PROPOSITION X. PROBLEM.

To construct a rectangular parallelopipedon which shall be equal in volume to a right parallelopipedon whose base is any parallelogram.

Let ABCD-M be a right parallelopipedon, having for its base the parallelogram ABCD.

Through the edges AI and BK pass the planes AQ and and BP, respectively perpendicular to the plane AK, the former meeting the face DL in OQ, and the latter meeting that face produced in NP: then will the polyedron AP be a rectangular parallelopipedon equal to the given parallelopipedon. It will be a rectangular parallelopipedon, because all of its

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faces are rectangles, and it will be equal to the given parallelopipedon, because the two may be regarded as having the common base AK (P. VI., C. 1), and an equal altitude AO (P. IX.).

Cor. 1. Since any oblique parallelopipedon may be changed into a right parallelopipedon, having the same base and altitude, (P. IX., Cor.); it follows, that any oblique parallelopipedon may be changed into a rectangular parallelopipedon, having an equal base, an equal altitude, and an equal volume.

Cor. 2. An oblique parallelopipedon is equal in volume to a rectangular parallelopipedon, having an equal base and an equal altitude.

Cor. 3. Any two parallelopipedons are equal in volume when they have equal báses and equal altitudes.

PROPOSITION XI. THEOREM.

Twoo rectangular parallelopipedons having a common lower base, are to each other as their altitudes.

Let the parallelopipedons AG and AL have the com mon lower base ABCD: then will they be to each other as their altitudes AE and AI

1o. Let the altitudes be commensurable, and suppose, for example, that AE is to AI, as 15 is to 8.

Conceive AE to be divided into 15 equal parts, of which AI will contain 8; through the points of division let planes be passed parallel to ABCD.

These planes will

and equal altitudes;

divide the parallelopipedon AG into 15 parallelopipedons, which have equal bases (P. II. C.) hence, they are equal (P. X., Cor. 3). Now, AG contains 15, and AL 8 of these equal parallelopipedons; hence, AG is to AL, as 15 is to 8, or as AE is to AI. In like manner, it be shown that AG is to AL, as AE is to AI, when the altitudes are to each other as any other whole numbers.

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2o. Let the altitudes be incommensurable.

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Now, if AG is not to AL, as AE is to AI, let us suppose that,

AG : AL :: AE AO,

in which AO is greater than AI.

Divide AE into equal parts, such that each shall be less than OI; there will be at least one point of division

But AO is greater than Am; hence, if the proportion is true, AL must be greater than P. On the contrary, it is less; consequently, the fourth term of the proportion cannot be greater than AI. In like manner, it may be shown that the fourth term cannot be less than AI; it is, therefore, equal to AI. In this case, therefore, AG is to AL, as AE is to AI.

Hence, in all cases, the given parallelopipedons are to each other as their altitudes; which was to be proved.

Sch. Any two rectangular parallelopipedons whose bases are equal in all their parts, are to each other as their altitudes.

PROPOSITION XII. THEOREM.

Two rectangular parallelopipedons having equal altitudes, are to each other as their bases.

Let the rectangular parallelopipedons AG and AK have the same altitude AE: then, will they be to each other as their bases.

For, place them as shown in the figure, and produce the

The parallelopipedons AQ and AK have the common base AL; they are therefore to each other as their altitudes AD and AM: hence,

vol. AQ: vol. AK :: AD : AM.

Multiplying these proportions, term by term (B. II., P. XII.), and omitting the common factor, vol. AQ, we have,

vol. AG: vol. AK :: AB × AD : AO × AM.

But AB × AD is equal to the area of the base ABCD· and AO × AM is equal to the area of the base AMNO hence, two rectangular parallelopipedons having equal alti tudes, are to each other as their bases; which was to be

proved.

PROPOSITION XIII. THEOREM.

Any two rectangular parallelopipedons are to each other as the products of their bases and altitudes; that is, as the products of their three dimensions.

Let AZ and AG be any two rectangular parallelopipedons: then will they be to each other as the products of their three dimensions.

For, place them as in the figure, and produce the faces necessary to complete the rectangular parallelopipedon AK. The parallelopipedons AZ and AK have a com

mon base AN; hence (P. XI.),

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have a common

The parallelopipedons AK and AG

altitude AE; hence (P. XII.),

vol. AK : vol. AG :: AMNO ABCD.

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Multiplying these proportions, term by term, and omitting the common factor, vol. AK, we have,

vol. AZ vol. AG: AMNO × AX: ABCD × AE;

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or, since AMNO is equal to AM × AO, and ABCD to AB × AD,

vol. AZ vol. AG: AM × 40 × AX: AB × AD × AE; which was to be proved.