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second figures of the root, and (c) the square of the second figure of the root; placing the tens-figure in each of the lines (b) and (c) exactly under the units-figure of the line above it. (See worked example below.)

5°. Multiply the true divisor by the second figure of the root, and subtract the result from the resolvend.

136. The cube root of decimal fractions is found precisely as that of whole numbers; only the periods of three figures are reckoned both ways from the decimal point. (Compare Art. 45.)

137. The cube root of a vulgar fraction is found by taking the cube root of the numerator and the denominator. (Compare Art. 46.)

[1. When the volume is given in several different denominations, as cub. yds., cub. ft., and cub. in., it must be reduced to its lowest denomination before the cube root can be extracted. The length of the side will be in the same denomination.

2. Volumes are expressed in cubic measure, areas in square measure, and lengths of sides in lineal measure.]

Ex. The volume of a cube is 15625 cubic yards; find the length of one of its sides.

(a) 2x2x3=12

2x5x3= 30

15,625(25

8

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Find the length of each side of the cubes whose volumes are

respectively :-
(1) 2197 c. in.
(2) 9261 c. in.
(3) 3375 c. in.

(4) 79507 c. in.

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(13) A cubical block of stone contains

is the length of its side?

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(14) A cubical box contains 110592 solid inches; find, in feet, the length of its side.

(15) Find the edge of a cube whose content is 1 c. yd. 7 c. ft. 567 c. in.

(16) A packing-case in the shape of a cube containing 12 c. ft. 1216 c. in. is exactly filled by 64 cubical boxes; find the length of the side of each box.

(17) A cubical vessel measures 5 ft. every way; find the length of another vessel of the same shape, that will hold eight times as much water.

(18) A cubic foot of water weighs 1,000 ozs.; find, in feet, the length of the side of a cubical cistern whose contents weigh 612 tons 10 cwts.

(19) The cost of a cube of metal at £3 6s. 8d. per c. ft. is £1706 13s. 4d. ; find the length of an edge.

(20) Supposing a cubical room to contain 29791 c. ft., the area of the floor.

find

138. CASE II. When the cube root has three figures. RULE.-Proceed as before until two figures have been obtained, then

6°. Bring down the next period, and find a trial divisor by multiplying the square of the number already obtained in the quotient (consisting of two figures) by 3.

[This can be more readily obtained, however, by placing the sum of (b) and (c) (Rule 4°) under the last true divisor, and adding together the three lines coupled by the bracket in the worked example below.] 7°. Repeat the process explained above (Rule 4o) to find the true divisor.

8°. Multiply the true divisor by the third figure of the root, and subtract the result from the resolvend.

Ex. Find the side of a cubical vessel containing 244 c. yds. 3 c. ft. 1377 c. in.

Here, 244 c. yds. 3 c. ft. 1377 c. in. =11390625 c. in.

2×2×3=12

2x2x3= 12
2x2=

11,390,625(225

8

3390

2648

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Thus the side is 225 in., or 18 ft. 9 in., or 6 yds. 9 in.

Ex. LVIII.

Find the length of each side of the cubes whose volumes are

respectively:-

(1) 244-140625 c. in.

(2) 1.953125 c. ft.

(3) 1498585 c. ft.

(4) 14675 c. yds.

(5) 614 c. ft. 216 c. in.

(6) 44 c. yds. 25 c. ft. 1088 c. in. (7) 347 c. yds. 2 c. ft. 1189 c. in. (8) 198 c. yds. 13 c. ft. 648 c. in.

(9) Find the edge of a cube whose content is 12.812904 c. ft.

(10) A cubical block of gold contains 113045 c. in.; find the length of a side.

(11) A cube contains 51 c. yds. 15 c. ft. 728 c. in.; find its edge.

(12) The cost of a cubical mass of metal at 1d. per c. in. is £5698 9s. 3d.; find the length of each side in feet and inches.

(13) Find the side of a cube whose volume is equal to the sum of the volumes of three cubes whose sides are 3 ft., 4 ft., and 5 ft. respectively.

(14) A cubical cistern is constructed to hold 775 gallons of water; find its length, allowing 6 gallons to the cubic foot.

139. The surface of a solid is its outside area.

140. The surface of a cube consists of six faces, each of which is a square; and the area of the whole surface is the sum of the areas of these squares, found by Art. 15.

