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300. In £16 7s. 8d. 2gr., how many dollars, cents and mills ?
Before dividing the pounds, as above, 7s. 8d. 2qr., must be reduced to a decimal of a pound, and annexed to £16. This may be done by Art. 143, or by inspection, thus, shillings being 20ths of a pound, every 23. will be 1 tenth of a pound: therefore write half the even number of shillings for the tenths £0.3. One shilling being 1 20th=£0.05; hence, for the odd shilling we write £0.05. Fart!ıings are 960ths of a pound, and if 960ths be increased by their 24th part, they are 1000th& Hence 8d. 2qr.(=34qr.+1=£0.035; and 16+0.370.05+0.035 =£16.385, which, divided as in the preceding example, give for English currency, $72.822, Can. $65.54, N. %. $40.962, &c. Hence,
30]. To change poun:ls, shillings, pence and farthings to Federal Money, and the reverse.
RULE. Reduce the shillings, &c. to the decimal of a pound; then, if it is English currency, divide by 0.225; if Canada, by
• 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 4, N. E. to. N. Y.. add 4, N. Y. to N. E. subtract 1, N. E. to Penn. add +, Penn. Co N. E. subtract , N. Y. to Penn. subtract fo, Penn. to N. Y. add th, N. E. to Can. subtract , Can. 'to N. E. add 4, &c.
EXCHANGE OF CURRENCIES.
TABLE 803. Of the most common gold and silver coins, containing their weight fineness, and intrinsic value in Federal Money.
1.000 0.500 0.250 0.100 1.111 0.655
92.90 451.62 386.18 418.47 418.47 450.90 225.45 432.93 216.46 265.68 504.20 162.70 443.80 301.90
Germany. a 16
Not The current values of several of the above cnins differ somewhat from ther wu..usic value, as expressed in the table
1. Mensüration of Superficies. 304. The area of a figure is the space contained within the bounds of Ho surfaee, without any regard to thiekness, and is estimated by the namoer 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 perpendiculas height, and the product will be the area.
1. What is the area of a 3. What is the area of a square wbove side is 5 feet? rhombus, whose length is 12
rods, and perpendicular height
Ans. 48 rocks.
Ans. 25 ft. MI
4. What is the area of a rhomboid 24 inches long, and 8 wide ?
Ans. 192 inches.
5. How many acres in 2. What is the area of a rectangular piece of ground, rectangle, whose length is 9, 56rods long, and 26 wide ? and breadth ift.? Ans. 36ft. 56X26: 160—9 . Ana . .. 306. To find the area of a triangle. (64) .
Rule 1.-Multiply the base by half the perpendicular height, apd t'ne 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 beparately ; multiply the half sum and the three remainden continually together, and the square root of the last product will be the area of the triangle.
301, 3u8. MENSURATION OF SUPERFICIES. 133
1. How many square feet 3. What is the area of a in a triangle, whose base is 40 triangle, whose three sides are feet, and height 30 fect? 13, 14 and 15 feet? 40 base.
13+14+1542 1535 perpend. height
and 42-221-half sum 21 21 21
13 14 15 and 21 X6X7X 40
rem. 8 7 6 600 feet. Ans.
Then 705612—84 feet, Ana. 2. The base of a triangle is 4. The three sides of a tri6.25 chains, and its height 5.20 angle are 16, 11 and 10 rode; chains; what is its area ? what is the area ? Ans. 16.25 square chains. I
Ans. 54.299 roda.
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.
1. One of the two parallel | 2. How many square feet sides of a trapezoid is 7.5 in a plank 12 feet 6 inches chairs, and the other 12.25, long, and at one end, 1 foot and the perpendicular distance and 3 inches, and, at the othen between them is 154 chains; 11 inches wide ? what is the area?
Ans. 1347 feet. 12.25 75
3. What is the area of a
piece of land 30 rods long. 2)16.75
and 20 rods wide at one end
and 18 rods at the other ? 9.875
Ans. 570 rodei 15.4
4. What is the area of a 39500
hall 32 feet long, and 22 feet 49375
wide at one end, and 20 at the 9875
other? Ans. 672 feet. 152.0750 st. chains. Ans.
808. To find the area of a trapezium, or an irregular polygon
RULL.Divide it into triangles, and then find the area al these triangles by Art. 306, and add them together.