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

of their radii. Let C denote the circumfer

squares

ence of the other circle, r its radius O B, A' its area; then

Inscribe within the given circles two regular polygons of the same number of sides; and, whatever be the number of sides, the perimeters of the polygons will be to each other as the radii OA and OB (Prop. IX.). Now, conceive the arcs subtending the sides of the polygon to be continually bisected, forming other inscribed polygons, until polygons are formed of an indefinite number of sides, and therefore having perimeters coinciding with the circumference of the circumscribed circles (Prop. XII. Cor.); and we shall have

C: CR: r.

Again, the areas of the inscribed polygons are to each other as O A2 to 0 B2 (Prop. IX.). But when the numO ber of sides of the polygons is indefinitely increased, the areas of the polygons become equal to the areas of the circles; hence we shall have

A: A':: R2: 2.

377. Cor. 1. The circumferences of circles are to each other as twice their radii, or as their diameters.

For, multiplying the terms of the second ratio in the first proportion by 2, we have

C: C2R: 2 r.

378. Cor. 2. The areas of circles are to each other as the squares of their diameters.

For, multiplying the second ratio of the second proportion by 4, or 2 squared, we have

A: A': 4 R2: 4r2.

PROPOSITION XIV.—THEOREM.

379. Similar arcs are to each other as their radii; and similar sectors are to each other as the squares of their radii.

Let AB, DE be similar arcs; ACB, DOE, similar sectors; and denote the radii CA and OD by R and r; then

will

and

AB:DE::R:r,

ACB: DOE :: R2: 72.

B

D

E

For, since the arcs are similar, the angle C is equal to the angle O (Art. 213). But the angle C is to four right angles as the arc AB is to the whole circumference described with the radius CA (Prop. XVII. Sch. 2, Bk. III.); and the angle O is to four right angles as the arc DE is to the circumference described with the radius OD. Hence, the arcs A B, D E are to each other as the circumferences of which they form a part. But these circumferences are to each other as their radii, C A, OD (Prop. XIII.); therefore

Arc A B: Arc DE:: R: r.

By like reasoning, the sectors AC B, DOE are to each

other as the whole circles of which they are a part; and these are as the squares of their radii (Prop. XIII.); therefore

Sector A CB: Sector DO E : : R2 : r2.

380. The area of a circle is equal to the product of the tircumference by half the radius.

Let C denote the circumference of the circle, whose centre is O, R its radius OA, and A its area; then will A = CXR.

For, inscribe in the circle any regular polygon, and from the centre draw OP perpendicular to one of the

A

P

sides. The area of the polygon, whatever be the number of sides, will be equal to its perimeter multiplied by half of O P (Prop. VIII.). Conceive the arcs subtending the sides of the polygon to be continually bisected, until a polygon is formed having an indefinite number of sides; its perimeter will be equal to the circumference of the circle (Prop. XII. Cor.), and OP be equal to the radius OA; therefore the area of the polygon is equal to that of the circle; hence

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

For, let C denote the circumference of the circle of which the sector DOE is a part, R its radius OD, and A its area; then we shall have (Prop. XVII. Sch. 2, Bk. III.),

Sector DOE: A :: Arc D E: C;

E

hence, since equimultiples of two magnitudes have the same ratio as the magnitudes themselves (Prop. IX. Bk. II.),

Sector DOE: A:: Arc DEXR: CXR.

But A, or the area of the whole circle, is equal to C×R; hence, the area of the sector DOE is equal to the arc DE × 1 R.

π

382. Cor. 2. Let the circumference of the circle whose diameter is unity be denoted by (which is called pi), the radius by R, and the diameter by D; and the circumference of any other circle by C, and its area by A. Then, since circumferences are to each other as their diameters (Prop. XIII. Cor. 1), we shall have,

therefore

CD: 1;

C=DX л= 2 R Xл.

Multiplying both numbers of this equation by R, we have

CX R R2 XT,

=

or A = Β' Χ π;

that is, the area of a circle is equal to the product of the · square of its radius by the constant number л.

383. Cor. 3. The circumference of every circle is equal to the product of its diameter, or twice its radius, by the constant number л.

384. Cor. 4. The constant number л denotes the ratio of the circumference of any circle to its diameter; for

385. Scholium 1. The exact numerical value of the ratio denoted by can be only approximately expressed. The approximate value found by Proposition XII. is 3.1415926; but, for most practical purposes, it is sufficiently accurate to take 3.1416. The symbol is the first letter of the Greek word πeρíμeтρov, perimetron, which signifies circumference.

386. Scholium 2. The QUADRATURE OF THE CIRCLE is the problem which requires the finding of a square equivalent in area to a circle having a given radius. Now, it has just been proved that a circle is equivalent to the rectangle contained by its circumference and half its radius; and this rectangle may be changed into a square, by finding a mean proportional between its length and its breadth (Prob. XXVI. Bk. V.). To square the circle, therefore, is to find the circumference when the radius is given; and for effecting this, it is enough to know the ratio of the circumference to its radius, or its diameter.

But this ratio has never been determined except approximately; but the approximation has been carried so far, that a knowledge of the exact ratio would afford no real advantage whatever beyond that of the approximate ratio. Professor Rutherford extended the approximation to 208 places of decimals, and Dr. Clausen to 250 places. The value of л, as developed to 208 places of decimals, is 3.14159265358979323846264338327950288419716939937 5105820974944592307816406286208998628034825342717 0679821480865132823066470938446095505822317253594 0812848473781392038633830215747399600825931259129

40183280651744.

Such an approximation is evidently equivalent to perfect correctness; the root of an imperfect power is in no case more accurately known.

PROPOSITION XVI. - PROBLEM.

387. To divide a circle into any number of equal parts by means of concentric circles.

Let it be proposed to divide the circle, whose centre is O, into a certain number of equal parts, three for instance, by means of concentric circles.

Draw the radius AO; divide AO into three equal parts, A B, BC, CO. Upon A O describe a semi-circumference,