa2: OK.

Again, since the triangles OaK, EAB are similar, having their sides respectively parallel,

Oa2 OK2:: AE2: EB2, hence,

P: P : : AE2: EB2,

or by division,

P: P-p AE AE2-EB2, or AB2.

་

But P is less than the square described on the diameter AE (Prop. VII. Cor.); therefore P-p is less than the square described on AB; that is, less than the given square on Q: hence the difference between the circumscribed and inscribed polygons may always be made less than a given surface.

Cor. 1. A circumscribed regular polygon, having a given number of sides, is greater than the circle, because the circle makes up but a part of the polygon: and for a like reason, the inscribed polygon is less than the circle. But by increasing the number of sides of the circumscribed polygon, the polygon is diminished (Prop. VII. Cor.), and therefore approaches to an equality with the circle; and as the number of sides of the inscribed polygon is increased, the polygon is increased (Prop. V. Sch.), and therefore approaches to an equality with the circle.

Now, if the number of sides of the polygons be indefinitely increased, the length of each side will be indefinitely small, and the polygons will ultimately become equal to each other, and equal also to the circle.

For, if they are not ultimately equal, let D represent their smallest difference.

Now, it has been proved in the proposition, that the difference between the circumscribed and inscribed polygons, can be made less than any assignable quantity: that is, less than D: hence the difference between the polygons is equal to D, and less than D at the same time, which is absurd: therefore, the polygons are ultimately equal. But when they are equal to each other, each must also be equal to the circle, since the circumscribed polygon cannot fall within the circle, nor the inscribed polygon without it.

Cor. 2. Since the circumscribed polygon has the same number of sides as the corresponding inscribed polygon, and since the two polygons are regular, they will be similar (Prop. I.); and therefore when they become equal, they will exactly coincide, and have a common perimeter. But as the sides of the circumscribed polygon cannot fall within the circle, nor the sides of the inscribed polygon without it, it follows that the perimeters of the polygons will unite on the circumference of the circle, and become equal to it.

Cor. 3. When the number of sides of the inscribed polygon is indefinitely increased, and the polygon coincides with the circle, the line OI, drawn from the centre O, perpendicular to the side of the polygon, will become a radius of the circle, and any portion of the polygon, as ABCO, will become the sector OAKBC, and the part of the perimeter AB+BC, will become the arc AKBC.

PROPOSITION IX. THEOREM.

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

Let there be the regular polygon GHIK, and ON, OT, radii of the inscribed circle. The triangle GOH will be measured by GH OT; the triangle OHI, by HI×ON: but ON OT; hence the two triangles taken together will be measured by (GH+HI) ×OT. And, by continuing the same operation for the other triangles, it will appear that the sum of them all, or the whole

I

N

H T

K

polygon, is measured by the sum of the bases GH, HI, &c. or the perimeter of the polygon, multiplied into OT, or 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 on one of the sides it is sometimes named the apothem of the polygon.

PROPOSITION X. THEOREM.

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 areas are to each other as the squares of those radii.

Let AB be the side of the one polygon, O the centre, and consequently a 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 polygon, o its centre, oa and od the radii of the circumscribed and the inscribed circles. The perimeters of

A D B 3 d

the two polygons are to each other as the sides AB and ab (Book IV. Prop. XXVII.): but the angles A and a are equal, being each half of the angle of the polygon; so also are the angles B and b; hence the triangles ABO, abo are similar, as are likewise the right angled triangles ADO, ado; hence AB: ab: AO ao :: DO: do; hence 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 these polygons are to each other as the squares of the homologous sides AB, ab; they are therefore 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.

PROPOSITION XI. THEOREM.

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

Let us designate the circumference of the circle whose radius is CA by circ. CA; and its area, by area CA: it is then to be shown that

circ. CA circ. OB:: CA: OB, and that

area CA area OB :: CA2 : OB2

Inscribe within the circles two regular polygons of the same number of sides. Then, whatever be the number of sides, their perimeters will be to each other as the radii CA and OB (Prop. X.). Now, if the arcs subtending the sides of the polygons be continually bisected, until the number of sides of the polygons shall be indefinitely increased, the perimeters of the polygons will become equal to the circumferences of the circumscribed circles (Prop. VIII. Cor. 2.), and we shall have

circ. CA circ. OB:: CA : OB.

Again, the areas of the inscribed polygons are to each other as CA2 to OB2 (Prop. X.). But when the number of sides of the polygons is indefinitely increased, the areas of the polygons become equal to the areas of the circles, each, to each, (Prop. VIII. Cor. 1.); hence we shall have

area CA area OB:: CA2 : OB2.

Cor. The similar arcs AB, DE are to each other as their radii AC, DO; and the similar sectors ACB, DOE, are to each other as the squares of their radii.

A

B

D

For, since the arcs are similar, the angle C is equal to the angle O (Book IV. Def. 3.) ; but C is to four right angles, as the arc AB is to the whole circumference described with the radius AC (Book III. Prop. XVII.); and O is to the four right angles, as the arc DE is to the circumference described with the radius OD: hence the arcs AB, DE, are to each other as the circumferences of which

they form part: but these circumferences are to each other as their radii AC, DO; hence

arc AB arc DE: AC: DO.

For a like reason, the sectors ACB, DOE are to each other as the whole circles; which again are as the squares of their radii; therefore

sect. ACB sect. DOE :: AC2; DO2.

PROPOSITION XII. THEOREM.

The area of a circle is equal to the product of its circumference by half the radius.

Let ACDE be a circle whose centre is O and radius OA: then will

area OA=40A x circ. OA.

For, inscribe in the circle any E regular polygon, and draw Oř perpendicular to one of its sides. Then the area of the polygon will be equal to OF, multiplied by the perimeter (Prop. IX.).

F

Now, let the number of sides of the polygon be indefinitely increased by continually bisecting the arcs which subtend the sides: the perimeter will then become equal to the circumference of the circle, the perpendicular OF will become equal to OA, and the area of the polygon to the area of the circle (Prop. VIII. Cor. 1. & 3.). But the expression for the area will then become

area OA=40A x circ. OA: consequently, the area of a circle is equal to the product of half the radius into the circumference.

Cor. 1. The area of a sector is equal to the arc of that sector multiplied by half its radius.

For, the sector ACE is to the whole circle as the arc AMB is to the whole circumference ABD (Book III. Prop. XVII. Sch. 2.), or as AMB AC is to ABDAC. But the whole circle is equal to ABD × AC; hence the sector ACB is measured by AMB x AC