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Denominate Numbers -- Surface Measure. 334. To determine the amount of carpet necessary for a given room several minor problems must be solved which can be best studied by means of an
1. How many yards of carpet must be purchased for a room 16 ft. by 20 ft. if the carpet is 1 yd. wide ?
(1) How many breadths will be necessary if the carpet is put down lengthwise of the room? How much must be cut off or turned under from one breadth in this case ?
(2) How many breadths will be necessary if the carpet is put down crosswise of the room? How much must be cut off or turned under from one breadth in this case ?
(3) Make two diagrams of the room on a scale of 1 inch to the foot and show the breadths of carpet in each case ?
(4) How many yards must be purchased in each case?
(5) If in the first case there is no waste in matching the figure and in the second case there is a waste of 8 inches on each breadth except the first,* which plan of putting down the carpet will require the greater number of yards?
(6) If the carpet costs 90¢ a yard and the conditions are as stated in No. 5, what is the cost of the carpet in each case ?
2. How many yards of carpet must be purchased for a room 16 ft. by 20 ft. if the carpet is of a yard wide and there is no waste in matching the figure?
3. How many yards of carpet must be purchased for a room that is 15 ft. 6 in. by 16 ft. 4 in. if the carpet is of a yard wide, is put down lengthwise of the room, and there is no waste in matching the figure?
(P. 372.) * Why except the first breadth ?
335. PLASTERING AND PAPERING.
1. How many square yards of plastering in a room (walls and ceiling) that is 15 ft. by 18 ft. and 12 ft. high, an allowance of 12 square yards being made for openings?
NOTE. — In estimating the cost of plastering, allowance is made for “ openings” (windows and doors) only when they are very large in proportion to the wall to be covered. Why are plasterers unwilling to deduct the entire area of all the openings ?
2. At 24$ a square yard how much will it cost to plaster a room that is 17 ft. by 20 ft. and 10 feet from the floor to the ceiling, deducting 16 square yards for openings?
3. How many “double rolls" of paper will be required for the walls of a room that is 14 ft. by 16 ft. and 11 ft. high above the baseboards, if an allowance of 1 full “double roll” is made for openings?
NOTE. - Wall paper is usually 18 inches wide. A "single roll” is 24 ft. long. A “ double roll” is 48 ft. long. In papering a room 11 ft. high it would be safe to count on 4 full strips from each “ double roll.” The remnant would be valueless unless it could be used over windows or doors. Since each strip is 18 inches wide, a “double roll” will cover 72 inches (6 ft.) of wall measured horizontally.
4. At 12¢ a "single roll,” how much will the paper cost for the walls of a room that is 12 ft. by 14 ft. and 7 ft. above the baseboards, if the area of the openings is equivalent to the surface of 2 “single rolls ” of paper ?
5. Find the cost, at 25$ a square yard, of plastering the walls of a room that is 48 ft. by 60 ft. and 18 feet high, deducting 30 square yards for openings.
336. FARM PROBLEMS. Find how many acres in
1. A piece of land 1 rod by 160 rods.
(a) Find the sum of the five results.
(b) Find the sum of the five results. 11. A piece of land 12 rods by 40 rods. 12. A piece of land 27 rods by 40 rods. 13. A piece of land 46 rods by 20 rods. 14. A piece of land 36 rods by 20 rods. 15. A piece of land 264 feet * by 20 rods.
(c) Find the sum of the five results. 16. A piece of land 1 rod by 1 mile. 17. A piece of land 11 rods by 1 mile. 18. A piece of land 66 feet * by 1 mile. 19. A piece of land 99 yards * by 1 mile. 20. A piece of land 198 feet * by 1 of a mile.
(d) Find the sum of the five results. 21. A piece of land of a mile long and as wide as the schoolroom.
* Change to rods.
337. FARM PROBLEMS.
1. A piece of land 1 foot wide and 43560 feet long is how
many acres ?
2. Change 43560 feet to miles.
3. A piece of land 1 foot wide must be how many miles in length to contain 1 acre ?
4. Some country roads are 66 feet wide. acres in 84 miles of such road?
5. How many acres in 1 mile of road that is 4 rods wide ?
6. A farmer walking behind a plow that makes a furrow 1 foot wide will travel how far in plowing 1 acre?
7. A farmer walking behind a plow that makes a furrow 16 inches wide will travel how far in plowing 1 acre?
8. If a mowing machine cuts a swath that averages 4 feet in width how far does it move in cutting 1 acre ?
9. If potatoes are planted in rows that are 3 feet apart (a) how many miles of row to each acre ?
(b) How many rods of row to each acre ? (c) If 4 rods of row on the average yield 1 bushel, what is the yield per acre ?
10. Strawberry plants are set in rows that are 2 feet apart. (a) How many miles of row to the acre? (b) How many rods of row to the acre? (c) How many feet of row to the acre ?
11. If corn is planted in rows 31 feet apart and if the "hills” are 31 feet apart in the row, how many hills to each acre ? *
* Think of each “hill
as occupying a piece of land 31 ft. by 3 ft.
1. Cut one half of a circular piece of paper as indicated in the diagram.
Observe that if the circle is cut into a very large number of parts and opened as shown in the figure, the circumference of the circle becomes, practically, a straight line.
NOTE. - Imagine the circle cut into an infinite number of parts and thus opened and the circumference to be a straight line.
Observe that a circle may be regarded as made up of an infinite number of triangles whose united bases equal the circumference and whose altitude equals the radius. Hence to find the area of a circle we have the following:
RULE I. Multiply the circumference by 1 of the diameter.
2. It has already been stated that if the diameter of a circle is 1, its circumference is 3.141592.* Hence the area of a circle whose diameter is 1 is (3.141592 x 4).785398.
3. A circle whose diameter is 2, is 4 times as large as a circle whose diameter is 1; a circle whose diameter is 3, 9 times as large, etc. Hence to find the area of a circle we have also the following:
RULE II. Multiply the square of the diameter by.785398.
4. The approximate area may be found by taking (or .78) of the square of the diameter. See Werner Arithmetic, Book II., page 256.
* See page 229, note.