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PROPOSITION XXVIII. THEOREM. 616. The bases of a cylinder are equal.

§ 614

§ 459

Let A B E and DCG be the bases of the cylinder A C.

We are to prove A BE= DCG.

Any sections A C and A G, embracing A D, an element of the cylinder, are 5). .. A B = DC and A E = DG.

§ 134 Now

BC is II to EG,

(each being II to AD).
Also
BC= EG,

§ 464 ... E C is a 0.

§ 136 . E B= G C,

§ 134 ..A E AB=AGDC.

§ 108 Apply the upper base to the lower base, so that DC will coincide with A B.

Then A GDC will coincide with A E A B, and point G will fall upon point E.

That is, any point G in the perimeter of the upper base will coincide with the point in the same element in the lower base. . i. the bases coincide, and are equal.

Q. E. D. 617. COROLLARY 1. Any two parallel sections A B C and A' B'C', cutting all the elements of a cylinder E F, are equal. For these sections are the bases of the cylinder A C'.

618. COR. 2. Any section of a cylinder parallel to the base is equal to the base.

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PROPOSITION XXIX. THEOREM. 619. The lateral area of a cylinder is equal to the product of the perimeter of a right section of the cylinder by an element of the surface. DI

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Let A B C D E be the base, and A A' any element of the

cylinder A C'; and let the curve abcde be any right
section of its surface.
Denote the perimeter of the right section by P,

and the lateral surface of the cylinder by S.
We are to prove S= P x A A'.

Inscribe in the cylinder a prism having the regular polygon A B C D E as its base.

The right section abcd e of this prism will be a regular polygon inscribed in the right section abcde of the cylinder. $ 604

Denote the lateral area of the prism by s,

and the perimeter of its right section by p.
Then
S = p X A A',

$ 524 (the lateral area of a prison is equal to the product of the perimeter of a right

section by a lateral edge). Now let the number of lateral faces of the inscribed prism be indefinitely increased,

the new edges continually bisecting the arcs in the right section.

Then p approaches P as its limit,

and : approaches S as its limit.
But, however great the number of faces,

s=p X A A.
..S=P X AA',

$ 199 Q. E. D.

620. COROLLARY 1. The lateral area of a right cylinder is equal to the product of the perimeter of its base by its altitude.

621. Cor. 2. Let a cylinder of revolution be generated by the rectangle whose sides are R and H revolving about the side H.

Then R is the radius of the base of the cylinder, and H the altitude of the cylinder.

The perimeter of the base is 2 R; § 381 hence,

S=2a R X H.
The area of each base is a R2 ;

$ 381 hence, the total area T of a cylinder of revolution is expressed by

T = 2 * R X H + 2 + R2 = 25 R (H + R). 622. Cor. 3. Let S, S' denote the lateral areas of two similar cylinders of revolution ;

T, T' their total areas ; R, R' the radii of their bases ; H, H' their altitudes. Since the generating rectangles are similar, we have H _R _ H+R

§ 266 H R H' + R'' .S 24 RH RUH H? R?

•ģ= 2R' H = R H =H12 = R12 and GT 2 * R (H + R) R (H + R H2 R2

T' = 2 * R' (II' + R") R (71' + R') =H2=R' 2'

That is, the lateral areas, or the total areas, of similar cylinders of revolution are to each other as the squares of their altitudes, or as the squares of the radii of their bases.

PROPOSITION XXX. THEOREM. 623. The volume of a cylinder is equal to the product of its base by its altitude.

Let V denote the volume of the cylinder AG, B its

base, and H its altitude.
We are to prove V=B X H.

Let V' denote the volume of the inscribed prism AG, B' its
base, and H will be its altitude.
Then
V = B' X H,

§ 543

(the volume of a prism is equal to the product of its base by its altitude).

Now, let the number of lateral faces of the inscribed prism be indefinitely increased, the new edges continually bisecting the arcs of the bases.

Then B' approaches B as its limit,

and V approaches V as its limit.
But however great the number of the lateral faces,

V = B' X H.

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$ 381

624. COROLLARY 1. Let V be the volume of a cylinder of revolution, R the radius of its base, and H its altitude.

Then the area of its base is a R2,

..V=R2 X H. 625. Cor. 2. Let V and V' be the volumes of two similar cylinders of revolution, R and R' the radii of their bases, H and H their altitudes. Since the generating rectangles are similar, we have

H _ R

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That is, the volumes of similar cylinders of revolution are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases.

Ex. 1. Required, the entire surface and volume of a cylinder of revolution whose altitude is 30 inches, and whose base is a circle of which the diameter is 20 inches.

2. Required, the volume of a right truncated triangular prism the area of whose base is 40 inches, and whose lateral edges are 10, 12, and 15 inches, respectively..

3. Let E denote an edge of a regular tetrahedron ; show that the altitude of the tetrahedron is equal to EV }; that the surface is equal to E? V 3; and that the volume is equal to 13V2.

4. Required, the number of quarts that a cylinder of revolution will contain whose height is 20 inches, and whose diameter is 12 inches.

5. Given S, the surface of a cube, find its edge, diagonal, and volume. What do these become when S= 54 ?

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