VOTE8.-1. The sides multiplied must be of the same denomination, and the product will be square units of that denomination, which may be reduced, if necessary, to higher denominations. 2. In dividing, the linear unit of the side must be of the same name as the square unit of the area, and the quotient will be linear units of the tame denomination. WRITTEN EXERCISES. 1. How many square feet in a floor 32 ft. long by 21 ft. wide? how many square yards ? SOLUTION.—To find the area, we multiply the length by the breadth, and we have 32X21=672 sq. ft.; reducing this to square yards, we have 743 sq. yd. 2. How many square yards.in the surface of a blackboard 27° ft. long by 4 ft. wide ? Ans. 12 sq. yd. 3. How many square yards in a garden 215 ft. long by 109 ft. wide ? Ans. 26038 sq yd. 4. What is the width of a room 25 feet long, whose floor contains 500 sq. ft. ? Ans. 20 ft. 5. A rectangle contains 4661 sq. ft., and one side is 16 ft 6 in. long; how long is the other side ? Ans. 28 ft. 3 in. 6. How many square feet in the sides of a room 18 ft long, 14 ft. 6 in, wide, and 9 ft. 6 in. high? Ans. 617. 7. A certain box is 3 ft. 6 in. long, 2 ft. 3 in. wide, and 1 ft. 4 in. high; how many square feet in its surface ? Ans. 31 sq. ft. 12 sq. in. 8. What is the surface of a cubical box, each of whose dimensions is 1 ft. 6 in. ? Ans. 131 sq. ft. THE TRIANGLE. 338. A Triangle is a plane surface having three sides and three angles; as, ABC. B D 339. The Base is the side upon which it seems to stand, as AB. The Altitude is a line perpendicular to the base, drawn from the angle opposite; as, CD. 340. A triangle which has its three sides equal is called equilateral; when two sides are equal it is called isosceles ; when its sides are unequal it is called scalene. Rule I.-- To find the area of a triangle, multiply the base by one-half of the altitude. Rule II.- To find the base or altitude of a triangle, divide the area by one-half the other dimension. WRITTEN EXERCISES 1. What is the area of a triangle whose base is 25 inches and altitude 18 inches ? SOLUTION.—To find the area, we multiply the base by one-half the altitude; 25x9=225; hence the area is 225 sq. in. 2. How many square feet in a triangle whose base is 18 ft. 6 in. and altitude 9 ft. 9 in. ? Ans. 90 sq. ft. 27 sq. in. 3. What is the area of the gable end of a house 29 ft. wide the ridge being 12 ft. higher than the top of the wall? Ans. 174 sq. ft. 4. The area of a triangular bed of flowers is 25 sq. ft., and ita base 10 ft. ; what is the altitude ? Ans. 5 ft. 5. The area of a triangular lot is 250 square yards, and its base is 250 ft. ; what is its altitude ? Ans. 18 ft. 6. The area of the gable of a house is 378 sq. ft., the base being 14 yards; what is the height of the ridge ? Ans. 18 ft. THE CIRCLE. 341. A Circle is a plane figure bounded by a curved line, every point of which is equally distant from a point within, called the rentre. 342. The Circumference of a circle is the bounding line; any part of the circumference, as BC, is an Arc. An arc of one-fourth of the circumference is called a Quadrant. 343. The Diameter is a line passing through the centre and terminating in the circumference; as, AB. The Radius is a line drawn from the centre to the circumference; as, OD. Rule 1.– To find the circumference of a circle, multiply the diameter by 3.1416. Rule 11.-To find the diameter of a circle, multiply the circumference by .3183. Rule III.- To find the area of a circle, multiply the cir. cumference by one-fourth of the diameter, or multiply the square of the radius by 3.1416. WRITTEN EXERCISES. 1. The diameter of a circle is 121 feet; what is its cirJamference ? SOLUTION.–To find the circumference, we multiply the diameter by 3.1416; 3.1416x 12 equals 39.27; hence the circumference equals 39.27 ft. 2. What is the circumference of the planet Venus, its diameter being about 7800 miles ? Ans. 24504.48 miles. 3. The distance round a circular pond is 500 feet; what is the distance across the pond? Ans. 159.15 ft. 4. How many times will a carriage wheel 4 ft. 6 in. in circumference revolve in driving 10 miles ? Ans. 11733 5. I have a circular flower-bed 50 feet in circumference; what is the area of the bed ? Ans. 198 sq. ft. 135 sq. in. 6. A cow is fastened to a stake by a rope 16 feet long; wbat space can she graze over ? Ans. 89.36+ sq. yd. SUPPLEMENTARY PROBLEMS. To be omitted unless otherwise directed. 