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Proposition VIII. THEOREM. 535. Two rectangular parallelopipeds having equal altitudes are to each other as their bases.

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Let a, b, and c, and a', b', c, be the three dimensions re

spectively of the two rectangular parallelopipeds
P and P.

P a Xb
We are to prove P=a' Xbo

Let Q be a third rectangular parallelopiped whose dimensions are a', b and c.

Now Q has the two dimensions b and c in common with P,
and the two dimensions a' and c in common with P'.

P a
Then
ē=a

$ 534 (two rectangular parallelopipeds which have two dimensions in common are

to each other as their third dimensions); and

$ 534 P =ū Multiply these two equalities together;

Pax 6 then Pa X6

Q. E. D.

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536. SCHOLIUM. This proposition may be stated thus : two rectangular parallelopipeds which have one dimension in common are to each other as the products of the other two dimensions.

PROPOSITION IX. THEOREM.

537. Any two rectangular parallelopipeds are to each other as the products of their three dimensions.

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Let a, b, c, and a,' b', c', be the three dimensions respec

tively of the two rectangular parallelopipeds P and P'. . .

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Let Q be a third rectangular parallelopiped whose dimensions are a, b, and d.

P с
Then

§ 534

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(two rectangular parallelopipeds which have two dimensions in common are

to each other as their third dimensions) ;

and

e a X6

§ 536 P= a XV (two rectangular parallelopipeds which have one dimension in common are to

each other as the products of their other two dimensions).

Multiply these equalities together ;

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then

Q. E. D.

PROPOSITION X. THEOREM. 538. The volume of a rectangular parallelopiped is equal to the product of its three dimensions, the unit of volume being a cube whose edge is the linear unit.

Let a, b, and c be the three dimensions of the rectan

gular parallelopiped P, and let the cube U be the
unit of volume.
We are to prove volume of P= a X6 X c.

Pa X 6 Xc
Ū=ixixi

§ 537
But
is the volume of P;

$ 500

.. the volume of P= a X 6 X c.

Q. E. D. 539. COROLLARY I. Since a cube is a rectangular parallelopiped having its three dimensions equal, the volume of a cube is equal to the third power of its edge.

540. Cor. II. The product a X b represents the base when c is the altitude; hence : The volume of a rectangular parallelopiped is equal to the product of its base by its altitude.

541. SCHOLIUM. When the three dimensions of the rertangular parallelopiped are each exactly divisible by the linear unit, this proposition is rendered evident by dividing the solid into cubes, each equal to the unit of volume. Thus, if the three edges which meet at a common vertex contain the linear unit 3, 4 and 5 times respectively, planes passed through the several points of division of the edges, and perpendicular to them, will divide the solid into cubes, each equal to the unit of volume; and there will evidently be 3 X 4 X 5 of these cubes.

Proposition XI. THEOREM. 542. The volume of any parallelopiped is equal to the product of its base by its altitude.

В Let A B C D-F be a parallelopiped having all its faces

oblique, and HR its altitude. We are to prove A B C D-F= A B C D X H R.

By making the right section HIJ N and completing the parallelopiped HIJ N-G L KM we have a right parallelopiped equivalent to, A B C D-F.

$ 529 (an oblique prism is equivalent to a right prism whose base is a right section

of the oblique prism and whose altitude is equal to a lateral edge of the oblique prism).

Through the edge I L make the right section I L P 0, and complete the right parallelopiped IL -H GQ R, and we have a rectangular parallelopiped equivalent to HIJ N-G L KM, § 529

and hence equivalent to A B C D-F.
Now DILG H = EFGH,

OOPQR=(O ILGH) = 0 JKMN; $ 530 and D ABCD= EFGH.

$ 500 ..O OPQR + O ABCD. Moreover, the three parallelopipeds have the common altitude HR. But OP QR-ILGH=0PQR X HR; $ 540

.. A B C D-F= A B C D X H R.

$ 322

Q. E. D.

PROPOSITION XII. THEOREM. 513. The volume of any prism is equal to the product of its base by its altitude.

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CASE I. When the base is a triangle. Let V denote the volume, B the base, and H the

altitude of the triangular prism A E C-E'. We are to prove V=B X H.

Upon the edges A E, EC, E E', construct parallelopiped A EC D-E'. Then A E C-E' = 1 AEC D-E',

$ 532 (the plane passed through two diagonally opposite edges of a parallelopiped

divides it into two cquivalent triangular prisms),
and
A EC= A EC D.

§ 133
But
A ECD-E' = 2 B X H,

$ 542 (the volume of any parallelopiped is equal to the product of its base by its

altitude).
.:V=į (2 B X H)= B X H.
CASE II. — When the base is a polygon of more than three sides.

Planes passed through the lateral edge A A', and the several diagonals of the base will divide the given prism into triangular prisms,

which have for a common altitude the altitude of the prism.

Hence, the volume of the entire prism is the product of the sum of their bases by the common altitude ; that is the entire base by the altitude of the prism.

Q. E. D. 544. COROLLARY. Prisms having equivalent bases are to each other as their altitudes ; prisms having equal altitudes are to each other as their bases; and any two prismis are to each other as the product of their bases and altitudes. Any two prisms having equivalent bases and equal altitudes are equivalent.

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