« PreviousContinue »
USEFUL PROBLEMS IN THE MENSURATION OF
SUPERFICES AND SOLIDS.
SECTION 1.-SUPERFICIES. The area of every plain surface is conceived to be made up of a certain number of squares, either greater or less, according to the measure by which the dimensions are taken, which is generally in inches, feet, yards, rods, &c. A square inch means a space an inch long and an inch broad, in which depth or thickness is not considered ; and so of square feet, yards, rods, &c. And the superficial contents, or area of any plain surface, is the number of square inches, feet, yards, rods, acres, &c., which it contains. PROBLEM 1.-To find the area of a Square.
RULE. Multiply the side of the square into itself, and the product will be the area, or contents.
EXAMPLES. 1. How many square feet of boards are contained in the floor of a room which is 16 feet square ?
Ans. 256, 2. How many acres are contained in a square field which measures 65 rods on each side ? [Reduce the whole number of square rods to acres.]
Ans. 26 acres 1 r. 25 rods. PROB. II.- To find the area of a parallelogram, or long square.
RULE. Multiply the length by the breadth, and the product will be the area.
1. How many square yards of ground are contained in a garden which is 126 feet long and 65 feet wide? [9 sq. feet=l sq. yard, therefore : the sq. feet by 9.]
Ans. 910 sq. yds. 2. How many acres are contained in a lot of land in the form of a long square, which is fifty-six rods in length and thirty-seven rods in width ?
Ans. 12 acres 3 r. 32 rods. 3. How many feet in a board or plank 18 feet long, and 1 foot 6 inches wide ?
By duodecimals, 18F O'X1F 6'=27ft. 0', Ans.
By decimals, 1ft. 6in.=1,5ft. Then 18X1,5=27, Ans. Or, multiply the length in feet by the breadth in inches, and divide the product by 12. Thus, 18ft.x18in.--12=27ft. Ans. as before.
4. How many square feet are contained in a piece of board 126 inches long and 16 inches wide ?
Note.-144 square inches, =12x12, make 1 square foot; therefore divide the inches by 144.
Ans. 14 sq. feet. 5. How much length, that is 9 inches wide, will make a square foot ? 1 sq. ft.=144 sq. in. ; therefore 144:9=16 inches, Ans.
6. How many rods in length must a piece of land be, which is 5 rods wide, to make an acre ?
1 acre=160 square rods, then 160-5=32 rods, Ans. 7. There is a piece of land in the form of a long square, which con. tains 6 acres. Its length is 120 rods ; I demand its width ?
6a.=960 sq. rods. Then 960---120=8 rods, Answer.
Prob. III.-To measure a Triangle.
Base D. A right-angled triangle is that which has one right angle, that is, square corner, as the angle A, Fig. 1; in which the side A B is called the base, the side A C the perpendicular, and the side B C the hypo. thenuse. Note. Both the base and perpendicular of a right-angled triangle are sometimes called the legs. ART. 1.- To find the area of a Triangle, (either right.
angled or oblique.)
RULE. Multiply the base by half the perpendicular, or multiply the perpen dicular by half the base, and the product will be the area.
Or, multiply the whole base by the whole perpendicular, and one half the product will be the area.
1. What is the area or contents of a triangle whose base is 36 feet, and perpendicular 24 feet ?
36X12=432 sq. feet, Ans. 2. In a triangular lot of land, whose base measures 514 rods and perpendicular 48 rods, how many acres ? Ans. 7a, 2r. 36 rods.
3. What is the area of a triangular field which measures on the base 65 rods, and the perpendicular from the corner opposite the base, to the base, is 27 rods 3
65 x13,5=877,5r.=5a. Ir. 37} rods.
ART. 2.-In every right-angled triangle, the square of the hypothenuse is equal to the sum of the squares of the other two sides. 1. Hence, when the legs are given, to find the hypothenuse.
RULE. Add the squares of the two legs together, and extract the square root of their sum.
2. When the hypothenuse and one leg is given, to find the other leg.
From the square of the hypothenuse subtract the square of the given leg, and the square root of the remainder will be the other leg.
1. There is an edifice whose height is 50 feet, and the width of the street running by it is 36 feet. What is the length of a ladder that will reach from the opposite side of the street to the top of the edifice ?
502=2500, and 362=1296.
Then V2500+1296 Ans. 61,6ft. + 2. Suppose the foot of a ladder, which is 32 feet in length, being placed on a level 16 feet from the bottom of a building, will just reach the top of the same; what is the height of the building ? (In this example, the length of the ladder is the hypothenuse.)
v32X32—16X16=27,7ft.+ Ans. 3. Two ships sail from the same port, one due east 60 miles, and the other due south 40 miles ; how far are they apart ? YIn this example, the legs are given to find the hypothenuse.)
Answer, 72,1+iniles. PROB. IV.-To measure Circles. Art. 1. The diameter of a circle given, to find the circumference.
