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EXAMPLES. 1. Required the diameter of a circle that will contain within its circumference the quantity of an acre of land. 1 acre=4840sq. yds. Then 1,4841=88,5+yds. Ans. 2. In the midst of a meadow abounding with feed,

For two acres, to tether my borse, I've agreed ;
How long must the rope be, that, feeding all round,
He

graze less or more than the two acres of ground?

Ans. 551 +yards. 3. A, B, and C, join to buy a grindstone, 36 inches in diameter, which cost $34, and towards which A paid $1}, B, $14, and C, 83}cts. The waste hole for the spindle

$ was 5 inches square. To what diameter ought the stone to be worn, when B and C begin severally to work with it allowing for the hole, and A first grinding down his share, next B, and then C?

29,324+inch. diameter where B begins to Ans. { "grind; and 19,013+in. diam. C begins. Note.-Twice the square of the side of a square, will bet he square of the diameter of its circumscribing circle.

PROBLEM V. The area of a circle given to find the circumference. RULE.-Divide the given area by ,07958-and the square root of the quotient is the required circumference.

EXAMPLES. 1. The expense of turfing a round plot, at 4 pence per square yard, was £2. gs. 9fd. ,8; what was its circumference!

Ans. 130+feet. 2. How many feet of boards will fence a round garden, containing just two acres, the fence five feet high; and what will be the expense at 64 mills per square foot ? Ans. 52314 +ft. boards ; and cost $32,69cts. 64m.t CASE 6.- To measure a sector of a

a circle. Definition.---A sector is a part of a circle, contained between an arch line and two radii, or semidiameters of the circle.

Rule. Find the length of the arch by saying, as 180 degrees are to the number of degrees in the arch, so is the radius, multiplied by 3,1416, to the length of the arch, which length, divided by 2, and multiplied by the radius, will become the required area,

EXAMPLE. Wbat is the area of the sector of a circle whose radius is 25 feet, its arch containing 125° ?* As 180° : 125° :: 25x3,1416 : 54,5416+ft. length of arch.

54,5416 Then,

-X25681,77ft. Ans.

2 RULE 2.-Find the area of a circle having the same radius; then say, as 360 degrees, (the number of degrees into which all circles are divided,] are to the area of the said circle, so is the number of degrees in the arch of the sector, to the area required.

EXAMPLE. Required the area of a sector of a circle whose arch contains 65 degrees, and radius 35 feet. Ans. 695368q. ft.

Case 7.To find the area of a segment of a circle. Definition.—A segment of a circle is any part of a circle cut off by a right line drawn across the circle, which does not pass through the centre, and is always greater or less than a semicircle.

Rule.-Find the area of the sector having the same arch as the segment, by Case 6; find also the area of the triangle formed by the chord of the segment and the radii of the sector, by Case 3; subtract the area of the latter from that of the former, and the remainder will be the area of the segment, when the segment is less than a semicircle: but the sum of the two areas is the answer, when it is greater.

EXAMPLE. What is the area of the segment, whose arcli contains 55° ; its chord 12,5 rods; the perpendicular of its triangle 16 rods; and its semidiameter 17,2 rods? First, find the area of a circle whose diameter is 34,4 rods As 7 : 22 :: 34,4 : 108,1+rods circumference. 108,1 34,4 Х

=929,66 area of the circle. 2

2 Then, as 360° : 929,66 ::55° : 142 +area of the sector.

* As we have not been able to obtain engravings to represent any of the figures in the preceding or subsequent Examples, the Teacher will, we trust, be so good as to draw them, for the Pupil, on paper or a slate.

a

And the chord=12,5 : the perpend. 16.
12,5x16

- 100 rods, area of the triangle.
2
Area of the sector=142
Area of the triangle=100

Area of the segment=42 rods. Ans. Note.-A regular polygon, whose sides and angles are all equal, may be measured by dividing it into triangles, finding the area of one, and multiplying this area by the number of triangles contained in the polygon. Case 8.—To describe, and find the area of an ellipse or

oval. RULE.-To deseribe an ellipse or oval, draw a line, set one foot of the dividers on the line, as a centre, and describe a circle ; move the dividers to some other point on the same line, (but not so far but that the dividers in forming a second circle may extend within the first, Jand describe a second circle of the same radius as the former; then, in the two points wbere the circles intersect, set the dividers to complete the sides of the oval; and through these intersecting points draw the line called the conjugate diameter, crossing the line first drawn called the transverse diameter, in the centre of the oval.

