244

From the similar triangles SAO, SDC, we have

AO DC SA : SD;

and by the similar triangles SAF, SDH,

AF

AF

hence
XIII, Cor. 1,) as

construction, AF

:

DH

SA : SD;

DH AO: DC, or (B. V, Prop. circ. AO is to circ. DC. But, by circ. AO; hence DH-circ. DC. Hence the triangle SAF, measured by AF× SA, is equal to the surface of the cone SAB which is measured by circ. AOXSA. For a like reason, the triangle SDH is equal to the surface of the cone SDE. Therefore the surface of the truncated cone ADEB is equal to that of the trapezium ADHF; but the latter (B. II, Prop. IV,) is measured by AD x

(AF+DH) Hence the

surface of the truncated cone ADEB is equal to its side AD multiplied by half the sum of the circumferences of its two bases.

Scholium. If a line AD, lying wholly on one side of the line OC, and in the same plane, make a revolution around OC, the surface described by AD will have for its circ. AO+ circ. DC. or AD x circ.

measure

AD × (circ.

2

DC,)

IK; the lines AO, DC, IK being perpendiculars, drawn from the extremities and from the middle of the axis OC. For, if AD and OC are produced till they meet in S, the surface described by AD is evidently that of a truncated cone having AO and DC for the radii of its bases, the vertex of the whole cone being S. Hence this sur

face will be measured as we have said.

This measure will always hold good, even when the

point D falls on S, and thus forms a whole cone; and also when the line AD is parallel to the axis, and thus forms a cylinder. In the first case, DC would be nothing; in the second, DC would be equal to AO and to IK.

But if the line AD is not in the same plane with OC, the surface described will be a hyperboloid of one sheet, whose properties cannot be fully explained without a knowledge of conic sections.

Cor. 1. Through I the middle point of AD, draw IKL parallel to AB, and IM parallel to AF: it may be shown, as above, that IM=circle IK; but the

trapezium ADHF ADX IM AD x circ. IK.

=

=

Hence it may also be asserted, that the surface of a truncated cone is equal to its side multiplied by the circumference of a section at equal distances from the two bases.

Cor. 2. The point I being the middle of AB, and IK a perpendicular drawn from the point I to the axis, the surface described by AB, F by the last proposition, will

M K. N

C

ОР

have for its measure AB × circ. IK. Draw AX parallel to the axis; the triangles ABX, OIK will have their sides perpendicular, each to each, namely, OI to AB, IK to AX, and OK to BX; hence these triangles are similar, and give the proportion AB : AX or MN :: OI : IK, or as circ. OI to circ. IK; hence AB x circ. IK=MN × circ. OI. Whence it is plain that the surface described by the partial polygon ABCD is measured by (MN+NP+PQ,) x circ. OI, or by MQ x circ. OI;

hence it is equal to the altitude multiplied by the circumference of the inscribed circle.

Cor. 3. If the whole polygon has an even number of sides, and if the axis FG passes through two opposite vertices F and G, the whole surface described by the revolution of the half polygon FACG will be equal to its axis FG multiplied by the circumference of the inscribed circle. This axis FG will, at the same time, be the diameter of the circumscribed circle.

PROPOSITION VIII.

THEOREM. Every section of a sphere, made by a plane, is a circle.

Let AMB be the section, made by a plane in the sphere whose centre is C. From the point C, draw CO perpendicularly to the plane AMB; and draw lines CM, CM to different points of the curve AMB, which terminates the section.

B

D

M

M

The oblique lines CM, CM, CB being equal, being radii of the sphere, they are equally distant from the perpendicular CO, (B. VI, Prop. v;) hence all the lines OM, MO, OB are equal; hence the section AMB is a circle, whose centre is O.

Cor. 1. If the section passes through the centre of the sphere, its radius will be the radius of the sphere; hence all great circles are equal.

Cor. 2. Two great circles always bisect each other; for their common intersection, passing through the centre, is a diameter.

Cor. 3. Every great circle divides the sphere and its surface into two equal parts; for, if the two hemispheres were separated, and afterwards placed on the common base, with their convexities turned the same way, the two surfaces would exactly coincide, no point of the one being nearer the centre than any point of the other.

Cor. 4. The centre of a small circle, and that of the sphere, are in the same straight line perpendicular to the plane of the little circle.

Cor. 5. Small circles are the less, the farther they lie from the centre of the sphere; for the greater CO is, the less is the chord AB, the diameter of the small circle AMB.

Cor. 6. An arc of a great circle may always be made to pass through any two given points in the surface of the sphere; for the two given points and the centre of the sphere make three points, which determine the position of a plane. But if the two given points were at the extremities of a diameter, these two points and the centre would then lie in one straight line, and an infinite number of great circles might be made to pass through the two given points.

(109.) Let NESW represent a section of the earth, supposed to be a sphere, made by a plane passing through the axis of rotation, so that the circumference NESW shall correspond with a meridian. Let A and B be two different places on this meridian, whose latitudes are denoted by the arcs EA, EB. From A and B, draw the lines AC, BD perpendicular to the axis NS.

general, degrees of longitude at different places of the earth, are as their distances from the axis of revolution. The length of a degree of longitude at latitude 60°, is just one half the length of a degree of longitude at the equator.

PROPOSITION IX.

THEOREM. The surface of a sphere is equal to its diameter multiplied by the circumference of a great circle.

It is first to be shown, that the diameter of a sphere, multiplied by the circumference of its great circle, cannot measure the surface of a larger sphere. If possible,

P

N

D

MA

BS E

let ABx circ. AC be the surface of the sphere whose radius is CD.

About the circle whose radius is CA, circumscribe a regular polygon having an even number of sides, so as not to meet the circumference whose radius is CD: let M and S be the two opposite vertices of this polygon;