=

chord of a great circle of the sphere, intended to be made tangent to it, the centre of it must lie in the perpendicular upon the middle D, of this chord, in the same plane, and therefore in the intersection of these two lines, that is in E; and the distances EA, and EB, will be equal; therefore, from the centre E, drawing the circle ABH, with the radius EA EB, it will be the great circle of a sphere which, by the revolution around one of its diameters, as around the line BI, or the centre E, will have no point common with the given sphere but the point B, that is it will be tangent to the given sphere; for if a plane be passed through B, perpendicular to the BI, it will lay fully outside of both spheres, because it is perpendicular to the end of the radius of both spheres equally.

Corol. 1. The two figures relating to this proposition, shew: that when the point of contact of the given sphere lies within the two tangents from the given point to the sphere, the tangent sphere lies without the given sphere, and when it lies in either of these points, or on the opposite side of them, in relation to the given point A, the sphere to be described, incloses the given sphere.

Corol. 2. If the given sphere is made to revolve around the radius AE, the point B, of the sphere described as above, will describe upon this sphere a small circle, which will be a parallel of the sphere.

CHAPTER V.

Application of Arithmetic to Solid Geometry.

$38. PROBLEM. To determine the solid content of a cube; and also its surface.

Determination. It is known by the elements of Arithmetic, that a ratio can be directly stated only between things of the same nature; therefore also the content of solids can be only compared to any one solid, taken as Unity of the kind. By the demonstration of § 34 the sides of two squares being n, and m, the cubes upon these sides will be to one another as n3 m3 ; either one of these being considered unity, or, what is the same thing, measuring the other, we derive from it the principle that the content of any cube whatever is the cube of the side of the same, or if n = 1, we have the cube upon m = m3.

In (fig. 120,) there will be in m, as many squares of lines = n, in the base of the larger square m, as sides n, in the sides of m, and upon these will be formed by the same n, taken in elevation as many cubes as squares in the base, repeated m, times; thence the bases being as n2: m2; the repetition of this ratio in the supposition, gives the ratio n3 m3.

:

The surface of the cube being the sum of the six surfaces which include the cube, the sum of the content of these, and in general, of all surfaces of solids, are to be calculated by the principles of Plane Geometry; these give in this individual case any of the squares representing the sides = m2; therefore, the whole surface of the cube = 6 m2.

$39. PROBLEM. To determine the solid content of a parallelepiped, the base and altitude of which are given.

Determination. By the propositions upon the equality of parallelepipeds upon equal bases, and between the same pa

rallel planes, we have seen that parallelepipeds, under any angle of the sides upon the plane of the base, are equal when their bases and altitudes are equal; therefore, any parallelepiped can be treated as having a base equal to the given base, and the planes upon these bases at right angle, limited by the parallel planes, determined by their altitude; therefore, the content of the parallelepiped, under whatever angle, will be equal to the product of its base into its altitude, or the perpendicular between the base and its opposite plane. So, that if b2 represent the parallelogramic base, and h = altitude, the content of the parallelepiped will be = b2.h C, being a quantity of three dimensions in the same manner as in the preceding problem for the cube, the m3; or the parallelepiped, of which all the sides are equal.

=

Scholium 1. The triangular prism having been proved to be half the parallelogram of equal base and altitude, when one of the parallelograms is taken for base, the solid content of a triangular prism will be the half of the product of the base into the altitude of the triangle, forming its opposite sides; that is, we would in that case b2 h for b, and h, as bases, and (height or) altitude of the

have =

solid.

2

Scholium 2. As it has been proved by the general equality of similar solids, of equal base and altitude, that prisms of any poligonal base are equal when their bases are equal, and they are between the same parallel planes, that is, of the same altitude, if any prism is on a triangular or poligonal base whatever, and its plane opposite to the base is equal and parallel to it, the solid content of any such prism, will be like that of any parallelogram, equal to the product of its base in its perpendicular altitude, or we shall in this case also have, under the preceding acceptations, b2 h, for the solid content of the prisms.

Corol. 1. All solids being in a certain determined Ratio to the parallelepiped, it is evident: that this Ratio acts in every case like a constant multiplier, either whole or fractional; and that therefore all what relates to the content of solids, is dependent entirely on this constant factor, and the principles of their generation, given in the definitions.

Corol. 2. In all these cases it is evident, that the surfaces of the solids will be equal to the sum of the superficial content of all the parallelograms, or triangles including the solid.

§ 40. PROBLEM. To determine the solid content of a cylinder, and of a cone.

Determination. In Plane Geometry we have determined the superficial content of a circle, and obtained a quantity of surface, in all respects in the same manner as of the parallelogram, or any rectilineal figure whatever; therefore, we obtain also for the calculations of Solid Geometry, a superficial base. of a prism of a circular figure, which, multiplied by its altitude, will give the solid content of the prism of a circular base, and of the altitude given by its perpendicular altitude.

I. If, therefore, we call r = radius of the circular base of a cylinder, and h-altitude of the same, we shall have, (by § 82, of Plane Geometry,) for the circumference of the circular base of the cylinder, = 2r. Where is the number determined in § 82, the superficial content of the circle will be C = 2, as before determined; and, therefore, any cylinder of the altitude h, will have for its solid content:

=

S=h.r2.x.

II. It has been proved, that the cone was the third part of the cylinder upon the same circular base and of the same altitude, therefore we have for the solid content of the cone the third part of the above content of the cylinder, under the same radius =r, and altitude h, or for the cone, h.r2..

=

P

=

3

§ 41. PROBLEM. To find the content of the curved surface of a rectangular cylinder; and of a cone the perpendicular from the vertex of which falls in the centre of the circular base.

Determination 1. The circumference of the circle of the ra=r, forming the base of the cylinder was = 2r; if there

dius

fore a plane is conceived to turn around this circle, terminated by a circle parallel to the base, and perpendicular upon the same, a parallelogram is formed, the base of which is the circumference of this circle and the altitude = h, that of the cylinder, this surface will therefore be determined like a parallelogram, and be the product of the circumference of the base into the altitude of the cylinder: or,

C = 2.h.r..

2. For the cone we have evidently the slent lines of the sides from the vertex to the circular base, all equal, therefore, radii of the same circle, as already evident from the principle of the generation of the cone, by which the hypothenuse is the side which describes this surface; therefore, the circumference of the base forms a part of the circumference of a circle, the radius of which is that slent side, and this circumference measures the arc, or the part of the circumference corresponding to it; this part of the content of the circle is, therefore, to be calculated upon the same principle as a whole circle would be, that is, by the product of the arc into half the corresponding radius, which is here half the slent side; if, therefore, this slent side is m, the side surface of the cone becomes, by the above:

C' = m.r.T.

3. If m, was not given, and only the altitude = h, it would either have to be calculated from this and the radius of the circle, or its value substituted in the place of m, here. For this

we have m2 = h2 + 2; therefore, m = (h2 + r2 )2 which by substitution, would transform the above into:

C'

= (h2 + p2 )3 r.5.

$42. PROBLEM. To determine the solid content of a truncated pyramid, or cone, that is formed by the base and a plane cutting it parallel to that base.

APPLIC.

Let A, be the vertex of a triangular pyramid upon the base BCD, or a, of a cone upon the circular base,