PROPOSITION I. THEOREM.

522. The sections of a prism made by parallel planes

Let the prism A D be intersected by the parallel planes GK, G' K.

We

=

e are to prove section G HIK L section G'H' I'K' L'.

GH, HI, IK, etc., are parallel respectively to G' H', H'I', I'K', etc.,

(the intersections of two || planes by a third plane are || lines).

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.. A GHI, HI K, etc., are equal respectively to G'H' I', H'I'K', etc.,

§ 462

(two not in the same plane, having their sides respectively parallel and lying in the same direction, are equal).

Also, sides GH, HI, IK, etc., are equal respectively to

G' H', H' I', I' K', etc.,

(lines comprehended between || lines are equal). ... section GHIKL = section G' H'I'K' L', (being mutually equiangular and equilateral).

§ 135

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Q. E. D.

523. COROLLARY. Any section of a prism parallel to the base is equal to the base; and all right sections of a prism are equal.

PROPOSITION II. THEOREM.

524. The lateral area of a prism is equal to the product of a lateral edge by the perimeter of the right section.

Let GHIKL be a right section of the prism A D'.

We are to prove lateral area of prism A D' = A A' × perimeter G HIK L.

Consider the lateral edges A A', B B', etc., to be the bases of the SA B', B C', etc., which form the convex surface of the prism.

Then the altitudes of these will be the GH, HI, IK, etc.,

and the area of each is the product of its base and alti

tude.

Now the bases of these

are all equal,

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§ 464

(|| lines comprehended between || planes are equal) ; and the sum of the altitudes GH, HI, I K, etc., is the perimeter

of the right section.

Hence, the sum of the areas of these is the product of a lateral edge A A' by the perimeter of the right section. That is, the lateral area of the prism is equal to the product of a lateral edge by the perimeter of a right section.

525. COROLLARY.

Q. E. D.

The lateral area of a right prism is equal

to the altitude multiplied by the perimeter of the base.

PROPOSITION III. THEOREM.

526. Two prisms are equal if the three faces including a trihedral angle of the one be respectively equal to the three corresponding faces including a trihedral angle of the other, and similarly placed.

Let A D, AG, AJ, be respectively equal to A' D', A'G', A'J', and similarly placed.

§ 492

(two trihedrals are equal, when the three face of the one are equal respectively to the three face of the other and are similarly placed).

Apply trihedral Z A to trihedral

A'.

Then the base A D will coincide with the base A' D',
face AG with A'G',

and face AJ with A'J';

.. FG will coincide with F' G', and FJ with F" J'.
.. the upper bases, FI and F'I', will coincide,
(being equal polygons, since they are equal to the equal lower bases).
.. the remaining edges will coincide,

(their extremities being the same points).

.. the prisms will coincide and be equal.

Q. E. D.

527. COROLLARY 1. Two truncated prisms are equal, if the three faces including a trihedral of the one be respectively equal to the three faces including a trihedral of the other, and be similarly placed.

528. COR. 2. Two right prisms having equal bases and altitudes are equal. If the faces be not similarly placed, if one be inverted, the faces will be similarly placed and the prisms can be made to coincide.

PROPOSITION IV. THEOREM.

529. An oblique prism is equivalent to a right prism whose bases are equal to right sections of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism.

Let A D' be an oblique prism, and FI a right section.

Complete the right prism F1', making its edges equal to

those of the oblique prism.

We are to prove oblique prism A D'≈right prism FI.

(two trihedrals are equal when three face of the one are respectively equal to three face of the other, and are similarly placed).

(two truncated prisms are equal when the three faces including a trihedral of the one are respectively equal to the three faces including a trihedral

of the other, and are similarly placed).

To each of these equal prisms add the prism F D'.
Then oblique prism A D' right prism FI'.

Q. E. D.

We are to prove faces A F and DG equal and parallel.

(two not in the same plane having their sides || and lying in the same

direction are equal).

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(if two not in the same plane have their sides || and lying in the same direction their planes are parallel).

allel.

In like manner we may prove A H and B G equal and par

Q. E. D.

531. SCHOLIUM. Any two opposite faces of a parallelopiped may be taken for bases, since they are equal and parallel parallelograms.