7. How many square feet are contained in a circle whose circumference is 18 17 yards? Ans. 236 459 square feet. 8. How much will the paving of a circular piece of ground come to at 2s. 8d. per square foot, if it be 126 feet round? Ans. 168l. 98. 1d. 9. The paving of a semicircular alcove with marble, at 2s. 6d. a-foot, came to 107.; what was the length of the semicircular arc?

Ans. 22 42 feet.

10. What is the area contained between two concentric circles whose radii are 2 feet 2 inches and 4 feet? Ans. 35 5174 square feet.

11. What is the area of the sector of a circle of radius 8 feet, if the arc of the sector contain 159 ? Ans. 88 802 square feet. 18 inches and radius of the Ans. 1 square feet.

12. The length of an arc of a sector is circle 2 feet; find the area of the sector.

13. Find the area of a sector whose arc of 40.2 is 8 feet long.

Ans. 45 6 square feet.

14. If the radius of a circle be 3 feet, what is the area of that segment of it whose height is 13 inches?

Ans. 3·476 square feet.

15. The chord of a segment is 2 feet and height 8 inches; find the area of the segment. Ans. 138 74 square inches.

MENSURATION OF SOLIDS.

THE PRISM AND PYRAMID.

8. To find the content of a cube, rectangular parallelopiped, or any prism. LET a, b, c be respectively the number of linear units contained in the three adjacent edges of a rectangular parallelopiped; then the cube described upon the unit with which these are compared will be the unit of volume or content. Now (Geo. of Solids) this unit of volume and the rectangular parallelopiped are to one another in a ratio compounded of their bases and altitudes. Hence proceeding, as in the analogous case for the rectangle (Art. 1), the content of the parallelopiped is found to be equal to a be, or the area of the base by the perpendicular height. And since (Geo. of Solids) prisms and parallelopipeds which have equal bases and altitudes are equal to one another, the volume of any prism, parallelopiped, or cylinder, is expressed by the area of its base into its perpendicular height; that is, in a prism or cylinder, if a be the area of its base, aud h its perpendicular height,

volume a2 h.

Cor. The content of a cube is found by cubing the length of one of its edges; for in this case a = b = c, and hence a be

=

a3.

9. To find the content of a pyramid or cone.

If a2 be the area of the base of the pyramid or cone, and h its perpendicular height, then

volume

=

a2 h.

This expression is deduced at once from the preceding Article and the Geometry of Solids.

10. To find the content of the frustum of a pyramid or cone.

Let ABG be the frustum of a square pyramid whose vertex is V and altitude V m. Let V m meet the other end of the frus

tum in the point n, and join n F, m B. Then (Geo. of Planes) B m and F n are parallel, as are also B A, E F. Put BA = a, EF = b, V n = c, and the height m n of the frustum h. Then by similar triangles chc: VB: VF::a: b, or a c = b (c + h). From this we get

bh

=

C =

a- b'

ah a-b

and by addition, c + h = a b'

Again, because the volume (V) of the frustum is equal to the difference of the volumes of the pyramids A ABV, EFV, we have, by Art. 9 and the preceding values of c and c + h,

a h

bh

a-b

α- -b
= h (a2 + ab + b2).

It will be seen by this formula that a and b are the areas of the two ends of the frustum, and a 6 the mean proportional between them.

A similar demonstration is applicable whatever be the figures of the ends. Cor. 1. If the base of the frustum of a pyramid be any regular

polygon of n sides, and a denote one side at the greater end and ʼn one side at the less; then the altitude of the frustum being h,

This is deduced in a similar way by means of Art. 5.

Cor. 2. If the frustum be that of a cone, the diameters of the two ends being d and d' then

11. The surfaces of solids bounded by plane figures (and the surfaces of cylinders and cones), are found by the appropriate methods given in the Mensuration of Planes. The following formula will be obvious :— (A). In a parallelopiped, if a, b, c be the edges,

(B). In a prism or cylinder, if p be the perimeter of either end, and h the height,

curve surface =

ph:

(C). In a pyramid or cone, if p be the perimeter of the base, and the slant height,

curve surface = pl: (D). In a frustum of a pyramid or cone, if Р and ters of the two ends, and the slant height,

curve surface } (p + q) l.

=

THE SPHERE.

be the perime

12. Before proceeding to the mensuration of the sphere, it is necessary to establish the two following lemmas :

Lemma 1.-Let A Z. BC, be two lines in the same plane not parallel to each other; Bb, Cc, perpendicular to the former, and MO perpendicular to the latter, M being the middle of BC: then if the plane in which are the lines A Z, BC, be supposed to revolve about AZ as an axis, the line B C will generate a conical surface which has for measure S = 2bc. MO.☎.

For produce CB to meet AZ in L, and join BO, CO; then by similar triangles,

Cc: Bb: LC: LB; hence (Euc. v. 18); Cc+ Bo: Bb:: LC+LB LB:: 2LM: LB, or (Cc+Bb) LB = 2 LM. Bb . . . . (1). Also, LM MO:: Lb: Bb, or LM.Bb-MO.Lb, and Lb: LB :: b c : B C, or Lb. BC= LB.bc. Whence by multiplication

LM. Bb. BC=MO.LB.bc Comparing (1) and (2),

(2).

