Area of a Circle.

OA, OB, OC, &c. to the vertices of the circumscribed polygon ABCD, &c. Draw the radii OM, ON, OP, &c. to the points of contact of the sides.

If, now, the sides AB BC, CD, &c. are taken for the bases of the triangles OAB, OBC, OCD, &c.; their altitudes, being the radii OM, ON, OP, &c., are all equal. The area of each of these triangles is, then, by § 251, half the product of its base AB, BC, CD, &c. by the common altitude OM.

The sum of the areas of the triangles, or the area of the polygon is, consequently, half the product of the sum of the sides, AB, BC, &c. by the common altitude OM; that is, half the product of the perimeter ABCD &c. of the polygon by the radius OM.

278. Corollary. Since a circle can, by § 229, be inscribed in any regular polygon, the area of the regular polygon is half the product of its perimeter by the radius of its inscribed circle.

279. Theorem. The area of a circle is half the product of its circumference by its radius.

Proof. For a circle is, by § 204, a regular polygon, and the radius of its inscribed circle is its own radius.

280. Corollary. If we use C, D, R, and π, as in § 237, and denote by A the area of a circle, we have CX R=1. 2π × RX R

A

282. Definition. A sector is a part of a circle com

An infinitely small Sector is a Triangle.

prehended between an arc and the two radii drawn to its extremities, as AOB (fig. 135).

283. Theorem. The area of a sector is half the product of its arc by its radius.

Proof. Suppose the arc AB (fig. 135) of the sector AOB divided into the infinitely small arcs AM, MN, NP, &c. Draw the radii OM, ON, OP, &c.

The sector AOB is divided into the infinitely small sectors AOM, MON, NOP, &c.; which may, by § 203, be considered as triangles, having for their basés AM, MN, NP, &c., and for their altitudes the radii OA, OM, ON, &c.

The sum of the areas of these triangles, or the area of the sector is, then, half the product of the sum of the bases AM, MN, NP, &c. by the common altitude OA; that is, half the product of the arc AB by the radius AO.

284. Corollary. The area of the segment ADB is found by subtracting the area of the triangle AOB from that of the sector AOB.

285. Scholium. In order that no doubt may exist with regard to the accuracy of the demonstrations of § 283, 279, and 271, it is important to show that the infinitely small quantities, which are neglected in considering an infinitely small sector as a triangle with a base equal to its arc and an altitude equal to its radius, come within the limitation of § 205.

Now, the difference between the infinitely small sector AOB (fig. 113), and the triangle AOB, is the segment ADB. But the segment ADB is less than the rectangle AEE'B; and, by § 242 and 251, the rectangle

AEE'B: the triangle AOBAB × CD: ABX OC

CD: OC =2 CD: OC;

Ratio of Similar Sectors and Segments.

and, therefore, as 2 CD is infinitely small in comparison with OC, the rectangle AEEB and the segment ADB must be infinitely small in comparison with the triangle AOB, and may be neglected by § 204; so that the sector AOB is equivalent to the triangle AOB.

The base of the triangle AOB is the chord AB, or, by § 203, the arc AB; and its altitude OC differs from the radius OD by the infinitely small quantity CD, which may be neglected.

Indeed,

The error arising from the neglect of these infinitely small quantities is altogether insensible, and cannot be rendered sensible by any magnifying process to which the mind can submit it; it is, then, no error at all. if there be an error, suppose it to be represented by A. Since the aggregate of the quanties neglected is infinitely small, that is, as small as we choose; we can choose it to be less than the error A; a manifest absurdity, for the error cannot be greater than the aggregate of the quanties neglected, and yet we cannot escape this absurdity so long as we suppose the error A to be of any magnitude whatever.

286. Definition. Similar sectors and similar segments are such as correspond to similar arcs.

287. Theorem. Similar sectors are to each other as the squares of their radii.

Proof. The similar sectors AOB, A'O'B' (fig. 136) are, by the same reasoning as in § 97, the same parts of their respective circles, which the angle 00' is of four right angles; and, therefore, they are to each other as these circles, or, by § 271, as the squares of the radii AO, A O'.

288. Theorem. Similar segments are to each other as the squares of their radii.

To find a Triangle equivalent to a given Polygon.

Proof. Let ADB, AD B' (fig. 136) be the similar segments. The triangles AOB and A'O'B' are similar, by § 179; for 0=0'; and, since AOBO and AO = B'O', we have

Hence, by § 266,

AO: AO' = BO: B'O'.

=

the triangle AOB: the triangle A O'BA02: A02; also, by the preceding article,

the sector AOB: the sector A' O'B' = A02: A 02, Hence, by the theory of proportions,

the sector AOB—the triangle AOB: the sector A'O'B' the triangle A OB'=AO2: A O2;

that is,

the segment ADB: the segment A' D'B'

289. Problem. To find a triangle equivalent to a given polygon.

Solution. Let ABCD &c. (fig. 137) be the given polygon. Join BD; through C, draw CM parallel to BD. Join DM, and AMDE &c. is a polygon equivalent to the given polygon, and having the number of its sides less by one.

In the same way, a polygon may be found equivalent to AMDE, and having the number of its sides less by one; and by continuing the process, the number of sides may be at last reduced to three, and a triangle is obtained equivalent to the given polygon.

Proof. a. The number of sides of AMDE &c. is less by one than that of ABCD &c.; for the two sides AM, MD are substituted for the three sides AB, BC, CD, the other sides remaining unchanged.

b. The polygon AMDE &c. is equivalent to ABCD

Quadrature of Polygon and Circle.

&c.; for the part ABDE &c. is common to both, and the triangles DBC, DBM are equivalent because they have the same base BD and the same altitude, by § 82, their vertices C and M being in the line CM parallel to this base.

290. Problem. To find a square equivalent to a given parallelogram.

Solution. Let B be the base and A the altitude of the given parallelogram. Find, by § 188, a mean proportional X between A and B, X is the side of the square sought.

Proof. For we have

and, therefore,

A: X-X: B,

X2AX B;

or, by §§ 242 and 243,

the square constructed upon X is equivalent to the given parallelogram.

291. Corollary. A square may be found equivalent to a given triangle, by taking for its side a mean proportional between the base and half the altitude of the triangle.

292. Corollary. A square may be found equivalent to a given circumscribed polygon, by taking for its side a mean proportional between the perimeter of the polygon and half the radius of the inscribed circle.

293. Corollary. A square may be found equivalent to a given circle, by taking for its side a mean proportional between the radius and half the circumference of the circle.

294. Corollary. In general the quadrature of any given polygon may be found, that is, a square may be