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Suppose 40 ft. Again suppose 60 ft. 40 8

X His body 20

body 30 His tail 19

tail 24

60 S His head 9

head 9

8 40

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10 2. A gentleman has 2 horses 4. A and B lay out equal shares and a saddle worth $50; if the in trade ; A gains $126, and B saddle be put on the first horse, loses $87, then A's money is his value will be double that of double that of B; what did each the second ; but if it be put on 1 lay out?

Ans. $300. the second, his value will be triple that of the first ; what is the 5. A and B have both the same value of each horse ?

income ; A saves 1 of his yearly, Ans. 1st horse 830, 2d 840. but B, by spending $50 per an

num more than A, at the end of 4 3. A man having driven his | years, finds himself $100 in swine to market, viz. hogs, sows debt ; what is their income, and and pigs, received for them all / what do they spend per annum?

50, being paid for each hog 18s. Ans. $125 their inc. per ann. for each sow 16s. and for each pig 2s. ; there were as many hogs as. sows, and for every sow 3 pigs; what was the number of each sort ? Ans. 25 hogs, 25 sows, 75 pigs.

B spends $150 } per ann.

QUESTIONS. 1. What is Position ?

4. What is the rule ? 2. Of how many kinds is it?

5 What does Double Position teach? 3. What does Single Position teach? | 6. What is the rule ?

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5. Permutation of Quantities.

Permutation of Quantities is a rule which shows how many different ways the order or position of any given number of things may be varied.

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Problem I. To find the number of permutations, or changes, that can be made of any

given number of things, all different from each other. RULE.*-Multiply all the terms of the natural series of numbers from 1 up to the given number, continually together, and the last product will be the answer required.

Examples. 1. How many changes can be 3. How many changes may be made of the letters in the word | rung on 6 bells ? Ans. 720. and ?

4. How many changes can be 1

made in the position of the 8 notes 2 a dn

of musick ? Ans. 40320.

Proof. 2 nd a

5. How many changes may be 3

rung on 12 bells, and how long

would they be in ringing, suppos6 Ans, or 1X2 X3=6 Ans. ing 10 changes to be rung in one

minute, and the year to consist of 2. How many days can 7 per 365 days, 5 hours and 49 min’ts ? sons be placed in a different posi- Ans. 479001600 changes, and 91 tion at dinner? Ans. 5040 days. I years, 26d. 22h. 41m. time.

Problem II. To find how many changes can be made out of a given number of different

things, by taking any given number at a time. Rule.-Take a series of numbers, beginning at the number of things given, and decreasing by 1 till the number of terms taken be equal to the number of things to be taken at a time, and the product of all these terms will be the answer.

Examples. 1. How many changes can be 2. How many words can be - rung on s bells out of 8 ?

made with 5 letters of the alpha8X7 X6=336 Ans. bet, supposing 24 letters in all,

and that a number of consonants alone will make a word ?

Ans. 5100480. QUESTIONS. 1. What is Permutation of Quanti- | changes that can be made of any ties?

number of things all different from 2. How do you find the number of each other?

* The reason of the rule may be shown thus: any thing, a, is capable of only one position, as d. Any two things, a and b, are capable of only two variations, as ab, ba ; when a number is expressed by 1x2. If there be three things, a, b, and c, then any two of these, leaving out the third, will have 1 X 2 variations ; and consequently, when the third is taken in, these will be 1*2*3 variations ; and so on as far as you please.

136

SECTION III.

or

B

1. Mensurartion of Superficies.

Definitions. 1. A point is position

the rectangle, square, without magnitude.

rhomboid & rhombus.

A 2. A line is length

11. A rectangle is a

11 without breadth

parellelogram having c thickness.

all its angles right. 3. A superficies,or sur

12. A square is a face, is a figure having 4

figure having four elength and breadth 3 qual sides, and all its

12 without thickness.

3 jangles right. 4. An angle is the

13. A rhomboid is opening between two

A an oblique angled palines, having different

B

crallelogram. directions, and meet.

