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.. E-A BC: E-ADC:: A B : DE.

In like manner E-A D C and E-D FC, regarded as having the common vertex E and their bases in the same plane DC, have a common altitude.

.. E-A DC: E-DFC::AADC:A DFC. § 575 But since the AADC and DFC have a common altitude, (the altitude of the trapezoid A C F D),

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(the section of a pyramid made by a plane || to the base is a polygon similar

to the base);

.. AB: DE:: AC: DF.

.. E-A BC: E-ADC:: E-A DC: E-D FC.

Now

E-ABCH X B,

and

E-D FCC-EDF=HX b.

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§ 573

§ 573

.. E-ADC= √ H × B × } H × b = } H √ B × b.

Q. E. D.

579. COROLLARY 1. Since the volume of the frustum is de noted by V, the lower base by B, the upper base by b, and the altitude by H,

we have V =

=

H × B + ¦ H × b + } H × √ B × b } H × (B + b + √B × b).

580. COR. 2. The frustum of any pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum.

For the frustum of any pyramid is equivalent to the corresponding frustum of a triangular pyramid having the same altitude and an equivalent base (§ 578); and the bases of the frustum of a triangular pyramid being both equivalent to the corresponding bases of the given frustum, a mean proportional between the triangular bases is equivalent to a mean proportional between their equivalents.

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581. 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.

D

F

Let A B C D E F be a truncated triangular prism whose base is A B C, and inclined section DE F.

We are to prove ABC-D E F≈ three pyramids, E-A BC, D-A BC and F-A BC.

Pass the planes AEC and DEC, dividing the truncated prism into the three pyramids E-A BC, E-AC D, and E-C D F. Now the pyramid E-A BC has the base ABC and the vertex E.

E-ACD

B-ACD,

$574

(for they have the same base ACD and the same altitude, since their vertices E and B are in the line E B || to the base AC D).

But pyramid B-A CD, which is equivalent to pyramid E-A CD, may be regarded as having the base ABC and the vertex D. Again,

E-CDFB-AC F,

for their bases CDF and AC F, in the same plane, are equivalent,

§ 325

(for the CDF and A CF have the common base CF and equal altitudes, their vertices lying in the line A D to CF).

Moreover, E-C D F and B-A C F have the same altitude, (since their vertices E and B are in the line EB to the plane of their bases ACDF).

But the pyramid B-A C F may be regarded as having the base A B C and the vertex F.

.. the truncated triangular prism A B C-D E F is equivalent to the three pyramids E-A BC, D-A BC, and F-A BC.

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582. COROLLARY 1. The volume of a truncated right triaugular prism is equal to the product of its base by one-third the sum of its lateral edges. For the lateral edges DA, EB, FC, being perpendicular to the base, are the altitudes of the three pyramids whose sum is equivalent to the truncated prism. And, since the volume of a pyramid is one-third the product of its base by its altitude, the sum of the volumes of these pyramids = ABCX (DA + EB + FC).

583. 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.

For let ABC-A'B'C' be any truncated triangular prism. Then the right section D E F divides it into two truncated right prisms whose volumes are D E F × 3 (A D + B E + C F') and DEFX (A'D+ B'E + CF).

Whence their sum is DEFX (A A' + B B' + C C').

EXERCISES.

1. Given a pyramid whose base is a rectangle 80 feet by 60 feet, and whose lateral edges are each 130 feet; find its volume, and its entire surface.

2. Given the frustum of a pyramid whose bases are heptagons; each side of the lower base being 10 feet, and of the upper base 6 feet, and the slant height 42 feet; find the convex surface in square yards.

3. Given a stick of timber 30 feet long, the greater end being 18 inches square, and the smaller end 15 inches square; find its volume in cubic feet.

4. Given a stone obelisk in the form of a regular quadrangular pyramid, having a side of its base equal to 25 decimetres, and its slant height 12 metres. The stone weighs 2.5 as much as water. What is its weight in kilogrammes ?

5. Given the frustum of a pyramid whose bases are squares; each side of the lower base being 35 decimetres, each side of the upper base 25 decimetres, and the altitude 15 metres; find its volume in steres.

6. Given a right hexagonal pyramid whose base is inscribed in a circle 30 feet in diameter, and whose altitude is 20 feet; find its convex surface, and its volume.

7. Given a right pentagonal pyramid whose base is inscribed in a circle 20 feet in diameter, and whose slant height is 30 feet; find its convex surface, and its volume.

8. Find the difference between the volume of the frustum of a pyramid, and the volume of a prism of the same altitude whose base is a section of the frustum parallel to its bases and equidistant from them.

9. Given a stick of timber 32 feet long, 18 inches wide, 15 inches thick at one end, and 12 inches at the other; find the number of cubic feet, and the number of feet board measure it contains. Find equivalents for the results in the metric system.

ON SIMILAR POLYHEDRONS.

584. DEF. Similar polyhedrons are polyhedrons which have the same form. They have, therefore, the same number of faces, respectively similar and similarly placed, and their corresponding polyhedral angles equal.

585. DEF. Homologous faces, lines, and angles of similar polyhedrons are faces, lines, and angles similarly placed.

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I. The homologous edges of similar polyhedrons are proportional.

Since the faces SA B, SA C, SBC and A B C are similar respectively to S'A' B', S'A' C', S'B' C' and A'B'C', we have

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II. Any two homologous faces of similar polyhedrons are proportional to the squares of any two homologous edges.

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III. The entire surfaces of two similar polyhedrons are proportional to the squares of any two homologous edges.

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