quently angle CED+CHD=ACD+DCB; CED= DCB; consequently CHD=ACD.

179. The radius of a circle may be drawn six times as a chord of the same circle.

Dem. (Fig. 107.) In a circle draw the chord AB equal to a radius of the circle. Draw the radii OA and OB. We shall thus have an equilateral triangle OAB, consequently the angle AOB = 4 R. A. = 60°. The angles about the centre O are together equal to 4 R. A. = 360°; consequently six angles each equal to AOB may be made about this centre. Equal angles at the centre intercept equal arcs; and the chords which subtend those arcs are equal (166 ;) consequently each of the chords of these 6 arcs is equal to AB.

180. Cor. To inscribe a regular hexagon in a circle, we have only to draw the radius six times as a chord. The angles of the hexagon will be equal; since each is measured by the half of an arc equal to & of the circumference. Thus the hexagon having equal sides and equal angles is a regular hexagon.

181. The preceding proposition enables us to solve a problem of great practical utility, viz., to find very nearly how often the radius or diameter of a circle is contained in the circumference of the same circle; that is the ratio which one bears to the other. Since a straight line can never coincide with a curved line, this cannot be found directly; nor with perfect accuracy; but by comparing the circumference of a circle with the perimeter of an inscribed regular polygon, the ratio has been found by approximation, as it is called. It is first compared with an inscribed hexagon. A chord being a straight line is less than the arc which it subtends; therefore the circumference is more than six times as great as a radius of the same circle, or more than 3 times as great as a diameter; that is, the diameter multiplied by 3 would not give us the length of the circumference. The smaller the chord the more nearly will it approach in magnitude to the arc it subtends; the greater the number of the sides of the inscribed polygon the more nearly will its perimeter approach to the circumference of the circumscribing circle; so that if the number of the sides were infinite the difference between the perimeter and circumference would be infinitely small. (Fig. 107.) From the centre O let fall a perpendicular upon the chord EF; thus the chord and the arc EF will both be bisected, at the points I and G respectively (165.) Draw GE and GF; we thus have 2 sides of an inscribed regular dodecagon. By means of the right-angled triangles OIF and GIF, by a process which will hereafter be explained (220,) we can find the magnitude of a side of this dodecagon. The side of the dodecagon is then bisected in like manner, and the magnitude of the side of a regular inscribed polygon of 24 sides is found. We thus approximate more and more nearly to the truth. The process has been continued until the perimeter of the inscribed polygon consisted of many thousand sides. The diameter of the circumscribed circle being taken equal to 1, the calculation of the perimeter of the inscribed polygon has been carried to 140 decimals; but the value 3.1415926 is near enough for all purposes; that is, the circumference of a circle is 3.1415926 times greater than its diameter. For ordinary purposes the expression 3.14

314

100

will be found sufficiently accurate. If the circumference is divided into 314 parts, the diameter will contain 100 of these parts; and by multiplying the dia

314 meter of any circle by we obtain the circumference 100

of that circle. For example, let the diameter of a circle

be 10 feet; the circumference is

10 × 314

100

=31 feet.

If the circumference is known, the diameter may be found by multiplying the circumference by 100 and dividing the product by 314. For example, if the circumference of a circle is 50 feet, the diameter is 50 × 100

314

15 feet.

8. MENSURATION OF CIRCLES.

182. The quadrature or squaring of the circle, that is, the finding the area of a circle, is a problem upon which a great deal of time has been wasted. It never can be solved exactly. We can obtain the approximate area by considering the circle as a regular polygon having an infinite number of sides. The area of such a polygon is found by multiplying its perimeter by half the radius of the inscribed circle. But the perimeter of a polygon having an infinite number of sides may be considered as coinciding with the circumference of the inscribed circle, since the difference will be infinitely small. Hence the radius of the inscribed circle and the radius of that to be measured will be equal; and therefore the area of a circle may be found by multiplying its circumference by half its radius.

For example, if the diameter of a circle is 8 feet, what is the area of the circle? Answer. The circum

ference of the circle will be

314 × 8
100

25 feet. Mul

tiply 25 feet by the radius or + the diameter, and we have 252×2=50% square feet for the area of the circle.

183. The process for finding the area of the sector of a circle, that is, the portion contained between an arc and the radii drawn from its extremities as OADB, (fig. 99,) is easily deduced from the preceding proposition. The arc may be considered as made up of infinitely small straight lines, from the extremities of which radii may be drawn. The sector will thus be divided into triangles, and its area will be equal to the sum of the products of the bases of the triangles by their altitudes. The triangles have a common altitude equal to the radius, and the sum of their bases is equal to the arc of the sector; therefore the area of the sector of a circle is equal to the product of its arc by half its radius.

184. The area of the segment ABD, (fig. 99,) is found by subtracting the area of the triangle AOB from the area of the sector OADB.

185. The area of an annular surface is found by subtracting the area of the smaller circle from that of the greater.

186. It is often required to measure a figure bounded in part by straight and in part by curved lines, as ABCDEH (fig. 108.) Draw the straight line HD. The area of the pentagon ABCDH is equal to the sum of the areas of the triangles into which it may be divided by the diagonals HB and HC. In the curve line DH take any points E, F, G, so that the parts DE, EF, &c., may without any material error be considered as straight lines. Draw HE and HF. Find the areas of the triangles HED, HFE, and HGF, and their sum added to the area of the pentagon will be the area of the entire figure.

V. SOLIDS.

1. OF SOLIDS IN GENERAL.

187. We have supposed a surface to be the space described by the motion of a line; in like manner a solid may be supposed to be produced by the motion of a surface in any direction but that of its length or breadth. The surface itself has extension in two directions, and the motion produces an extension in a third direction, viz., in thickness; thus a solid has three dimensions of extension, viz., length, breadth, and thickness. As lines are terminated by points, and surfaces by lines, so solids are terminated by surfaces. These surfaces may be either plane or curved. For example, a triangle, quadrilateral, or other rectilineal figure, moved in a direction perpendicular to itself, generates the prism, a solid bounded by plane surfaces. A right-angled triangle, revolving on one of those sides which include the right angle, generates a cone, a solid bounded by one curved and one plane surface. A rectangle revolving on one of its sides generates a cylinder, a solid bounded by one curved and two plane surfaces. A circle or a semi-circle revolving on its diameter generates the sphere, a solid bounded by one curved surface. The three solids last named, viz., the cone, the cylinder,