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 DOE:: R2: r2.

PROPOSITION XV. - THEOREM.

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 OP (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

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

Sector DOE: A:: Arc DEXR: C×R.

But A, or the area of the whole circle, is equal to CXR; hence, the area of the sector DOE is equal to the arc DEXR.

π

382. Cor. 2. Let the circumference of the circle whose diameter is unity be denoted by n (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

C: D::: 1;

C=DXл= 2 RX л.

Multiplying both numbers of this equation by R, we have CX R R2X", =

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 7 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 suf ficiently accurate to take = 3.1416. The symbol is the first letter of the Greek word πерíμетρov, 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.

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,

For join A E, AD; then the angle A D O, being in a semicircle, is a right angle (Prop. XVIII. Cor. 2, Bk. III.); hence the triangles DA O, D CO are similar, and consequently are to each other as the squares of their homologous sides; that is,

consequently, since circles are to each other as the squares of their radii (Prop. XIII.), it follows that the circle whose radius is OA, is to that whose radius is OD, as OA to OC; that is to say, the latter is one third of the former.

In the same manner, by means of the right-angled triangles E AO, EBO, it may be proved that the circle whose radius is O E, is two thirds that whose radius is O A. Hence, the smaller circle and the two surrounding annular spaces are all equal.

NOTE. This useful problem was first solved by Dr. Hutton, the justly distinguished English mathematician.

PLANES.

BOOK VII.

- DIEDRAL AND POLYEDRAL ANGLES.

DEFINITIONS.

388. A STRAIGHT line is perpendicular to a plane, when it is perpendicular to every straight line which it meets in that plane. Conversely, the plane, in the

same case, is perpendicular to the line.

A

M

B

The foot of the perpendicular is the point in which it meets the plane.

Thus the straight line A B is perpendicular to the plane MN; the plane MN is perpendicular to the straight line AB; and B is the foot of the perpendicular A B.

389. A line is parallel to a plane when it cannot meet the plane, however far both of them may be produced. Conversely, the plane, in the same case, is parallel to the line.

390. Two planes are parallel to each other, when they cannot meet, however far both of them may be produced.

391. A DIEDRAL ANGLE is an

angle formed by the intersection of two planes, and is measured by the inclination of two straight lines drawn from any point in the line of intersection, perpendicular to that line, one being drawn in each plane.

B

M

N