M

E

N

I

Q

P

Ο

H

A

D

K

L

B

C

solid AG: solid AQ:: ABCD ×AE: AIKL× AP.

If, instead of the base ABCD, we put its equal ABX AD, and instead of AIKL, we put its equal AIX AL, we shall have

solid AG: solid AQ:: AB × AD × AE : AIX AL× AP.

Therefore any two rectangular parallelopipeds, etc.

Scholium. Hence a rectangular par

allelopiped is measured by the product of its base and altitude, or the product of its three dimensions.

It should be remembered that, by the product of two or more lines, we understand the product of the numbers which represent those lines; and these numbers depend upon the linear unit employed, which may be assumed at pleasure. If we take a foot as the unit of measure, then the number of feet in the length of the base, multiplied by the number of feet in its breadth, will give the number of square feet in the base. If we multiply this product by the number of feet in the altitude, it will give the number of cubic feet in the parallelopiped. If we take an inch as the unit of measure, we shall obtain in the same manner the number of cubic inches in the parallelopiped.

The volume of a prism is measured by the product of its base by its altitude.

For any parallelopiped is equivalent to a rectangular parallelopiped, having the same altitude and an equivalent base (Pr. 7). But the volume of the latter is measured by the product of its base by its altitude; therefore the volume of the former is also measured by the product of its base by its altitude.

Now a triangular prism is half of a parallelopiped having the same altitude and a double base (Pr. 5). But the volume of the latter is measured by the product of its base by its altitude; hence a triangular prism is measured by the product of its base by its altitude.

But any prism can be divided into as many triangular prisms of the same altitude as there are triangles in the polygon which forms its base.

Also, the volume of each of these triangular prisms is measured by the product of its base by its altitude; and, since they all have the same altitude, the sum of these prisms will be measured by the sum of the triangles which form the bases, multiplied by the common altitude. Therefore the volume of any prism is measured by the product of its base by its altitude.

Cor. If two prisms have equal altitudes, the products of the bases by the altitudes will be as the bases (B. II., Pr. 10); hence prisms having equal altitudes are to each other as their bases. For the same reason, prisms having equivalent bases are to each other as their altitudes; and any two prisms are to each other as the products of their bases and altitudes.

PROPOSITION XII. THEOREM.

Similar prisms are to each other as the cubes of their homolo.gous edges.

Let ABCDE-F, abcde-f be two sim

ilar prisms; then will the prism AD- F
F be to the prism ad-ƒ as AB3 to ab3,
or as AF3 to aƒ3.

For the solids are to each other as
the products of their bases and alti-
tudes (Pr. 11, Cor.); that is, as ABCDE
× AF to abcdex af. But, since the A
prisms are similar, the bases are simi-
lar figures, and are to each other as

B

C

a

the squares of their homologous sides; that is, as AB2 to al2. Therefore we have

solid FD: solid fd:: AB2 × AF : ab2 × af.

But, since RF and bf are similar figures, their homologous sides are proportional; that is,

AB2: ab2: AF2: af2.
AF: af:: AF: af.

Therefore (B. II., Pr. 12),

AB2 × AF: ab2 ×aƒ:: AF3: aƒ3 :: AB3 : al3.

Hence (B. II., Pr. 4) we have

solid FD: solid fd:: AB3: ab3 :: AF3: aƒ3.

Therefore similar prisms, etc.

If a pyramid be cut by a plane parallel to its base,

1st. The edges and the altitude will be divided proportionally. 2d. The section will be a polygon similar to the base.

Let A-BCDEF be a pyramid cut by a plane bedef parallel to its base, and let AH be its altitude; then will the edges AB, AC, AD, etc., with the altitude AH, be divided proportionally in b, c, d, e, f, h, and the section bedef will be similar to BCDEF.

First. Since the planes FBC, fbc are parallel, their sections FB, fb, with a third plane AFB, are parallel (B. VII., Pr. 12); therefore the triangles AFB, Afb are similar, and we have the proportion

AF: Af:: AB: Ab.

