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(6.) Every Prism on a triangular Base can be divided into three equal Pyramids. Let ABE, DFC be the prism; draw a plane through DE and E, cutting off the pyramid E, DFC, and through ED and B draw a plane dividing the remainder into two pyramids E, ABD and E, BCD. Now since these two latter have their vertexes coincident at E, and since their bases ABD, BDE are manifestly equal being halves of the parallelogram AC, the two pyramids E, ABD and E, BDE are equal. Now the pyramid E, ABD is clearly the same as the pyramid D, ABE. But the base ABE = base DEF, and the perpendicular from E on DFE, is equal to that from D on ABE, since these planes are parallel; hence the pyramid D, ABE = pyramid E, DFE, and therefore the three pyramids are equal.

COR. 1.-Hence if a pyramid and a prism have equal triangular bases and are of equal altitudes, the pyramid is one third part of the prism.

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Fig. 38.

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COR. 2.-Let A area of base, height of pyramid. Then the volume of the

prism on base A and of height h = Ah. And .. the volume of pyramid =

Ah

COR. 3.-If we have a pyramid on a polygonal base of which the area is A, it can be divided into triangles whose areas we will suppose to be A, A2 A,.... so that A=A, + A2 + Ag + ... Now if h be the height of the pyramid, then its volume being clearly the sum of the pyramids whose bases are A, A, Ag.... and height h, will equal

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COR. 4.-This is true, however great the number of sides there are to the polygon, and hence is true in the limit. Now if we suppose the polygon to be regular, its limit is the circumscribing circle, and the limit of the pyramid is the circumscribing

Ah
3

cone. volume of cone = where A is the area of the circular base.

F

D

E

(7.) To determine the Volume of the Frustum of a right Prism on a triangular Base. Let ABC, DEF be the frustum, where ABC is perpendicular to the edges. Let A = area of ABC, and let h, h2 he be the edges AF, BE, CD respectively, and V the required volume. Join FC and suppose plane to pass through FCE, cutting off the pyramid C, FED, and another through ECA cutting off the pyramids E, ABC, E, ACF, these three make up the volume V. Join FB, DB, and DA. Now volume of

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is the triangle FCD, and we have already seen that the perpendiculars from E

and B on the plane ACDF are equal, .. the pyramid E, CDF

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=

the pyramid B, ACD,

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Cor: If the prism be ABC FED, in which neither of the Ex ends is perpendicular to the edges, take abc an area whose plane is perpendicular to the edges, and let A

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h, BD = h2 CF xbD

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volume required

=

=

area of abc.

h3

X 3

X2 CF
Y2 cc = Y3

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(8.) To determine the Volume of a Frustum of a Right Prism on a base which

is a Parallelogram.

3

Let ABCDEFGH be the frustum in question; let h1 h2 h, h, be the edges at A. B. C. D. respectively; draw AC, and if a plane E, pass through AC and E it divides AB.

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If we had divided the figure by a plane passing through BD, we should have had

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Similarly if we had divided at C and D, we should have had respectively

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Now let Sh1 + h2 + h2 + h ̧. Then adding the four values of V we have

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C

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(9). To find the Volume of a Frustum of a right Prism on a regular Pentagonal Base.

1 2 3

Let P, P, P, P, P, be the base, and let h ̧ h2 h ̧ h, h, be the edges corresponding to these angles. Join P, P, and P, P1, and let the areas

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3

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3 4

and the area of P1 P2 P1 be A2. Let V be the required volume, then this volume consists of three portions of prisms of the same kind as in article (7) and hence

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Now divide the base by lines drawn through P2 the areas will be the same as before, and hence

V = 11 (h2 + h2 + h ̧ ) + 12 (h2+h ̧+hs) +
A2

A1 3

hz

3

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Fig. 42.

base by lines drawn
we obtain

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3

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3

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4

h1 + h2 + h ̧ + h + h5 (S + S + S) +

A2

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(S + S ; S) + 11 (S + $ r S)

(2A1 + A2) S

A. S.

•. V = Ak, + k + kg + hạ + k

5

The student can easily prove that a similar formula is true in the case of a prism on a hexagonal base, or indeed on any base

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BCNM perpendicular to the plane of the rectangle, these divide the given figure into
three, viz., AEKDLG, BDN, and BMFCHN.
Let AB = a AD = h AK = x DL =
Then (a) the figure BDN is of the same kind as in article 8.
DB = ah, if therefore V, is its volume,

=x, EK

=

b GL =

b, MF =

c NH: = C1
Now the area

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(b.) Let V2 be the volume of the figure AEKDLG, draw planes through ED,K and GD,K, dividing the figure into three pyramids, DGL,K, AEK,D, and DEG,K.

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Now the volume of GED,K is to that of ADE,K as GED is to ADE, or as AE to GD, or since triangles AEK, GDL are similar, these volumes are as AK to DL, or as EK to GL.

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btc 2 a +

by to

+

2

2

and (x + 0,) ( +6+C), which equals
***** { } (a+*)(+ **)}

' ! (
{ 0 + a +%%

1

+

+

1
b c

b +

2 is clearly four times the area of a section made by a plane parallel to ABFE and DEHG, and half way between them.

Hence the solid content of a prismoid is found by the following rule : “ To the areas of the ends add four times the area of the mean section, multiply this sum by ith of the height of the figure ; this product is the volume required.”

In practical cases it will generally happen that b = c, and b, = cz. This does not affect the enunciation of the rule, but simplifies the formula, which becomes

h V

{ 3 a (5€ + $7) + 27 (26, + b) + < (26 +8,)}.

1

(11.) To find the Solid Content of a Railway Cutting. In the last article the figure is very nearly that of a portion of a railway cutting, in which ABCD is the road, AG and BH, the sloping sides of the embankment, and hence the solid content required can be found by means of the rule given in the last article. It is to be observed that the rule requires E G and FH to be straight lines, cr, as an approximation, to be very nearly straight lines, or EH to be a plane, which is not true if the cutting is a long one. For this case we derive the following rule from the above formula.

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Let ABC represent a section made by a vertical plane of the hill to be cut through, AB the level of the road, and suppose sections of the cutting to be made by planes perpendicular to AB, at equal distances along that line, viz., at M, M, M, ... M2n + 1, let the terminal sections at A and B be a and b, and the sections at P,M, P,M,.. Pi P2 Pz ....P2n P2n + 1, the number of sections being odd, and let the common distance between the sections be h. Now by last article the volume of portion

h AM,P.D is

3

(a + 4 pi + P2)

AMA(a + 4 pi + px) or

2

2

and the volume of the portion P, M,M.P, is . (P. + 4 P3 + Px), and so on,

h and the volume of Pan M2n BC is

3

(P2 + 4 22 +1 + b).

2

n

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