Note.-The diameter is a right line drawn across the centre of a cir'cle, dividing it into two equal parts.
RULE. As 7 is to 22, or more accurately, as 1 is to 3,14159, so is the diam. eter to the circumference. Art. 2. The circumference given, to find the diameter.
RULE. As 22 is to 7, or as 3,14159 is to 1, so is the circumference to the diameter.
EXAMPLES. 1. What is the circumference of a wheel or circle whose diameter is 5 feet?
7:22 :: 5 .... Ans. 15,7+feet. 2. What is the diameter of a circle whose circumference is 110 feet?
22: 7 :: 110 .... Ans. 35. ART. 3. To find the area of a circle.
RULE. Multiply half the diameter by half the circumference, and the pro duct will be the area. Or, multiply the square of the diameter by ,7854, and the product will be the area.
1. Required the area of a circle whose diameter is 21 inches, and circumference 66 inches ? circumference 66=33 inches, and 1 diameter 21=10,5 inches ;
Then, 33x10,5=346,5. Ans. 346,5 inches. 2. What is the area or contents of a circle whose diameter is 20 rods? (By the second method.] 20x20=400, square of the di: ameter. Then 400 X,7854=314,16 rods 1 acre 3r. 34,16 rods.
3. If the area of a circle be 78,54 sq. feet, what is the diameter ?
Note.—The area, divided by ,7854, will show the square of the di. ameter ; then the square root of that will be the diameter required. Thus, 78,54:-,7854=V100=10 feet, diameter.
Art. 4. To find the area of a globe or ball. The area of the surface of a globe or ball is 4 times as much as the area of a circle of the same diameter.
Therefore, multiply the whole circumference into the whole diame. ter, and the product will be the area.
1. How many square inches of paper will it take to cover à globe that is 12 inches in diameter ?
7:22 :: 12 : the circumference, 37,7+ inches. Then 37,7x12 =452,4 sq. in., and 452,4 -144=3,14 sq. feet.
SECTION II. SOLIDS. A solid body is that which has length, breadth, and thickness. Sol. ids are generally estimated by the solid inch, solid foot, &c.
PROB. I.—To find the solidity of a Cube. A cube is a solid having six equal sides, each of which is an exact square.
RULE. Multiply the side by itself, and that product by the same side, gives the solid contents.
1. How many solid inches are contained in a cube 12 inches (=1 foot) long, 12 inches wide, and 12 inches thick ?
12x12x12=1728 solid inches, which are equal to 1 solid foot. 2. Suppose a cellar to be dug which shall be 11 feet 6 inches every way, in length, breadth and depth ; how many solid feet of earth must be thrown out of the same ?
11ft. 6in.=11,5ft. then 11,5X11,5X11,5=1520,875 solid ft. Ans. PROB. II.—To find the contents of any regular solid of three dimensions, length, breadth and thickness, as a piece of timber squared, whose length is more than its breadth and depth.
RULE. Multiply the length, breadth and thickness continually together, and the product will be the contents.
EXAMPLES. 1. If a square piece of timber be 1 foot 6 inches broad and 9 inches thick, and 8 feet long, how many solid feet does it contain ? 1. By decimals.
2. By duodecimals.
breadth =1 6
1 l' 6"
length X8 0
Ans. 9 0 0
Ans. 9,000=9 solid ft. Or, 1ft. 6in.=18in., and 8ft.=96in.; then 18in.X9in.X96in,= 15552 solid inches; and 15552:-1728=9 solid feet, Answer.
Note.—The breadth in inches, muliplied by the depth in inches, and that product multiplied by the length in feet, and the last product di. vided by 144, will give the solid contents in feet, &c.
2. What are the solid contents of a stick of timber which is 22 feet long, 15 inches broad, and 9 inches thick ?
in. in. ft.
Thus, 15X9X22=2970:-144=20,625ft., Ans. 3. How many solid feet are contained in a piece of timber 14in. broad, 9 inches thick, and 8 feet long ?
Ans. 7 feet. PROB. III.—The breadth and thickness of a piece of tim. ber given in inches, to find how much in length will make a solid foot.
RULE. Divide 1728, (the number of inches in a solid foot,) by the product of the breadth and depth, the quotient will be the length making a solid foot
1. If a piece of timber be 14 inches broad and 9 inches deep, how much length will make a solid foot ?
Thus, 14X9=126, and 1728:-126=13,7+ inches, Ans. 2. If a piece of timber be 18 inches broad and 14 inches deep, how much length will make a solid foot ?
18X14=252; then 1728-_252=6,85+ inches, Ans.
PROB. 19.–To measure a Cylinder. A cylinder is a round body, whose bases are circles like a round col. umn or stick of timber, of equal bigness from end to end.
RULE. Multiply the square of the diameter of the end by ,7854, which gives the area of the base ; then multiply the area of the base by the length, and the product will be the solid contents. Or,