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15. In $255.406 how 16. Change £240 159. N. inany pounds, shillings, E. to the several other curpence and farthings ?

rencies. ( $76 12s. 5d. N. E.

(£321 03. Od. N. Y.

£300 18ś. 9d. Penn. Ane £102 8s. 3d. N. Y.

Ans. £93. 15s, 6%d. Penn.

£200 12s. 6d. Can. (£63 17s. 011. Can.

($802.50 Fed. Mon.

TABLE
303. Of the most common gold and silver coins, containing their weigh,
Bineness, and intrinsic value in Federal Money.
Country. Names of coius. | Weight. Fineness. 1 Value.
GOLD COINS. Grs.

Dolls.
U. States. Eagle.

270

22 110.000 Half Eagle. 135.

22

5.000
Quarter Eagle. 67.5

2.50
England.
Guinea.
1 129.44

4.666
Half Guinea. 64.72

2.333
75. piece.
43.15

1.556
France.
Louis d'or (old) 125,51

4.440
Louis d'or (new) 117.66

4.171
199.25

7.051
Spain.
Pistole (old)
104.62

3.773
Pistole (new) 104.62

3.685 Germany. Ducat.

53.85

2.088
Austria.
Souvrain.
85.50

3.074
Portugal. Joanese.

221.40

7.981 N. Crusade.

15.57 21 01 0.551 SILVER COINS.

oz. pwt. U. States. Dollar.

416. 10 14

1.000 Half Dollar. 208.

10 14 10.500 Quarter Dollar. / 104.

10 14 0.250 41.6

10 14 0.100
England. Crown.

464.50 11 2 1.111
Half Crown. 232.25 11 2 0.556
Shilling.
92.90

0.222
France. Crown.

451.62 10 17% 1.06
5 franc piece. 386.18 10 16 0.898
Spain. Dollar (old) 418.47

11 0

10.991 Dollar (oew] 418.47

10 15 0.972 Germany. Rix dollar con.) 450.90

1.037 Florin (do.) 225.45

10 13

0.518 Rix dol.(conv.) 432.93

0.926 Florin (do.) 216.46

0.463 Portugal. New Crusade. 265.63 10 16 0.615 Holland. Ducatoon.

504.20 11 5 | 1.222 Gilder or flor. 162.70 10189 0.775 Rix dollar.

443.80

10 112! 1.009 Goldgiider, 301.90 | 8 53 10.602

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NOTE.-The current values of several of the above coins differ come. *hat from tbeir intrinsic value, as expressed in the table.

SECTION II.

MENSURATION.

I. 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 lie so many square incbes, square feet, square yards, or square rods, &c.

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

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

1. What is the area of al 3. What is the area of a square whose side is 5 feet ? rhombus, whose length is 12

rods, and perpendicular height 4?

Aos. 48 rods.
Aps. 25 fi.

i 4. What is the area of a.
rhomboid 24 inches long, and
8 wide ? Aas. 192 inches.

5. How many acres in a

rectangular piece of ground, 2. What is the area of a | 56 rods long, and 26 wide ? rectangle whose length is 9,1 56 X 26-:-160=95 Ans. and breadth 4 feet ? Ans. 36ft,

306. To find the area of a triangle. 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 togeth- . er, and take half the sum: from the half sum subtract each, side separately ; multiply the half sum and the three remalo.. ders continually together, and the square-root of the last prer duct will be the area of the triangle.

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 feet? 1 13, 14 and 15 feet ? 40 base.

13+14+15=42 15=1 perpend. height. and 42--2321=half sum.

21 21 21 200

13 14 15 and 21x6x7x8 40

[=7056.

rem.8 7 6 1 600 feet, Ans.

Then 7056=84 feet, Ans. 2. The base of a triangle is.

1 6.25 chains, and its height 5.20 I

4. The three sides of a tri

A mhart chains ; what is its area?

angle are 16, 11 and 10 rods ; Ans. 16.25 square chains.

what is the area ? i

Ans. 54.299 rods. 307. To find the area of a trapezoid. 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
chains and the other 12.25, long, and at one end 1 foot and
and the perpendicular distance | 3 inches, and at the other 11
between them is 15.4 chains ; 1 inches wide ?
what is the area?

