1

0.25; if N. E., by 0.3; if N. Y., by 0.4; if Penn., by 0.375, and if Georgia, by 0.23;-the quotient will be their value in dollars, cents and mills. And to change Federal Money into the above currencies, multiply it by the preceding decimals, and the product will be the answer in pounds and decimal parts.

3. In £91, how many dollars ?

9. Reduce £25 15s. N. E.,

£91 E. $404.444. to Federal Money.

Can. $364. N. E., $303.333.
N. Y. $227.50, &e. Ans.

4. Reduce £125, N. E. to Federal Money.

Ans. $416.666. 5. Change $100 to each of the foregoing currencies.

$100 £22 10s. Eng.=£25 Can. £30 N. E.-£40 N. Y.

Ans. $85.833.

10. In £227 17s. 54d. N. E., how many dollars, cents and mills?

Ans. $759 57cts. 3m.

11. In $1.612, now many shillings, pence and farthings? Ans. 123. 104d. N. Y. S 9s. Ed. N. E.

12. Reduce £33 13s. N. Y., to Federal Money.

302. The following rules, founded on the relative value of the several currencies, may sometimes be of use:—

To change Eng. currency to N. E. add, N. E. to N. Y. add, N. Y. to N. E. subtract 4, N. E. to Penn. add 1, Penn. to N. E. subtract, N. Y. to Penn. subtract, Penn. to N. Y. add, N. E. to Can. subtract &, Can. to N. E. add д, &c.

TABLE

303. Of the most common gold and silver coins, containing their weight fineness, and intrinsic value in Federal Money.

Country. Names of coins. | Weight. | Fineness. | Value.

NOTE. The current values of several of the above coins differ somewhat

from their intrinsic value, as expressed in the table.

SECTION II.

MENSURATION.

1. Mensuration of Superficies.

304. The area of a figure is the space contained within the bounds of its surface, without any regard to thickness, and is estimated by the number of squares contained in the same; the side of those squares being either an inch, a foot, a yard, a rod, &c. Hence the area is said to be so many square inches, square feet, square yards, or square rods, &c.

305. To find the area of a parallelogram (65), whether it be a square, a rectangle, a rhombus, or a rhomboid.

RULE.-Multiply the length by the breadth, or perpendicular height, and the product will be the area.

306. To find the area of a triangle. (64)

RULE 1.-Multiply the base by half the perpendicular height, and the product will be the area.

RULE 2.-If the three sides only are given, add these together and take half the sum; from the half sum subtract each side separately; multiply the half sum and the three remainders continually together, and the square root of the last product will be the area of the triangle.

307. To find the area of a trapezoid. (65)

RULE.-Multiply half the sum of the two parallel sides by the perpendicular distance between them, and the product will be

the area.

308. To find the area of a trapezium, or an irregular polygon,

RULE.-Divide it into triangles, and then find the area of these triangles by Art. 306, and add them together.

134

MENSURATION OF SUPERFICIES

i. A trapezium is divided into two triangles, by a diagonal 42 rods long, and the perpendiculars let fall from the opposite angles of the two triangles, are 18 rods and 16 rods; what is the area of the trapezium?

309, 310

2. What is the area of a trapezium whose diagonal is 108 feet, and the perpendiculars 561 and 603 feet?

Ans. 6347 feet.

3. How many square yards in a trapezium whose diagonal is 65 feet, and the perpendiculars let fall upon it 28 and 33.5 feet?

Ans. 222

309. To find the diameter and circumference of a circle, either from the other. (67)

RULE 1.--As 7 is to 22, so is the diameter to the circumference, and as 22 is to 7, so is the circumference to the diameter.

RULE 2.—As 113 is to 355, so is the diameter to the circumference, and as 355 is to 113, so is the circumference to the di

ameter.

RULE 3.-As 1 is to 3.1416, so is the diameter to the circumference, and as 3.1416 is to 1, so is the circumference to the 'diameter.

1. What is the circumference of a circle whose diameter is 14 feet?

By Rule 1.

As 7:22: 14: 44, Ans.
By Rule 2.

As 113: 355:: 14:4311, Ans.
By Rule 3.

As 1 : 3.1416:: 14:43.9824, Ans.

2. Supposing the diameter of the earth to be 7958 miles, what is its circumference?

Ans. 25000.8528 miles.

3. What is the diameter of a circle whose circumference is 50 rods?

By Rule 1.

As 22:7:50: 15.9090, Ans.
By Rule 2.

As 355 113: 50: 15.9155, Ans.
By Rule 3.

As 3.1416: 1:50: 15.9156, Ans.

4. Supposing the circumference of the earth to be 25000 miles, what is its diameter? Ans. 7957 nearly.

310. To find the area of a circle.

RULE. Multiply half the circumference by half the diameter, or the square of the diameter by .7854,—or the square of the circumference by .07958,-the product will be the area.