141. To find the whole surface of a cube.

RULE. Multiply the square of the side by 6, and the product will be the area of the whole surface. Ex. Finċ. the whole surface of a cube whose side is 4 ft. Area of each face-4x4-16 sq. ft.

Area of whole surface=16×6=96 sq. ft.

Ex. LIX.

Find the area of the whole surface of the cubes which have the following lengths :

(1) 2 ft. 3 in.

(2) 1 ft. 9 in.

(3) 1.25 ft.
(4) 3.75 yds.

(5) 221 ft.
(6) 15 yds.

(7) What is the cost of hewing all the faces of a cubical block of stone whose side is 2 ft. 6 in., at 6d. per sq. foot? (8) A cubical block of stone contains 389017 solid feet; what is the superficial content of one of its sides?

[Find the length of an edge by Art. 134, then its square will give the area of each face, Art. 15.]

(9) The surface of a cube contains 13.5 sq. ft.; find the length of an edge.

[Area of six faces=13.5, area of one face=13'5÷6. The length of an edge may then be found by Art. 43.]

(10) The area of the whole surface of a cube is 204 sq. ft. 24 sq. in.; find the volume.

(11) The cost of polishing a cubical block of granite at 3s. 4d. per sq. ft. is £20 5s.; find its solid content.

(12) A cubical box contains 15625 cubic inches; find the expense of gilding the top and sides, at three-halfpence per square inch.

(13) How much sheet-lead is used in lining the sides and bottom of a cubical vessel containing 729 c. ft. of water?

[The surface to be covered with lead consists of five faces only.]

(14) A cubical cistern measures 5 ft. every way; find the cost of lining it with lead, 7 lbs. to the square foot, at £42 per ton.

(15) A cubical cistern holds 18 c. ft. 1664 c. in. of water; find how many sq. ft. of lead will be required for lining its sides and bottom.

(16) If a cubical cistern requires 180 sq. ft. of lead for lining its sides and bottom, find the number of cubic feet of water it will hold.

(17) A cubical box with a lid is covered all over with sheetlead weighing 4 lbs. per sq. ft., and 384 lbs. of lead are used; what is the length of each side of the box?

THE RECTANGULAR PARALLELOPIPED.

142. A Rectangular Parallelopiped is a solid bounded by six rectangular faces, each of them being equal and parallel to its opposite face.

The annexed figure re- F presents a parallelopiped in which it will be seen the following faces are equal and parallel :

(1) ABCD and FGHEG (front and back);

(2) FADE and GBCH (top and bottom);

B

A

(3) FGBA and E HCD (two ends).

H

E

D

A D is the length, A F the breadth, A B the height or depth; and G B C H the base.

Suppose we have a rectangular parallelopiped which is 4 in. long, 3 in. broad, and 2 in. deep. Let it be cut by planes an inch apart parallel to the faces, as in the annexed figure; it is thus divided into 24 equal solids, each of which is a cube, being an inch long, an inch broad, and an inch thick.

Such a cube is called a cubic inch; and as the parallelopiped contains 24 of them, its volume is said to be 24 cubic inches.

The number of cubic units in each horizontal row will obviously be equal to the number of square units in the base; and the number of these rows will be equal to the number of linear inches in the depth; therefore the number of cubic inches in the whole figure will be the number of square inches in the base multiplied by the number of linear inches in the depth or height.

143. To find the volume of a rectangular parallelopiped.

RULE. Multiply the area of the base by the height, and the product will be the volume.

Ex. A rectangular block of granite is 7 ft. 6 in. long, 3 ft. 4 in. broad, and 2 ft. 3 in. thick; what is its cost, at 6s. 9d. per cubic foot?

Here, 7 ft. 6 in. 7 ft.; 3 ft. 4 in. -33 ft.; 2 ft. 3 in.=21 ft.
Area of base 71 × 33=25 sq. ft. (Art. 50.)

Solid content =25×21-56 c. ft.
Cost 6s. 9d. × 561 = £18 19s. 8d.

Ex. LX.

Find the number of cubic feet and inches in rectangular parallelopipeds which have the following dimensions :

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(1) Length, 2 ft. 6 in.; breadth, 1 ft. 4 in.; depth, 4 ft. 2 in. Length, 4 ft. 8 in.; breadth, 2 ft. 9 in.; depth, 1 ft. 6 in.

(3) Length, 3.75 ft.; breadth, 2-5 ft.; depth, 1.25 ft. (4) Length, 128 ft.; breadth, 53 ft.; depth, 21 ft.

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