RULE.-To find the area of an ellipse, multiply the transverse, or longest diameter, by the conjugate, or shortest diameter, and their product by 97854 ; and the kast product is the area required.

EXAMPLE. If the transverse diameter of as oval fish pond be 34 rods, and the conjugate diameter be 24 rods, what is its area ?

34 x 24x,7854=640,8864 rods. Ans. CASE 9.-To find the area of a globe or sphere. Definition.-A sphere or globe is a round solid body, in the middle or centre of which is an imaginary point, from which every part of the surface is equally distant. An apple, or a ball used by children in some of their pastimes, may be called a sphere or globe.

RULE.Multiply the circumference by the diameter, and the product will be the area, or surface.

EXAMPLES. 1. What is the superficial content of the earth, if it be 360 degrees in circumference, and every degree measure 694 miles ? 360x694=25020 circumf. 355 : 113 : : 25020 : 7964+

diameter. And 25020x7964–199259280 area in squa. miles. Ang.

2. If the moon's diameter be 2180 miles, what is her area ?

Ans. 14928640+square miles. SECTION II.-OF SOLIDS. Solids are measured by the solid inch, foot, or yard, &c. 1728 of these inches, that is 12x12x12, make 1 cubic or solid foot.

CASE I.-To measure a Cube. Definition.--A cube is a solid of six equal sides, each of wbich is an exact square.

RULE.—Multiply the side by itself, and that product by the same side, and this last product will be the solid content of the cube.

EXAMPLES. 1. If the side of the cubie block be 18 inches, or 1 foot and 6 inches, how many solid feet does it contain ? Ift. 6in.=1,5ft. and 1,5x1,5x1,5=3,375 solid ft. Ans.

in. in. in. Or, 18x18x18

=3,375 as before.

1728 2. Suppose a cellar is to be dug which shall contain 12 feet every way, in length, breadth, and depth ; how many solid feet of earth must be taken out to complete it.

Ans. 1728 sol. ft.

Case 2.-To find the content of any regular solid, of

three dimensions, length, breadth, and thickness, such as a piece of square timber, whose length is more than its breadth and depth.

RULE.—Multiply the breadth by the depth or thick ness, and that product by the length; the last product is the solid content.

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EXAMPLES. 1. How nany solid feet are there in a piece of square timber that is i foot and 6 inches, or 18inches broad, 9 inches thick, and 9 feet, or 108 inches long?

1ft. 6in.=1,5 foot. ,75x1,5x9=10,125 sol. ft. Ans. 9 inches=,75 foot.

in. in. in. Or, 18x9x108

=10,125 as before. 1728 In measuring timber, however, you may multiply the breadth in inches by the depth in inches, and that product by the length in feet : divide this last product by 144, and the quotient will be the solid content in feet, &c.

2. How many solid feet does a piece of square timber, or a block of marble, contain, if it be 16 inches broad, 11 inches thick, and 20 feet long? 16x11x20=3520, and 3520: 144=24,4+sol. ft. Ans.

3. If a stick of square timber be 15 inches broad, 8 inches thick, and 25 feet long, how many solid feet are in it ?

Ans. 20,8+ feet. CASE 3.- When the breadth and thickness of a piece of

square timber are given in inches, to find how much in length will make a solid foot.

RULE.--Divide 1728 by the product of the breadth and depth, and the quotient will be the length, making a solid foot.

EXAMPLES 1. In a piece of square timber 11 inches broad and 8 inches deep, what length will make a solid foot ?

11x8=88)172(19,6+inches. Ans. 2. In a piece of square timber 18 inches broad and 14 inches deep, what length will make a solid foot ?

Ans. 6,8+inches. CASE 4.- To measure a cylinder. Definition.--A cylinder is a round body whose bases or ends are circles, like a round column or stick of timber, of equal bigness from end to end.

RULE.-Multiply the square of the diameter of the base or end by ,7854, which will give the area of the base ; then multiply the area of the base by the length, and the product will be the solid contenta

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