M

(Cc+ Bb) BC

=

2 MO. bc

.

(3).

Hence (Art. 11),

C

B

S = (Cc+ Bb) BC.

* = 2bc.MO.,

is the measure of the surface generated by BC in a

complete revolution round A Z.

* It will be obvious that the surface generated by BC is that of a conical frustum,

the radii of whose ends are Cc and Bb.

Lemma 2.-The volume (V) generated by the triangle BOC, in revolving round A Z, has for measure the expression

V = f 0 M2. bc. w.

Since the volume V is evidently equal to the difference between the cone generated by the triangle LCC and the cones generated by CO c, BbO, Bb L, we have, by Art. 9,

w

-

V=Lc.Cc2. Oc.Cc. (Ob.Bb2. + Lb.Bb. w)
= } –
LO (Cc Bb)LO (Cc+ Bb) (Cc - Bb).... (4).
But by similar triangles, and Euc. v. 17,

Cc-Bb Bb:: BC: LB, or Cc-Bb: BC:: Bb: LB:: MO: LO;
hence
(Cc - Bb) LO=BC. MO.

This gives by means of (3),

(Cc Bb) (Cc+ Bb) LO = 2 M 02. b c ....(5).

Whence by (4) and (5),

V=MO2. be. w.

13. To find the surfaces of a spherical segment, zone, and sphere.

M

A

H

The segment of the sphere may be conceived to be generated by the revolution of the arc AD about the diameter A Z. Inscribe in this arc and circumscribe about it two regular polygons, ABCD, A'B'C' D', similar to each other. Represent by A and A' the areas of the surfaces of revolution generated by the inscribed and circumscribed polygons in turning round AZ; by r and R, the perpendiculars O G, OH, on the sides A B, A' B', from the centre O; and by h and h', the distances Ad, A' d' (Dd, c D' d', etc., being perpendiculars on A Z). Then the whole surface generated by the inscribed polygon, by Lemma 1, is (the perpendiculars O G, OM, etc., being equal), A=2(Ab+be+cd) GO.w=2h.r.w..(1). Similarly, A' 2 h'. R. A' R h' Consequently,

A

=

=

....

.(2).

Now by increasing the number of sides of the polygons, all the quantities in these equations will vary with the exception of R, the radius of the circle; and these equations will still hold good whatever be the number of sides. Moreover, the number of sides of the polygons may be taken so large that r and R, as well as h and h', may differ by less than any given quantity. Hence the value to which the expression (3) approximates as the number of sides of the polygons is increased, is

If it can now be shown that the surface of the segment is always comprehended between A and A', whatever be the number of sides of the polygons, it will obviously follow that the expression for the area of the spherical segment will be the same as that of the inscribed polygon (1),

when R is written for r; for the resulting expression for A in (1) is what the inscribed and circumscribed polygons approximate to, as we increase the number of sides.

Now the surface of the segment is evidently greater than A, since every convex surface is less than any other surface which wholly envelopes it. It is moreover less than A'; for if we draw to the point D the tangent DK, it is clear that the surface generated by A'B'C' K will be greater than the surface of the segment. But the surface generated by A'B'C' K is less than A'; for these surfaces have a common part, produced by the revolution of A' B' C'K; and the surfaces of the frustums of the cones generated by D K, D' K, are (Art. 11), respectively, (Kk + Dd) w KD and (Kk + D' d' ) ∞ K D', of which D d < D'd', and KD < K D'. Hence the surface of the segment lies between A and A'. Consequently if S be the surface of the segment, h its height, and R the radius of the sphere,

or the area of the spherical segment equals the circumference of a great circle multiplied by the height of the segment (Art. 6).

Cor. 1. The surface of a zone b B C D d is equal to the circumference of a great circle multiplied by its height. For the zone is the difference of the two segments A B b, AD d, and hence it has for measure the circumference of a great circle multiplied by Ad - Ab, or by b d.

Cor. 2. The surface of a sphere is equal to the circumference of a great circle multiplied by its diameter; for it may be regarded as a segment whose height is equal to the diameter. Hence, in a sphere whose radius is r,

14. To find the volumes of a sphere and spherical sector.

The volume of the sector may be conceived to be generated by the revolution of the sector A CDO (fig. to Art. 13) round the radius A O. Then if v denote the volume generated by the polygonal sector A B C DO, we have by lemma 2, Art. 12,

in which is the perpendicular on each of the equal sides A B, B C, etc., from the centre O, and h, as in last Article.

Reasoning with these equations as in the preceding Article, it is easily shown that by increasing the number of sides of the polygons, v and v' may be made to differ by less than any assignable quantity, and that the volume of the sector is always comprehended between v and v'. Consequently if V be the volume of the sector of a sphere whose height is h and radius of sphere R,

V

R2 h w.