7. If the equatorial diameter of the earth is is 7925.75 miles, what are its circumference and the length of a degree of longitude at the equator ? Ans. 24899.536+ miles; 69.16+miles. 8. A circular flower-bed being divided into four equal parts by lines drawn from the centre, one section was planted with tulips; what was the area of the tulip-bed, its outer edge being 7 feet? Ans. 15.5967 sq. ft. MEASUREMENT OF LAND. 344. The Unit of Measure of land is the Acre, which is sometimes divided into square rods and sometimes into square chains. Hundredths of an acre are also frequently used. Government lands are divided by parallels and meridians into townships, which contain 36 square miles or sections, and each section is sul divided into quarter-sections. Hence, 640 acres make a section, and 160 acres a quarter-section. The quarter-sections are still further subdivided into half-quarter-sections, quarter-quarter-sections, and lots. Lots are often of irregular form on account of natural boundaries, but contain, as near as may be, a quarter-quarter-section. NOTE.—The pupil will remember that rods multiplied by rods give square rods, chains by chains give square chains; also, that 1 acre = 10 square chains or 160 square rods. WRITTEN EXERCISES. 1. How many square rods in a grass plat 65 ft. long and 15 ft. wide ? SOLUTION.—The area equals 65 x 15, or 975 sq. ft.; reducing to square rods, we have 3 lž sq. rd. 2. How many acres in a rectangular meadow 725 rods long and 400 rods wide ? Ans. 1812 A. 80 P. 3. What is the value of a farm 208.7 rods long and 120 rods wide, at $814 an acre ? Ans. $12795.913. 4. Mr. A bought 64 A. 116 P. of land for $3.50 per square rod, and sold it for $3.75 per square rod; what did he gain? Ans. $2589. 5. A rectangular pond is. 200 rd. 17 yd. long, and 150 rd. 15 yd. wide; required its area. Ans. 193 A. 137 P. 153 sq. yd. 6. I have a field 16.5 ch. long and 9.75 ch. wide; how much land does it contain ? Ans. 16 A. 14 P. 7. Mr. Wilson's farm contains 163 A. 3 ch., and its length is 71 ch. ; how many rods of fence would be required to surround it ? Ans. 752 rd. SUPPLEMENTARY PROBLEMS. To be omitted unless othermise directed. 8. If a township is equally divided among 480 families, how many acres does each family receive, and what part of a section does each receive? Ans. 48 acres; * of a section. 9. How many rails are required to fence a quarter-quarter-section, the fence being 5 rails high, and each rail 8 ft. long; and what will be the cost at $35 per thousand rails ? Ans. 3300 rails; $115.50. 10. A field 80 rods long contains 15 acres, while another field of the same width contains 9 acres; what is the length of the latter field ? Ans. 48 rods. NOTE.-In Ex. 9, the area is 6400 sq. rd., and each side is 80 rd. 11. How much less will it cost to fence a field 72 rods square than a rectangular field 3 times as long and f as wide, if fencing cost $2.50 a rod ? Ans. $480. 12. A mechanic having a lot of ground 50 rods square, planted 3 acres with corn, 200 square rods with vegetables, 15 rods square with flowers, and the remainder he kept to pasture his cow; how much of the lot was pasture ? Ans. 9 A. 155 P. COST OF ARTIFICERS' WORK. 345. By Artificers' Work we mean plastering, painting, papering, paving, stone-cutting, etc. 346. Plastering, painting, papering, paving, and ceiling are estimated by the square foot or square yard. Roofing, flooring, partitioning, slating, etc., generally by the square, wbich consists of 100 square feet, but sometimes by the square foot or yard. 347. Shingles, which commonly measure 18 in. by 4in, are estimated by the thousand or bundle. 1000 are generally allowed to a square of 100 sq. ft. WRITTEN EXERCISES. 1. What will be the expense of paving a sidewalk 303 ft. long and 7 ft. wide, at $2.25 per square yard? SOLUTION.—The area equals 303x73, or 22721 sq. ft., which equals 252} sq. yd.; hence the cost is $2.25X252), or $568.12). 2. What will it cost to plaster a school-room 40 ft. long, 20 ft. wide, and 10 ft. high, at $0.36 a square yard ? Ans. $80. 3. What is the cost of wainscoting a room 28 ft. long by 15 ft. 4 in. wide, to a height of 4 ft. 3 in. at $0.45 per square yard ? Ans. $18.413. 4. What is the cost of slating a roof 52 ft. 10 in. long, each side being 20 ft. wide, at $15.25 per square ? Ans. $322.285. 5. A frame house is 50 ft. long, 28 ft. wide, and 35 ft. high; what will be the expense of outside painting at $12.25 per square ? Ans. $668.85 |