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595. Cor. 2. The volumes of two pyramids are to each other as the products of their bases and altitudes.

596. CoR. 3. Pyramids having equivalent bases are to each other as their altitudes, and those having equal altitudes are to each other as their bases.

597. Cor. 4. Pyramids having equal altitudes and equivalent bases are equivalent.

598. Scholium. The volume of any polyedron may be obtained by dividing it into pyramids and finding their volumes separately.

PROPOSITION XVIII. THEOREM

599. The frustum of a triangular pyramid is equivalent to the sum of three pyramids each having the same altitude as the frustum and whose bases are the upper and lower bases of the frustum, and the mean proportional between these bases.

Hyp. ABC-DEF is a frustum of any triangular pyramid, with bases ABC [B], and DEF [1], and altitude H.

To prove ABC-DEF equivalent to the sum of three pyramids whose common altitude is H, and whose bases are ABC and DEF, and a mean proportional between them.

Proof. Through the vertices C, D, B, and C, D, E, pass planes, dividing the frustum into three pyramids.

The pyramids D-CAB and C-FDE have the common altitude H, and for bases respectively ABC and DEF, the lower and upper bases of the frustum.

It remains to show that C-DEB is equivalent to a pyramid whose altitude is H, and whose base is a mean proportional between ABC and DEF [VB x b].

Pyramids C-ABD and C-BDE have their bases in the plane ABED.

... C-ABD: C-BDE = A ABD: ABDE. (Why ?) A ABD: ABDE AB: DE.

(Why ?) Hence, C-ABD: C-BDE AB: DE. (1) (Why ?)

The pyramid C-DBE may be read D-BCE, and we have the pyramids D-CEF and D-BCE with bases in the plane FEBC and vertices at D.

.:. D-BCE: D-CEF::A BCE:A CEF.

But

A BCE: A CEF::BC: EF.

Hence, D-BCE: D-CEF::BC: EF.

(2) The A ABC and DEF are similar, hence,

AB: DE BC: EF. .:. C-ABD: C-BDE D-BCE: D-CEF. But C-ABD= } H x B, and D-CEF= { H xb. (Why?)

::. C-BED= V H x B x H xb= { H VB xb. Hence the frustum of the pyramid is equivalent to the sum of three pyramids, whose bases are the upper and lower bases of the frustum, and the mean proportional between these bases, and having the altitude, H.

600. Cor. 1. Since the volume of a pyramid is equal to the base multiplied by one-third of its altitude, we have Vol. of frustum of a triangular pyramid = { H(B+b+VBxb).

601. Cor. 2. The volume of the frustum of any pyramid is equal to one-third of its altitude multiplied by the sum of its bases and a mean proportional between them.

Ex. 1109. Find the volume of a pyramid whose base is 60, and whose altitude is 5.

Ex. 1110. Find the altitude of a pyramid whose volume is 200, and whose base is a rectangle having the sides 5 and 12 respectively.

Ex. 1111. The three sides of the base of a pyramid are respectively 10, 17, and 21. Find the volume if the altitude is 5.

Ex. 1112. The three sides of the base of a pyramid are respectively 9, 10, 17. Find the volume if a lateral edge is 20, and its projection upon the base equals 12.

Ex. 1113. A lateral edge of a pyramid equals 10, and its inclination to the base is 30°. Find the area of the base if the volume of the pyramid is

100.

Ex. 1114. The base of a pyramid is a rhombus whose diagonals are respectively 10 and 12. Find the volume if the altitude is 6.

Ex. 1115. The diagonals of a parallelopiped divide the figure into six equivalent pyramids.

Ex. 1116. If any point within a parallelopiped be joined to the 8 vertices, 6 pyramids are formed, of which the sum of any opposite two is equal to the sum of any other opposite two.

Ex. 1117. Each edge of a triangular pyramid is equal to 10. Find the volume.

Ex. 1118. The perimeter of the triangular base of a regular pyramid is 40. Find the volume if the altitude is 12.

Ex. 1119. The base of a pyramid is a parallelogram of base 10 and altitude 8. Find the volume if a lateral edge is equal to 6, and forms with the base an angle of 45°.

Ex. 1120. The base of a pyramid is a rectangle having sides respectively equal to a and b. A lateral edge is equal to c, and is inclined to the base 30°. Find the volume.

Ex. 1121. Find the altitude of a pyramid of base b, equivalent to another pyramid of base a and altitude h.

Ex. 1122. Find the volume of the frustum of a triangular pyramid, if the lower base is 9, the upper 4, and the altitude 5.

Ex. 1123. Find the volume of the frustum of a regular quadrangular prism, if the edge of the lower base is 7, of the upper 6, and if the altitude is 8.

Ex. 1124. The upper base of a frustum of a pyramid is 2, the lower 18. Find the altitude if the volume is 260.

PROPOSITION XIX. THEOREM 602. A truncated triangular prism is equivalent to the sum of three pyramids whose common base is the base of the prism and whose vertices are the three vertices of the inclined section.

[blocks in formation]

Hyp. ABC-HKG is a truncated triangular prism of which ABC is the base.

To prove ABC-HGK - H-ABC + K-ABC + G-ABC.

Proof. Pass planes through H, B, C, and through H, K, C, forming the pyramids H-ABC, H-BCK, and H-CKG. H-ABC is evidently one of the required pyramids.

H-BCK may be read C-HBK.

C-HBK = C-AKB. (Equivalent bases and the same altitude.) C-AKB may

be read K-ABC.

[blocks in formation]

.. ABC-HKG - H-ABC + K-ABC + G-ABC.

Q.E.D.

603. Cor. 1. The volume of a truncated right triangular prism is equal to the product of its base by one-third the sum of its lateral edges.

Hint. — The lateral edges are the altitudes of the three pyramids which form the truncated prism.

604. Cor. 2. The volume of any truncated triangular prism is equal to the product of its right section by one-third the sum of its lateral edges.

Hint. — The right section divides the truncated prism so that two truncated right prisms are formed.

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