Ans. 134 feet.
- 12.25
7.5

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 rods. 15.4

4. What is the area of a hall. 39500

32 feet long, and 22 feet wide 49375

at one end, and 20 at the oth- · 9875

| er?

Aps. 672 feet. 152.0750 sq. ch’s. Ans. 308. To find the area of a trapezium, or an irregular polygon.

Rule.—Divide it into triangles, and then find the area of i these triangles by, Art. 306, and add them together.

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309, 310. MENSURATION OF SUPERFICIES. 189

1. A trapezium is divided 1 2. What is the area of a into two triangles, by a diago. trapezium whose diagonal nal 42 rods long, and the per- is 1081 feet, and the perpendiculars let fall from the i pendiculars 564 and 60: opposite angles of the two triangles, are 18 rods and 16 rods,

feet? Ans. 63171 feet. what is the area of the trape 3. How many square yards zium ?

in a trapezium whose diagonal 42 42 336

| is 65 feet, aod the perpendicu98 378

lars let fall upon it 28 apu,

33.5 feet? 378 336 714 rods, Ans.

Ans. 2221! yus.

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

Rule 1. As 7 is to 22, so is the diameter to the circumference, and as 22 is to 7, £o 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 circuinference to the diameter.

RULE 3. A3 1 is to 3.1416, so is the diameter to the circunference, and as 3.1416 is to 1, so is the circumference to the diameter.

1. What is the circumfer. | 3. What is the diameter of ence of a circle whose diame. a circle whose circumferenoe ter is 14 feet?

Jis 50 rods ?
By Rule 1.

By Rile 1.
As 7: 22 :: 14:44, Ang. As 22:7:50: 15.9090, Ans.
By Rule 2.

By Rule 2. As 113: 355 :: 14:43111. As 355:113::50: 15 9155, Ing. By Rule 3.

By Rule 3. As 1:3.1416 :: 14:43 9824 | As 3.1416: 1 ::50:15.9156, A.

2. Supposing the diameter 1 4. Supposing the circumferof the earth to be 7958 miles, egce of the earth to be 25000 what is its circumference ? | miles, what is its diameter ? Ans. 25000.8528 miles.

Ans. 7957 nearly.

310. To find the area of a circle. RULE.-Multiply half the circumference by half the diame. ter, or the square of the diameter by .7834,-or the square of the circumference by .07958,-the product will be the area,

1. What is the area of a 1 3. What is che area of a circircle whose diameter is 7 cle whose diameter is 10 rods, and circumference 22 feet? and circumference 31.416 ? 11= circumference.

Ans. 78.54 rods. 3.5=1 diameter.

4. How many square chains 55

in a circular field, whose cir

cumference is 44 chains, and 33

diameter 14? Ans. 154 chains.

38.5 feet, Ans.

..5. How many square feet in 2. What is the area of a cir. ( a circle whose circumference cle whose diameter is l, and is 63 feet? circumference 3.1416 ?.

Ads. 315 feet. . Ans. .7854.

- 311. The area of a circle given to find the diameter and circumference.

RULE.-1. Divide the area by .7854, and the square root of the quotient will be the diameter.

2. Divide the area by .07958, and the square root of the quotient will be the circumference.

J. What is the diameter of a 3. demand the length of a circle whose area is 154 rods ? | rope to be tied to a horse's neck

that he may graze upon 7854 .7854)154.0000(196(14 rods. : square feet of pew feed every 7854 )

day, for'4 days, one end of the

rope being each day fastened 75465 2496

to the same stake? 70686 96

Ist circle contains 7854 - : feet--.7854=10000, and a 47740

| 10000=100 dianı. ;2=50 47124

I feet, the 1st rope. ad circle 616

| contains 15708-77854= 20

000, and .v 20000=1411 or 2. The area of a circle is ! 70% leet, second rope, &c. 78.5 feet; what is its circum

Ist rope 50 feet.

2 - 700 feet. ference ? Aos. 31.4 feet..

3 - 86% feel.

> Ans.

4 - 100 feet.) 312. To find the area of an oval, or ellipsis. RULE:-Multiply the longest and shortest diameters together, and the product by 7854; the last product will be the area.

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