Here, again, the product is 434 square feet + (3 × 12) +11 inches, or 434 square feet, 47 square inches. And this manner of estimating the inches must be observed in all cases where two dimensions in feet and inches are thus multiplied together.

3. A certain board is 28 ft. 10 in. 6" long, and 3 ft. 2 in. 4" wide; how many square feet does it contain? Ans. 92 ft. 2 in. 10" 6".

4. A partition is 82 ft. 6 in. by 13 ft. 3 in.; how many square feet does it contain? Ans. 1093 ft. 1 in. 6".

5. If a floor be 79 ft. 8 in. by 38 ft. 11 in.; how many square feet are therein? Ans. 3100 ft. 4 in. 4".

6. There is a yard of 21 ft. 6 in. by 17 ft. 6 in., which is to be paved with stones 1 ft. 6 in. square; how many stones are necessary for the purpose? Ans. 167+ 7. Suppose the dimensions of a bale of goods to be 7 ft. 6 in., 3 ft. 3 in., and 1 ft. 10 in.; what is the solid content? Ans. 44 ft. 8 in. 3".

SHIPS' TONNAGE.

By a law of the Congress of the United States of America, the tonnage of a ship is to be found in the following

manner.

RULE.

If the vessel be double-decked, take the length thereof, from the fore part of the main stem to the after part of the stern post above the upper deck; the breadth thereof at the broadest part above the main walls, half of which breadth shall be accounted the depth of such vessel; then deduct from the length three-fifths of the breadth, multiply the remainder by the breadth, and the product by the depth; divide the last product by ninety-five, and the quotient will be the true content or tonnage of such vessel.

If the vessel be single-decked, take the length and breadth as above directed, in respect to a double-decked vessel, and deduct from the length three-fifths of the breadth; and taking the depth from the under side of the deck plank to the ceiling in the hold, multiply and divide as aforesaid, the quotient will be the tonnage.

NOTE.-Carpenters, in finding the tonnage, multiply the the length of the reel by the breadth of the main beam, and the depth of the hold in feet, and divide the product by 95; the quotient is the tonnage for a single-decked vessel. In double-decked vessels, half the breadth is taken for the depth.

EXAMPLES.

Suppose the length of a double-decked vessel is 80 feet, and the breadth 24 feet; what is the tonnage, of the breadth 24 ft.-14.4 and of 24-12 the depth?

80
14.4

65.6 × 24 = 1574.4

12 the depth.

95)18892.8(198.8 Ans.

2. Required the legal tonnage of a double-decked vessel, the length being 110.5 feet, and breadth 30.6 feet.

3. What is the tonnage of a single-decked vessel, which is 76.4 feet in length, 28.6 feet in breadth, and 12.3 feet in depth? Ans. 219.362 tons. NOTE. The tonnage of ships of war is found by taking the continued product of the length, breadth, and depth, and dividing the product by 100.

4. What is the tonnage of a sloop of war, length 97 feet, breadth 31 feet, and depth 15 feet? Ans. 466.085 tons.

MENSURATION OF SURFACES.

THE following rules are given to find the areas, or superficial contents of the following figures.

RULE.

Multiply the base by the height.

NOTE. This rule is also applicable to the rhombus, or rhomboid, the height being taken perpendicular to the base.

EXAMPLES.

1. What is the area of a square, the sides whereof are 35.25 chains? Ans. 124 acres, 1 rood, 1 perch. 2. What is the area of a square whose side is 14 feet? Ans. 196 square feet. 3. What is the area of a board whose length is 14 feet, 6 inches, and breadth 4 feet 9 inches?

Ans. 68 feet, 10 inches. 4. What is the area of a rectangular piece of ground of which the length is 1375 links, and breadth 950?

An. 13 acres, 10 perches.

1. The base of a triangle is 28.2 yards, and the perpendicular height 18.4 yards, what is the area?

Ans. 259.44 sq. yds. 2. What is the area of a triangle whose base is 18 feet, 4 inches, and perpendicular height 11 feet 10 inches? Ans. 108 feet, 5 inches+ 3. What is the area of a triangle, whose base is 625 links, and perpendicular height 520 links?

S162500 square links, or
Ans. 1 acre, 2 roods, 20 perches.

4. How many square yards in a triangle whose base is 49 feet, and perpendicular height 25 feet?

Ans. 685 or 68.7361.

OF THE CIRCLE.

Diameter

Circumference

Having the diameter to find the circumference, or having the circumference to find the diameter.

RULES.

1. As 7 is to 22, so is the diameter to the circumference. Or as 113 is to 355, so is the diameter to the circumference.

Or as 1 is to 3.1416, so is the diameter to the circumference.

ter.

2. As 22 is to 7, so is the circumference to the diameter. As 355 is to 113 so is the circumference to the diameter. As 3.1416 is to 1, so is the circumference to the diame

EXAMPLES.

1. If the diameter of a circle be 22.6 feet, what is the circumference? Ans. 71 feet. 2. If the diameter of the earth be 7970 miles, what is its circumference, supposing it to be a perfect sphere? Ans. 25038.5 miles.

3. If the circumference of a circle be 12 feet, what is the diameter? Ans. 3.819708.

To find the area of a circle.

RULES.

1. Multiply the square of the diameter by .7854. 2. Multiply the square of the circumference by .079578. 3. Multiply circumference by the diameter.

EXAMPLES.

1. If the diameter of a circle be 1, and its circumference 3.1416, what is the area? Ans. .7854.

2. If the diameter of a circle is 22.6 feet, what is the area? Ans. 401.1509.

4. If the circumference of a circle be 71 feet, what is the area? Ans. 401.1627.

To find the area of long irregular figures.

RULE.

Measure the breadths at both ends, and at several places at equal distances. Add together all these intermediate breadths, and half the two extremes, which sum multiply by the length, and divide by the number of parts for the

area.

EXAMPLES.

1. The breadths of an irregular field I find to be as follows: at the ends, 820 links and 860 links, and the three intermediate breadths 740, 920, and 1020 links, the whole length being 390 links required the area.

NOTE. AS 100000 square links is equal to one acre, if five figures be cut off on the right hand for decimals, the rest will be acres.

2. The length of an irregular field is 840 links, and the breadths at six equidistant places, 174, 206, 142, 165, 201, 244 links: what is the area? Ans. 15 acres+

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