14. A rhombus is an ing in a point.

A

equal sided rhomboid. 5. A right angle is

15. A trapezium is a formed by one right 5

4 sided figure whicho line falling perpendi

C

D

has neither pair of its cularly upon another.

opposite sides parallel. 6. An oblique angle is

16. A trapezoid has formed by two oblique

16 A А one pair of its opposite A lines, and may be ei

sides parallel

c ther greater or less

17. A circle is a fig

B than a right angle.

ure bounded by a con7. A triangle is a fig

7

tinued curve line, calure having 3 sides A

led the circumference, and 3 angles.

every part of which is 4 8. A right angled A 'B equally distant from triangle has one right

a point within called

D angle.

the centre. 9. An oblique angled

18. The radius of a

BA triangle has all its 4

circle is a right line angles oblique.

from the centre to 10. A parallelogram

the circumference. is a four sided figure

19. The diameter of a which has both pair of

circle is a right line its opposite sides par

extending thro' the allel, of which there

centre, & terminated are 4 varieties, viz.

by the circumference.

B

The area of any figure is the space contained within the bounds of its surface, without any regard to thickness, and is estimated by the number of squares con tained 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 squ inches, square feet, square yards, or equare rods, &c.

Problem I.

To find the area of a parallelogram, 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.

Examples. 1. What is the area of a square 3. What is the area of a rhomwhose side is 5 feet ? 5

perpendicular height 4 ? 5

Ans. 18 rods.

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Ans. 25 feet.

4. What is the area of a rhom5

boid 24 inches long, and 8 wide?

Ans. 192 inches. 5

5. How many acres in a rec2. What is the area of a rec- ! tangular piece of ground 56 rods tangle whose length is 9, and long, and 26 wide? breadth 4 feet?: Ans. 36 feet. 56 x 26-160=96 acres, Ans.

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Problem IT.

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 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.

Examples. 1. How many square feet in a

3. What is the area of a triantriangle whose base is 40 feet, gle whose three sides are 13, 14 and height 30 feet?

and 15 feet ? 40 base.

13+14+15=42 15=1 perpendicular height. and 42-2=21=sum.

21 21 21 200

13 14 15 and 21 X 6 X 7 X 8 40

[=7056

Rem. 8 6 600 feet. Ans.

Then 7056) —84 feet, Ans. 2. The base of a triangle is 6.25 chains, and its height 5:20

4. The three sides of a triangle chains; what is its area? are 16, 11 and 10 rods; what is Ans. 16.25 5 sq.

chains. the area ? Ans. 54.299 rods.

; Problem III.

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.

Examples. 1. One of the two parallel 2. How many square feet in a sides of a trapezoid is 7.5 chains plank 12 feet 6 inches long, and and the other 12.25, and the at one end, 1 foot 3 inches, and perpendicular distance between at the other 11 inches wire ? them is 15.4 chains; what is the

Ans. 1324 feet. area?

12.25
7.5

3. What is the area of a piece

of land 30 rods long and 20 rods 2)19.75

wide, at one end, and 18 rods at 9.875

the other? Ans. 570 rods. 15.4

39500 49375 9875

4. What is the area of a hall 32 feet long, and 22 feet wide at one end, and 20 at the other?

Ans. 672 feet.

152.0750 sq. chains, Ans.

Problem IV.

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 Problem 11. and add them together.

Examples. 1. A trapezium is divided into 2. What is the area of a trapetwo triangles, by a diagonal 42 zium whose diagonal is 1084 feet, rods long, and the perpendiculars and the perpendiculars 564 and let fall from the opposite angles | 608 feet? Ans. 6347 feet. of the two triangles, are 18 rods and 16 rods; what is the area of the trapezium?

3. How many square yards in 42 42 336

a trapezium whose diagonal is 9 8 378

65 feet, and the perpendiculars

let fall upon.it 28 and 33.5 feet P. 378 336 704 rods, Ans.

Ans. 2227 yds.

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