Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[ocr errors]

3. How many acres are in a field of oblong form, whose length is 14,5 chains, and breadth 9,75 chains ?

Aps. 14ac. Oroo. 22rods. NOTE.—The Gunter's chain is 66 feet, or 4 rods, long, and contains 100 links. Therefore, if dimensions be given in chains and decimals, point off from the product one more decimal place than are contained in both factors and it will be acres and decimals of an acre; if in chains and links, do the same, because links are hundredths of chains, and, therefore, the same as decimals of them. Or, as 1 chain wide, and 10 chains long, or 10 square chains, or 100000 square links, make an acre, it is the same as if you divide the links in the area by 100000. 4. If a board or plank be 21 feet long, and 18 inches

a broad, how many square feet are contained in it?

18 inches =1,5 foot. And 21 x1,5=31,5 feet. Ans.

Or, in measuring boards, you may multiply the length in feet by the breadth in inches, and divide the product by 12; the quotient will give the answer in square feet, &c.

21 x 18 Thus, in the preceding example, =313 sq. feet as before.

12 5. If a board be 8 inches wide, how much in length will make a foot square ?

RULE.—Divide 144 by the width; thus, 8)144

Ans. 18 inch. 6. If a piece of land be 5 rods wide, how many rods in length will make an acre ?

RULE.--Divide 160 by the width, and the quotient will be the length required; thus,

5) 160

[ocr errors]

32 rods in length. Ans. Note.—When a board, or any other surface, is wider at one end than the other, but yet is of a true taper, you may take the breadth in the middle, or add the width of both ends together, and halve the sum, for the mean width : then multiply the said mean breadth in either case, by the length; the product is the answer, or area sought.

7. How many square feet in a board 10 feet long, and 13 inches wide at one end, and 9 inches wide at the other ?

13+9

11 inches mean width.
2

ft. in.
10x11

-=91ft. Ans.

12 8. How many acres are in a lot of land which is 40 rods long, and 30 rods wide at one end, and 20 rods wide at the other ? 30+20

=25 rods mean width. 2 Then, 25 x 40

-=64 acres. Ans.

160 9. If a farm lie 250 rods on the road, and, at one end, be 75 rods wide, and, at the other, 55 rods wide, how many acres does it contain ? Ans. 101ac. 2roo. 10ro.

Case 3.-To measure the surface of a triangle. Definition.-A triangle is any three cornered figure which is bounded by three right lines."

Rule.-Multiply the base of the given triangle into half its perpendicular height, or half the base into the whole perpendicular, and the product will be the area.

EXAMPLES. 1. Required the area of a triangle whose base or longest side is 32 inches, and the perpendicular height 14 inches. 142-7=} the perpend. and 32x7=224 sq. in. Ans.

2. There is a triangular or three cornered lot of land, whose base or longest side is 51} rods; the perpendicular, from the corner opposite to the base, measures 44 rods; how many acres does it contain ? 44+2 22-half perpendicular. And 51,5 x 22

=7 acres, 13 rods. Ans.

160 * A triangle may be either right-angled or oblique ; in either

the teacher can easily give the scholar a just idea of the base and perpendicular, by marking it down on slate or paper, &c. In a right-angled triangle, the longest of the two legs which include the right-angle, is called the base; but in such as are oblique, the longest of the three sides is so called.

W

case,

[ocr errors]

3. If a piece of land lie in the form of a right-angled triangle, its base being 37 rods, and the perpendicular line being 24 rods, how many acres are in it?

Ans. 2,3617+ acres. 4. If the base of a triangular field be 7 chains and 50 links, and the perpendicular 4 chains and 25 links, how much does it contain ?

Ans. lac. 2roo. 15rods. Joists and Planks are measured by the following Rule. Find the area of one side of the joist or plank, by one of the preceding rules; then multiply it by the thickness in inches; and the last product will be the superficial content.

EXAMPLES. 1. What is the area, or superficial content, or board measure, of a joist, 20 feet long, 4 inches wide, and 3 inches thick?

20X4

-X3=20 feet. Ans.

12 2. If a plank be 32 feet long, 17in. wide, and 3in. thick, what is the board measure of it ? Ans. 136 feet.

