Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

PROPOSITION XV. THEOREM. 569. The lateral area of a regular pyramid is equal to one-half the product of the perimeter of its base by its slant height.

Let V-ABCDE be a regular pyramid, and VH its

slant height. We are to prove the sum of the faces V AB, VBC, etc. = 1 (A B + BC, etc.) X V H. Now AB=BC=C D, etc.,

§ 363 (being sides of a regular polygon). VA= VB= V C, etc.,

$ 450 (oblique lines drawn from any point in a I to a plane at equal distances

from the foot of the I are equal). ..A VAB, V B C, etc. are equal isosceles A, § 108 whose bases are the sides of the regular polygon and whose common altitude is the slant height V H.

Now the area of one of these A, as V A B,= { base A B X altitude V H,

§ 324 .. the sum of the areas of these A, that is, the lateral area of the pyramid, is equal to the sum of their bases (A B + BC + C D, etc.) X V H.

Q. E. D. 570. COROLLARY 1. The lateral area of the frustum of a regular pyramid, being composed of trapezoids which have for their common altitude the slant height of the frustum, is equal to one-half the sum of the perimeters of the bases multiplied by the slant height of the frustum.

571. Cor. 2. The dihedral angles formed by the intersections of the lateral faces of a regular pyramid are all equal. § 492

PROPOSITION XVI. THEOREM. 572. Two triangular pyramids having equivalent bases and equal altitudes are equivalent.

B

Let S-A BC and S'-A' B'C' be two triangular pyramids

having equivalent bases A B C and A' B'C' situated
in the same plane, and a common altitude A X.
We are to prove S-A B C = S-A' B'C'.
Divide the altitude A X into a number of equal parts,

and through the points of division pass planes || to the planes of their bases, intersecting the two pyramids.

In the pyramids S-A BC and S'-A' B'C' inscribe prisms whose upper bases are the sections D E F, G H I, etc., D' E' F', G'H' I', etc.

The corresponding sections are equivalent, $ 568 (if two pyramids have equal altitudes and equivalent bases, sections made by

planes |to their bases and at equal distances from their vertices are equivalent).

is the corresponding prisms are equivalent, $ 544 (prisms having equivalent bases and equal altitudes are equivalent).

Denote the sum of the prisms inscribed in the pyramid S-A B C, and the sum of the corresponding prisms inscribed in the pyramid S-A' B' C" by V and V' respectively. Then

V=V. Now let the number of equal parts into which the altitude A X is divided be indefinitely increased ;

The volumes V and V are always equal, and approach to the pyramids S-A B C and S'-A' B'C' respectively as their limits. Hence S-A B C = S-A' B'C'.

§ 199 Q. E. D.

PROPOSITION XVII. THEOREM. 573. The volume of a triangular pyramid is equal to onethird of the product of its base and altitude.

[merged small][ocr errors][merged small]

Let S-ABC be a triangular pyramid, and Hits altitude.

We are to prove S-A BC= } A B C X H. On the base A B C construct a prism A B C-SE D, having its lateral edges II to S B and its altitude equal to that of the pyramid.

. The prism will be composed of the triangular pyramid S A B C and the quadrangular pyramid S-A C D E.

Through S A and S D pass a plane S A D. This plane divides the quadrangular pyramid into the two triangular pyramids, S-A C D and S-A ED, which have the same altitude and equal bases.

§ 133 .. S-AC DE S-A ED,

$ 572 (two triangular pyramids having equivalent bases and equal altitudes are

equivalent). Now the pyramid S-A E D may be regarded as having ESD for its base and A for its vertex.

.:. pyramid S-A ED pyramid S-ABC, $ 572 . (having equal bases S E D and A B C and the same altitude).

.. the three pyramids into which the prism A B C-S E D is : divided are equivalent.

.. pyramid S-A B C is equivalent to } of the prism.

But the volume of the prism is equal to the product of its base and altitude ;

$ 543 ::. SA BC= } ABC X H.

Q. E. D.

PROPOSITION XVIII. THEOREM. 574. The volume of any pyramid is equal to one-third the product of its base and altitude.

Let S-A B C D E be any pyramid.
We are to prove S-A B C D E = { A B C D E X SO.

Through the edge SD, and the diagonals of the base DA, DB, pass planes.

These divide the pyramid into triangular pyramids, whose bases are the triangles which compose the base of the pyramid,

and whose common altitude is the altitude SO of the pyramid.

The volume of the given pyramid is equal to the sum of the volumes of the triangular pyramids.

But the sum of the volumes of the triangular pyramids is equal to f the sum of their bases multiplied by their common altitude,

$ 573 (the volume of a triangular pyramid is equal to one-third the product of its

base and altitude), that is, the volume of the pyramid S-A B C D E = } A B C D E X SO.

Q. E. D.

575. COROLLARY. Pyramids having equivalent bases are to each other as their altitudes ; pyramids having equal altitudes are to each other as their bases. Any two pyramids are to each other as the products of their bases and altitudes.

576. SCHOLIUM. The volume of any polyhedron may be found by dividing it into pyramids, and computing the volumes of these pyramids separately.

PROPOSITION XIX. THEOREM. 577. Two tetrahedrons having a trihedral angle of the one equal to a trihedral angle of the other are to each other as the products of the three edges of these trihedral angles.

Let V and V denote the volumes of the two tetra

hedrons D-A BC, D'-A B'C', having the trihedral A of the one equal to the trihedral A of the other.

V ABX AC XAD
We are to prove vi = RXA CIXAD

Place the tetrahedrons so that their equal trihedral As shall be in coincidence.

Consider A B C and A B'C' the bases of the two tetrahedrons, and from D and D' draw D O and D' O' I to the base ABC.

V ABC X DO A BC - DO
Now

3575 ViA B'C' X D'O=A B'C' * D'Or (any two pyramids are to each other as the products of their bases and

altitudes).

A B C A B XAC
But
A B C = A B'XAC''

$ 341
DO AD
and

$ 278
D'O=A DI
(being homologous sides of the similar A ADO and A D' O).

. V A B X AC XAD
"T1 = A B' X A C'X A DI

Q. E. D.

« ΠροηγούμενηΣυνέχεια »