inches in a link = 594 in. 594 inches 12=49 feet 6 in.

17+ rods=17.5×6=105 rods. [75 links x 7.92 the number of 105 rodsx 16.5 the number of feet in a rod-1732.5 feet. OPERATION.

17.5 rods, length of bridge

6 stringpieces.

OPERATION.

7.92 inches in a link.

75 links.

In 1 Mile how many links? 1 mile = 80 chains. 80 chains x 100 the number of links in a chain = 8000 Links.

By adding or annexing two ciphers, multiplies by 100.

× 80 chains in a mile, by 100 links in a chain.

Prod. 8000 links in a mile.

LESSON 8.

Ans. 594 in.-49 feet 6 in.

LESSON 9.

From Albany to New-York is a distance of 165 miles; how many rods and links? 165×3 the furlongs in a mile= 1320 fur. 1320x40 the rods in a furlong=52800 R. 52800 x25 the links in a rod = 1320000 Links.

OPERATION.

165 Miles.

8 Furlongs in a mile.

1320 Furlongs.

40 Rods in a furlong.

1320000 Links in 165 miles. Answer, 1320000 Links.

LESSON 10.

In 125 rods how many

How many inches and feet chains and links?

in 75 links ?

Answer, 31 ch. 25 Links.

OF LAND AND SQUARE MEASURE.

RULE 1.

because 100 links make 1

To bring rods into chains, chain, the same as 100 cents

divide by 4.

4)125

make 1 dollar.

LESSON 12.

A carpenter has 68 feet of joist, which he wishes to cut into braces of 4 feet and 3 inches long; how many braces may he have? Answer, 16 braces. 68×12=816 inches for a dividend :

4 ft.x12=48+3 in.=51 inches for a divisor.

LAND AND SQUARE MEASURE,

CALLED MENSURATION OF SUPERFICIES.

144 square inches make 1 square foot.

1 square mile.

Let the first business be to investigate and prove our Table.

OF LAND AND SQUARE MEASURE.

LESSON 1.

Suppose we have 72 boards, each 12 inches wide, and 14 feet long; how many feet in the whole? Answer, 1008 ft.

72 boards.

14 feet in each board.

288

72

1008

Thus we have 1008 feet in

the whole; but reduce these

square feet into inches, and

divide them by 144 the num-144)193536(1344 feet.

ber of inches in a foot; see if the result will be 1008.

1008 × 12 × 12=145152

144 1008 for proof.

144

.495

432

1008 ft.

12in. long

LESSON 2. How many square feet in 72 boards, each 14 feet long

and 16 inches wide?

Answer, 1344 feet.
72 boards.
14 feet long.

238

72

LESSON 3.

If the roof of a house be 14 feet from the ridge to the edge of the eaves, and from gable end to gable end 21 feet; how many superficial feet including both sides?

14 x 21

Answer, 588 feet. 291 for one side,

294 × 2 = 588.

LESSON 4.

If a barn-floor be 35 feet long and 21 feet wide, how many planks 14 feet long by 15 inches wide, will lay the

1008 feet 12 in. by 16. floor? Answer, 42 planks.

OF LAND AND SQUARE MEASURE.

35 x 21=735 feet: 735 x 12 x 1105840 inches for a dividend.

14 × 12 × 15 = 2520 inches for a divisor.

35 feet the length of the
21 the width.

14 feet in each plank 12 by

12

168
15

[floor. 840

[15 inches.

In 42 planks, each 14 feet in length and 15 inches wide, how many feet? Answer, 735 feet. Observe the planks are 14 feet long, and every foot is 15 inches wide. 15 inches make 1 foot 3 inches, or 14 foot. Multiply 1 ft. 3 in. by 14, the product will be the contents of one plank. Multiply by 7 and by 2. Carry

at 12 in the inches.

1 ft. 3 in.

7

8 9

2

17 6 contents in each plank.

Multiply 17 ft. 6 in. by 42, the product will be the whole number of feet in 42 planks. Multiply by 7 and by 6.

A SHORT OPERATION.

ft. in. 17 6 42 planks by 14 and of the product added.

7

14

OF LAND AND SQUARE MEASURE.

LESSON 5, PERFORMED DECIMALLY.

Each foot of the plank contains 1 foot 3 inches, or 1

foot, or, 1.25 foot.

1.25×14=17.50; that is, 174 feet in each plank.

We will not attend to many of these short rules, till our minds are better informed; whereby we can scan the origin and nature of the work to be performed.

LESSON 6.

A paver undertook to lay a pavement 10 rods by 27 feet; how many square yards must be allowed him in settlement? Ans. 495 yards.

5 yards or 5.5 yards=1 rod.

10 rodsx5.5=55 yards, the length of the pavement. 27 feet 9 yards, the breadth.

Multiply the length by the breadth, the product will be the superficial contents.

55 yards the length.

9 yards the breadth.

495 square yards.

NOTE. In all superficies multiply the length by the breadth: the product will give the superficial contents or area.

The multipliers must be of one denomination: if the length be in rods, and the breadth be in feet; reduce the length to feet.

LESSON 7.

A spadiard digs a garden 8 rods in length, and 66 feet in breadth; what is the area or superficial contents in square yards? Ans. 968 square yards.