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XIX. THE RECTANGULAR PARALLELOPIPED.

Definitions.—The rectangular parallelopiped is a solid contained by six rectangular faces, each of them being equal and parallel to its opposite face.

parallel; thus the face ABCD is equal and parallel to the face EFGH, and the face ADFE to the face BCGH; and so on. EFGH is also called the base; and the faces AEFD, BGHC are frequently called ends.

But base, ends, &c. are all included under the general term faces.

Also, AB or EG is called the length; AD or EF the breadth; and AE or DF the height or depth.

The whole surface consists of the areas of the two ends AEFD, BCHG, and four side faces AEGB, DFHC, ABCD, EFGH.

RULES.-(1) To find the volume of a rectangular parallelopiped.

Multiply the length by the breadth, and this product by the height.

Or, multiply the area of the base by the height.

(2) To find the whole surface.

Multiply the perimeter of the end (which is =2 width+2 depth) by the length, and the product is the area of the side faces; to which add

twice the area of one end, and the sum is the whole surface.

Or, add together the areas of the two ends and four sides, and the sum is the whole surface.

Note 1.-To find one dimension, either the length, breadth, or depth, when the volume and the other dimensions are given.-Divide the volume by the product of the two given dimensions, and the quotient is the required dimension.

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C

B

Note 2.-A parallelopiped may also be rhomboidal or oblique (fig. 2); in which case the edges are not perpendicular to each other—that is, all the faces are not rectangles. The annexed figure represents an oblique parallelopiped, in which some of the side faces, as ABCD, F EFGH, &c. are rect

angles, but others, as

Fig. 2.

G

AEGB, DFHC, &c. are rhomboids.

The volume of an oblique parallelopiped may be found thus:-Multiply the length AB by the breadth AD, and this product by the perpendicular height AK (and not by AE).

Note 3.-To increase or diminish proportionally the dimensions of a rectangular parallelopiped, so as to effect a corresponding increase or diminution of the volume.

Supposing it is required to find the dimensions of the parallelopiped (fig. 4) whose volume is twice that of the given parallelopiped (fig. 3), we may proceed thus: Since the

volume FGKL is twice the volume ABED, and that the volume of F G K L=GH × F G × GL, and also the volume of ABED=BCX ABX BE, we have,

GH X FG X GL 2 X BCX ABX BE;

that is, length x breadth x depth (of the required figure) =length 3/2 × breadth 3/2 × depth 3/2 (of given figure). Hence, if the volume is to be twice as large, then multiply each given dimension by 3/2.

If the volume is to be three times as large, then multiply each given dimension by 3/3.

If the volume is to be four times as large, then multiply each given dimension by 3/4; and so on.

Note 4.-To find the cubic feet or cubic inches of the material used in the construction of a rectangular cistern or box.

First find the volume of a rectangular parallelopiped whose length, breadth, and depth are the same as the external length, breadth, and depth of the cistern or box; then find the volume of another rectangular parallelopiped whose length, breadth, and depth are the internal length, breadth, and depth of the cistern or box.

Then the difference of these two volumes gives the number of cubic feet or cubic inches of the material.

N.B.-If the material employed is uniformly thick, then: external length =internal length + 2 thickness of material.

external breadth=internal breadth + 2 thickness

of material.

external depth internal depth + thickness of

material (if the vessel is without a lid).

=internal depth+2 thickness of material (if the vessel has

a lid).

Example 1.-Find the volume of a rectangular parallelopiped whose length is 6 ft., breadth 4 ft. 6 in., and height 3 ft. 6 in.

The volume 6 ft. x 41 ft. x 31 ft.=6 × 2 × 7=182 cub. ft. =941 cub. ft.=94 cub. ft. 864 cub. in.

Example 2.-A rectangular cistern is 24 ft. long and 15 feet broad; find its depth that it may hold 4320 cub. ft. of water.

By Note 1:

The depth-volume divided by the product of the two given dimensions.

=4320+(24 x 15)=4320-360=12 ft.

Example 3.-A box without a lid, in the shape of a rectangular parallelopiped, is made of wood 1 in. thick; externally the length is 3 ft. 6 in., the breadth 3 ft., and the depth 2 ft. Find the number of cubic inches of wood used

in its construction.

The external dimensions are 42 in., 36 in., and 24 in.; therefore the volume of the rectangular parallelopiped having these dimensions is 36288 cub. in.

Again, the internal dimensions are (42-2) in., (36-2) in., and (24—1) in., or 40, 34 and 23 in.; therefore the volume of the rectangular parallelopiped having these dimensions is 31280 cub. in.

=

Hence, volume of wood 36288-31280=5008 cub. in. =2 cub. ft. 1552 cub. in.

Formulæ

I. Volume=length × breadth x height. II. Whole surface= {2(breadth+height) × length}+2(breadth x height).

Or, sum of the areas of the six rectangular faces.

III. Required dimension

EXAMPLES.

:

Find the volume, and also the whole surface, of rectangular parallelopipeds with the following dimensions :(1) Length 10 in., breadth 8 in., and depth 6 in. (2) Length 2 ft. 4 in., breadth 2 ft., and height 1 ft. 8 in. (3) Length 4 ft. 8 in., breadth 3 ft. 10 in., and height 3 ft. (4) Length 6 yds. 2 ft., breadth 5 yds. 1 ft., and height 2 yds. 2 ft.

(5) Length 5 yds. 1 ft. 6 in., breadth 4 yds. 2 ft. 3 in., and height 3 yds. 2 ft. 6 in.

(6) Length 6.75 ft., breadth 5.5 ft., and height 4.25 ft.

Find the volume of the following rectangular parallelopipeds, when the area of the base and the height are, respectively

(7) Area of base 97 sq. in., and height 10 in.

(8) Area of base 3 sq. ft. 20 sq. in., and height 1 ft. 3 in. (9) Area of base 11 sq. ft. 40 sq. in., and height 5 ft. 3 in. Find the height or depth of a rectangular parallelopiped, when its volume, length, and breadth are, respectively(10) Volume 378 cub. in., length 9 in., and breadth 7 in. (11) Volume 1680 cub. in., length 1 ft. 2 in., and breadth 1 ft.

(12) Volume 3 cub. ft. 864 cub. in., length 2 ft., and

breadth 1 ft. 6 in.

(13) Volume 1164 cub. ft. 1008 cub. in., length 12 ft. 6 in., and breadth 10 ft. 9 in.

(14) If the area of the bottom of a box is 27 sq. ft. 72 sq. in.; find what must be its depth, so that the volume of the box may be 123 cub. ft. 1296 cub. in.

(15) Find the price of a balk of timber 39 ft. 6 in. long, and 3 ft. 7 in. thick each way, at 2s. 6d. a cub. ft.

(16) A tank is 30 ft. 9 in. long, 16 ft. 7 in. wide, and 6 ft. 4 in. deep; find how much water it will hold in cubic feet and inches.