AB: Ab:: AC: Ac,

and so for the other edges. Therefore the edges AB, AC, etc., are cut proportionally in b, c, etc. Also, since BH and bh are parallel, we have AH: Ah:: AB: Ab.

Secondly. Because fb is parallel to FB, bc to BC, cd to CD, etc., the angle fbe is equal to FBC (B. VII., Pr. 15), the angle bed is equal to BCD, and so on. Moreover, since the triangles AFB, Afb are similar, we have FB: fb:: AB: Ab.

And because the triangles ABC, Abc are similar, we have

AB: Ab:: BC: bc.

Therefore, by equality of ratios (B. II., Pr. 4),

For the same reason,

FB: fb:: BC: bc.

BC: bc:: CD: cd, and so on.

Therefore the polygons BCDEF, bedef have their angles equal each to each, and their homologous sides proportional; hence they are similar. Therefore, if a pyramid, etc.

Cor. 1. If two pyramids having the same altitude, and their bases situated in the same plane, are cut by a plane parallel to their bases, the sections will be to each other as the bases.

Let A-BCDEF, A-MNO be two pyramids having the same altitude, and their bases situated in the same plane; if these pyramids are cut by a plane parallel to the bases, the sections bcdef, mno will be to each other as the bases BCDEF, MNO.

But, since bcdef and mno are in the same plane, we have
AB: Ab:: AM: Am (B. VII., Pr. 16);

consequently, BCDEF: bcdef:: MNO: mno.

Cor. 2. If the bases BCDEF, MNO are equivalent, the sections bcdef, mno will also be equivalent.

The lateral surface of a regular pyramid is equal to the product of the perimeter of its base by half its slant height. Let A-BDE be a regular pyramid whose base is the polygon BCDEF, and its slant height AH; then will its lateral surface be equal to the perimeter BC+CD+DE, etc., multiplied by half of AH.

F

H

E

The triangles AFB, ABC, ACD, etc., are all equal, for the sides FB, BC, CD, etc., are all equal (Def. 15); and, since the oblique lines AF, AB, AC, etc., are all at equal distances from the perpendicular, they are equal to each other (B. VII., Pr. 5). Hence the altitudes of these several triangles are equal.

B

C

But the area of the triangle AFB is equal to FB multiplied by half of AH; and the same is true of the other triangles ABC, ACD, etc. Hence the sum of the triangles is equal to the sum of the bases FB, BC, CD, DE, EF multiplied by half the common altitude AH; that is, the lateral surface of the pyramid is equal to the perimeter of its base multiplied by half the slant height.

Cor. 1. The lateral surface of a frustum of a regular pyramid is equal to the sum of the perimeters of its two bases multiplied by half its slant height.

H

B

C

D

Each side of a frustum of a regular pyramid, as FBbf, is a trapezoid (Pr. 13). Now the area of this trapezoid is equal to the sum of its parallel sides FB, fb, multiplied by half its altitude Hh (B. IV., Pr. 7). But the altitude of each of these trapezoids is the same; therefore the area of all the trapezoids, or the lateral surface of the frustum, is equal to the sum of the perimeters of the two bases multiplied by half the slant height.

Cor. 2. If the frustum is cut by a plane parallel to the bases, and at equal distances from them, this plane must bisect the edges Bb, Cc, etc. (B. IV., Pr. 16); and the area of each trapezoid is equal to its altitude multiplied by the line which joins the middle points of its two inclined sides (B. IV., Pr. 7, Cor.). Hence the lateral surface of a frustum of a pyramid is equal to its slant height multiplied by the perimeter of a section at equal distances between the two bases.

PROPOSITION XV. THEOREM.

Two triangular pyramids having equivalent bases and equal altitudes are equivalent.

Let A-BCD, a-bcd be two triangular pyramids having equiva lent bases BCD, bed, supposed to be situated in the same plane,