it. The contrary would be the consequence of our hypothesis; accordingly this hypothesis cannot be maintained; therefore the circumference of the base of a cone multiplied by the half of its side cannot be the measure of the surface of a greater cone.

2. We say also, that this same product cannot be the measure of the surface of a less cone. For let BO be the radius of the base of the given cone, and, if it be possible, let circ. BO × 1 SB be the surface of a cone of which S is the vertex, and AO, less than OB, the radius of the base,

The same construction being supposed as above, the surface of the pyramid SMNPT will always be equal to the perimeter MNPT multiplied by SA, Now the perimeter MNPT is less than circ. BO, and SA is less than SB; therefore for this double reason the convex surface of the pyramid is less than

circ. BOX SB.

which, by hypothesis, is the surface of a cone of which AO is the radius of the base; consequently the surface of the pyramid would be less than that of the inscribed cone. But on the contrary it is greater; for by applying the base of the pyramid to that of an equal pyramid, and the base of the cone to that of an equal cone, the surface of the two pyramids will enclose that of the two cones, and consequently will be greater. Therefore it is impossible that the circumference of the base of a given cone multiplied by the half of its side should be the measure of the surface of a less cone.

We conclude then, that the convex surface of a cone is equal to the circumference of the base multiplied by half of its side. 529. Scholium. Let L be the side of a cone, and R the radius of the base, the circumference of this base will be 2 R, and the surface of the cone will have for its measure 2 R× L, oг л RL.

π

THEOREM.

530. The convex surface of the frustum of a cone ADEB (fig. 261) is equal to its side AD multiplied by the half sum of the Fig. 261. circumferences of the two bases AB, DE.

Demonstration. In the plane SAB, which passes through the axis SO, draw perpendicularly to SA the line AF, equal to the circumference which has for its radius AO; join SF, and draw DH parallel to AF.

Geom.

24

On account of the similar triangles SAO, SDC,

AO: DC:: SA: SD;

and, on account of the similar triangles SAF, SDH, AF: DH:: SA: SD;

whence AF: DH:: AO: DC :: circ. AO: circ. DC (287). But, by construction, AF circ. AO; consequently

DH circ. DC.

This being premised, the triangle SAF, which has for its measure AF × ¦ SA, is equal to the surface of a cone SAB, which has for its measure circ. AO × SA. For a similar reason the triangle SDH is equal to the surface of the cone SDE. Whence the surface of the frustum ADEB is equal to that of the trapezoid ADHF. This has for its measure AD X

(AF+DH) (178).

2

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

531. Corollary. Through the point I, the middle of AD, draw IKL parallel to AB, and IM parallel to AF; it may be shown as above that IM circ. IK. But the trapezoid (179)

=

ADHF = AD × IM = AD × circ. IK.

Hence we conclude further that the surface of the frustum of a cone is equal to its side multiplied by the circumference of a section made at equal distances from the two bases.

532. Scholium. If a line AD, situated entirely on the same side of the line OC and in the same plane, make a revolution about OC, the surface described by AD will have for its measure circ. AO+ circ. DC or AD x circ. IK; the lines AO, 2

ᎯᎠ x

(circ.

DC),

DC, IK, being perpendiculars let fall from the extremities and from the middle of the line AD upon the axis OC.

For, if we produce AD and OC till they meet in S, it is evident that the surface described by AD is that of the frustum of a cone, of which OA and DC are the radii of the bases, the entire cone having for its vertex the point S. Therefore this surface will have the measure stated.

This measure would always be correct, although the point D should fall upon S, which would give an entire cone, and also when the line AD is parallel to the axis, which would give a cylinder. In the first case DC would be nothing, in the second DC would be equal to AO and to IK.

LEMMA.

533. Let AB, BC, CD (fig. 262), be several successive sides of a Fig. 262. regular polygon, O its centre, and OI the radius of the inscribed circle; if we suppose the portion of the polygon ABCD, situated entirely on the same side of the diameter FG, to make a revolution about this diameter, the surface described by ABCD will have for its measure MQ × circ. OI, MQ being the altitude of this surface, or the part of the axis comprehended between the extreme perpendiculars AM, DQ.

