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PROPOSITION XVII. THEOREM.

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

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Let S-ABC be a triangular pyramid, and H its altitude. We are to prove S-A B C

А В СХ Н.

On the base ABC construct a prism ABC-SED, having its lateral edges to SB and its altitude equal to that of the pyramid. The prism will be composed of the triangular pyramid SABC and the quadrangular pyramid S-A C D E.

Through SA and SD pass a plane SA 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.

.. S-ACD S-A E D,

equivalent).

§ 133

§ 572

(two triangular pyramids having equivalent bases and equal altitudes are

Now the pyramid S-A ED may be regarded as having ESD for its base and A for its vertex.

.. pyramid S-A ED≈ pyramid S-A BC,
(having equal bases SED and ABC and the same altitude).

§ 572

.. the three pyramids into which the prism A B C-S ED 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

.. SABC

=

3

А ВСХ Н.

Q. E. D.

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

S

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Let S-ABCDE be any pyramid.

We are to prove S-A B C D E

=

ABCDEX 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 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-ABCDE = ABCDEX SO.

Q. E. D.

144

3

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.

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Let V and V denote the volumes of the two tetrahedrons D-A BC, D'-A B' C', having the trihedral A of the one equal to the trihedral A of the other.

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

Consider ABC and A B'C' the bases of the two tetrahe

drons,

and from D and D' draw DO and D' O' to the base ABC.

Now

V
X
V A B C X D' O' A B' C' D'O'

ABCX DO ABC DO

=

=

§ 575 (any two pyramids are to each other as the products of their bases and

altitudes).

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(being homologous sides of the similar ▲ A DO and A D'0').

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$341

$ 278

Q. E. D.

EXERCISES.

1. Given a cubical tank holding one ton of water; find its length in feet, if a cubic foot of water weigh 1000 ounces.

2. At 17 cents a square foot, what is the cost of lining with zinc a rectangular cistern 5 ft. 7 in. long, 3 ft. 11 in. broad, 2 ft. 81 in. deep?

3. Find the side of a cubical block of cast iron weighing a ton, if iron weigh 7.2 as much as water, and a cubic foot of water weigh 1000 ounces.

4. How many cubic yards of gravel will be required for a walk surrounding a rectangular lawn 200 yards long, and 100 yards wide; the walk to be 3 feet wide and the gravel 3 inches deep?

5. The volume of a rectangular solid is the sum of two cubes whose edges are 10 inches and 2 inches respectively, and the area of its base is the difference between 2 squares whose sides are 13 feet and 13 feet respectively; find its altitude in feet.

6. A rectangular cistern whose length is equal to its breadth is 22 decimetres deep, and contains 10 tonneaux of water; find its length.

7. Given a regular prism whose base is a regular hexagon inscribed in a circle 6 metres in diameter, and whose altitude is 8.7 metres; find the number of kilolitres it will contain, if the thickness of the walls be 1 decimetre.

8. A pond whose area is 11 hectares, 21 ares, 22.2 centares, is covered with ice 21 centimetres thick. What is the weight of this body of ice in kilogrammes, the weight of ice being 92% that of water.

9. Given two hollow oblique prisms, whose interior dimensions are as follows: the area of a right section of the first is 18 sq. ft., of the second 2.1 sq. metres; a lateral edge of the first is 9 ft., of the second 2.1 metres; find the volume of each in cubic yards, cubic metres, cubic decimetres, and cubic centimetres; find the capacity of each in gallons and litres, in bushels and hectolitres; and find the weight of water in pounds and in kilogrammes, required to fill each prism.

PROPOSITION XX. THEOREM.

578. The frustum of a triangular 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 two bases of the frustum.

D

B

C

Let B and b denote the lower and upper bases of the frustum ABC-DE F, and H its altitude.

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

Now the pyramid E-A B C has for its altitude H, the altitude of the frustum, and for its base B, the lower base of the frustum.

And the pyramid C-E D F has for its altitude H, the altitude of the frustum, and for its base b, the upper base of the frustum. Hence, it only remains

To prove E-A DC equivalent to a pyramid, having for its altitude H, and for its base √B × b.

E-A BC and E-A D C, regarded as having the common vertex C, and their bases in the same plane BD, have a common altitude.

.. E-A BC: E-ADC::▲ AEB:^ AED. § 575 (pyramids having equal altitudes are to each other as their bases). Now since the AA E B and A E D have a common altitude, (that is, the altitude of the trapezoid A B ED),

we have ▲ AEB:▲ AED::AB: DE,

§ 326

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