NOTE.-There are some numbers, the sum of whose squares makes a perfect square ; such are 3 and 4, the sum of whose squares is 25, the square root of which is 5; consequently, when one leg of a right-angled triangle is 3, and the other 4, the hypotenuse must be 5. And if 3, 4, and 6, be multiplied by any other pumbers each by the same, the products will be sides of true right-angled triangles. Multiplying them by 2, gives 6, 8, and 10-by 3, gives 9, 12 and 15—by 4, gives 12, 16 and 20, &c. ; all which are sides of right-angled triangles. Hence architects in setting off the corners of buildings, commonly measure 6 feet on one side, and 8 feet on the other; then, laying a 10 feet pole across from those two points, it makes the corner a true rightangle. RULE 2.-To find the area of any triangle when the three

sides only are given. RULE.—From half the sum of the three sides subtract each side severally; multiply these three remainders and the said half sum continually together; then the square root of the last product will be the area of the triangle.

EXAMPLE. Suppose I have a triangular fish-pond, whose three sides measure 400, 318, and 312yds. ; what quantity of ground does it cover ?

Ans. 10 acres, 3roods, 8+rods. S

a

CASB 4.-To measure irregular surfaces. RULE.-Divide the figure or plane into triangles, by drawing diagonal lines from one angle to another; then measure all the triangles, by either of the rules in Case 3; and the sum of their several areas will be the area of the given figure.

EXAMPLE. If a piece of ground be divided into two triangles by a diagonal line drawn through it measuring 30 rods, and two perpendiculars be let fall, one measuring 8 rods, and the other 14 rods; how many acres does it contain ?

Ans. 215 acres.

CASE 5.- To measure a circle. Definition.--A circle is a figure bounded by a curve or circular line, every part of which is equally distant from the middle or centre. The curre line is called the periphery or circumference ; a line drawn, from one side to the other, through the centre, is called the diameter ; and a line drawn, from the centre to the circumference, is called the semidiameter, (half diameter,) or radius.

PROBLEM I. The diameter given to find the circumference. RULE.-As 7 are to 22, so is the given diameter to the circumference ; or, more exactly, as 113 are to 355, so is the diameter to the circumference, &c.

EXAMPLES. 1. What is the circumference of a wheel whose diameter is 4 feet ?

As 7 : 22 ::4:12,57 +feet the circumference. Ans.

2. What is the circumference of a circle whose diameter is 35 rods ? As 7:22 :: 35 : 110 rods. Ans.

NOTE.-To find the diameter, when the circumference is given, reverse the foregoing rule, and say, as 22 are to 7, so is the given circumference to the required diameter; or, as 355 are to 113, so is the circumference to the diameter.

3. What is the diameter of a circle whose circumference is 110 rods?

As 22:7::110: 35 rods the diam. Ano.

PROBLEM. II.

To find the area of a circle. RULE.—Multiply half the diameter by half the circumference, and the product is the area; or, if the diameter alone is given, multiply the square of the diameter by ,785398 or, which is near enough, by ,7854—and the product will be the area.

EXAMPLES. 1. Required the area, or superficial content of a circle whose diameter is 12 rods, and circumference 37,7 rods.

18,85=half the circumference.

6=half the diameter.

а

113,10 area in square rods. Ans. 2. What is the superficial content of a circular garden whose diameter is 11 rods ? By the second method.

11x11=121. ,7854x121=95,0334 rods. Ans. 3. What will be the cost of a circular platform to the curb of a round well, at 104 cents per square foot; if the diameter of the well be 42 inches, and the breadth of the platform be 144 inches?

Ans. $1,87cts. +

[ocr errors]

PROBLEM III. To find the area of a circle when the circumference alone is given.

Rule.—Multiply the square of the circumference by ,079577, or, which is near enough, by ,07958—and the product will be the area. Note.-,

.-,785398 is the area of a circle whose diameter is 1, and ,079577 is the area of a circle whose circumference is 1.

EXAMPLE. What is the area of a circle whose circumference is 30 rods? 30x30x,07958=71,62200 rods. Ans.

PROBLEM IV. The area of a circle given to find the diameter. RULE.-Divide the area by ,7954-and the square root of the quotient is the required diameter.

« ΠροηγούμενηΣυνέχεια »