Demonstration. The point I being the middle of AB, and IK being a perpendicular to the axis let fall from the point I, the surface described by AB will have for its measure AB × circ. IK (532). Draw AX parallel to the axis, the triangles ABX, OIK, will have their sides perpendicular each to each, namely, Ol to AB, IK to AX, and OK to BX; consequently these triangles will be similar, and will give the proportion

AB: AX or MN:: OI: IK:: circ. OI: circ. IK, therefore AB × circ. IK = MN × circ. OI. Whence it will be perceived that the surface described by AB is equal to its altitude MN multiplied by the circumference of the inscribed circle. Likewise the surface described by BCNP x circ. OI, the surface described by CDPQ x circ. OI. Accordingly the surface described by the portion of the polygon ABCD has for its measure (MN + NP + PQ) × circ. OI, or MQ × circ. OI; therefore this surface is equal to its altitude multiplied by the circumference of the inscribed circle.

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

THEOREM.

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

Demonstration. 1. We say that the diameter of a sphere multiplied by the circumference of a great circle cannot be the

measure of the surface of a greater sphere.

For, if it be possiFig. 263. ble, let AB x circ. AC (fig. 263) be the surface of a sphere whose radius is CD.

About the circle, whose radius is CA, circumscribe a regular polygon of an even number of sides, which shall not meet the circumference of the circle whose radius is CD; let M and 8 be two opposite vertices of this polygon; and about the diameter MS let the semipolygon MPS be made to revolve. The surface described by this polygon will have for its measure

MSX circ. AC (534);

but MS is greater than AB; therefore the surface described by the polygon is greater than AB × circ. AC, and consequently greater than the surface of the sphere whose radius is CD. On the contrary the surface of the sphere is greater than the surface described by the polygon, since the first encloses the second on all sides. Therefore the diameter of a sphere multiplied by the circumference of a great circle cannot be the measure of the surface of a greater sphere.

2. We say also, that this same product cannot be the measure of the surface of a less sphere. For, if it be possible, let

DE x circ. CD

be the surface of a sphere whose radius is CA. The same construction being supposed as in the first case, the surface of the solid generated by the polygon will always be equal to

MS x circ. AC.

But MS is less than DE, and circ. AC less than circ. CD; therefore for these two reasons the surface of the solid generated by the polygon would be less than DE × circ. CD, and consequently less than the surface of the sphere whose radius is AC. But on the contrary the surface described by the polygon is greater than the surface of the sphere whose radius is AC, since the first surface encloses the second; therefore the diameter of a sphere multiplied by the circumference of a great circle cannot be the measure of the surface of a less sphere.

We conclude then, that the surface of a sphere is equal to the diameter multiplied by the circumference of a great circle.

536. Corollary. The surface of a great circle is measured by multiplying its circumference by half of the radius or a fourth of the diameter; therefore the surface of a sphere is four times that of a great circle.

537. Scholium. The surface of a sphere being thus measured and compared with plane surfaces, it will be easy to obtain the absolute value of lunary surfaces and spherical triangles, the ratio of which to the entire surface of the sphere has already been determined.

In the first place the lunary surface, whose angle is A (fig. 276), Fig. 276. is to the surface of the sphere, as the angle A is to four right angles (493), or as the arc of a great circle, which measures the angle A, is to the circumference of this same great circle. But the surface of the sphere is equal to this circumference multiplied by the diameter; therefore the lunary surface is equal to the arc, which measures the angle of this surface, multiplied by the diameter.

In the second place, every spherical triangle is equivalent to a lunary surface whose angle is equal to half of the excess of the sum of its three angles over two right angles (503). Let P, Q, R, be the arcs of a great circle which measure the three angles of a spherical triangle; let C be the circumference of a great circle and D its diameter; the spherical triangle will be equivalent to the lunary surface whose angle has for its measure P+Q+R―C, and consequently its surface will be

2

Thus, in the case of the triangle of three right angles, each of the arcs P, Q, R, is equal to C, and their sum is 3 C, the excess of this sum over C is C, and the half of this excess is C; therefore the surface of a triangle of three right angles= } C× D which is the eighth part of the whole surface of the sphere.

The measure of spherical polygons follows immediately from that of triangles, and it is moreover entirely determined by the proposition of art. 505, since the unit of measure, which is the triangle of three right angles, has just been estimated on a plane surface.

THEOREM.

538. The surface of any spherical zone is equal to the altitude of this zone multiplied by the circumference of a great circle.

Demonstration. Let EF (fig. 269) be any arc, either less or Fig. 269. greater than a quadrant, and let FG be drawn perpendicular to