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RULE.

265. The volume of a sphere is equal to one-sixth the cube of its diameter, multiplied by 3.1416.

WRITTEN EXERCISES.

1. Find the volume of a sphere whose radius is 10 in.

2. The outer diameter of a spherical shell is 12 in. and the inner diameter is 8 in.; find the contents.

3. Find the weight of a ball of gold 8 in. in diameter, if a cubic foot weighs 1204 lb.

4. Find the weight of a cannon-ball of cast iron 15 in. in diameter, if a cubic foot weighs 450 lb.

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WRITTEN EXERCISES. 1. Find the side of the largest cube that can be cut from a sphere 15 in. in diameter.

It is readily seen that

A B? + BC? + CD2 A D?. But A B = BC = CD;
then
3 A B’ = A D

225,
A B2 = 75,
AB - V75 = 8.66+ in.

15o =

2. Find the side of the largest cube that can be cut from a sphere 30 in. in diameter.

PROBLEMS IN MENSURATION. 1. How many acres in a triangle whose base is 40 chains and altitude 36 chains ?

2. How many square yards in a triangle whose sides are each 50 feet?

3. How many acres in a triangle whose sides are respectively 20, 30, and 40 chains ?

4. The hypotenuse of a right triangle is 312 feet and the base is 282 feet; what is the perpendicular?

5. Find the diagonal of a square whose side is unity. 6. Find the diagonal of a cube whose side is unity.

7. How many acres in a parallelogram whose base is 161.5 rods and altitude 106.3 rods?

8. The area of a rectangle is 1872 square rods, and the length is to the width as 4 is to 3; required the length of the sides.

9. The area of a circle is 2 A. 3 R. 15 P.; what is the circumference ?

10. Find the side of a square whose area is equal to the area of a circle 40 feet in diameter.

11. Find the length of an arc of 35° in a circle whose radius is 40 rods.

12. Find the area of the largest possible square that can be cut from a circle 62 feet in circumference.

13. Find the diameter of a circle whose area is numerically equal to 10 times its circumference.

14. How many square feet in the bottom and convex surface of a scrap-basket whose lower base is 9 inches in diameter, upper base 12 inches in diameter, and altitude 10 inches?

15. A piece of timber 90 feet long is 6 by 8 inches at the smaller end and 18 by 24 inches at the larger end; how many cubic feet does it contain ?

16. Find the diameter of a sphere whose surface and volume are numerically equal.

17. Find the diagonal of a parallelopiped whose base is 5 feet by 7 feet, and altitude 16 feet.

18. Required the entire surface of a square pyramid whose altitude is 36 inches and side of the base 30 inches.

19. Required the dimensions of a cubical bin which will hold 1000 bushels of grain.

20. Find the contents of a cone whose altitude is 60 feet and the radius of the base 20 feet.

21. Find the volume of the frustum of a cone whose altitude is 24 feet, the radius of the lower base being 8 feet and of the upper

base 6 feet. 22. If a conical hay-stack 24 feet in diameter and 20 feet high contains 12 tons, what are the dimensions of a similar stack which contains 20 tons ?

23. A circular mirror 24 inches in diameter is surrounded by a frame 3 inches wide; what is the cost of the frame, at $2 a square foot ?

24. A garden containing 74 square feet less than one-fourth of an acre is planted with hyacinths in squares 6 inches apart; how many plants will be required, if the outer rows are 6 inches from the edge ?

25. The largest possible square piece of timber of uniform size is cut from a log 40 feet long, 15 inches in diameter at one end, and 18 inches at the other; how much is wasted ?

26. A cylinder is 6 feet in diameter and 8 feet high ; what are the dimensions of a similar cylinder whose surface is 64 times as much ?

27. How much material will be required to make a hollow sphere whose outside diameter is 15 inches and thickness 1 inch?

28. What is the value of a gold sphere 4 inches in diameter, if a sphere of silver 1 inch in diameter is worth $5, and the value of gold is to the value of silver as 16 to 1?

SECTION XII.

PROGRESSIONS.

ARITHMETICAL PROGRESSION. 267. An Arithmetical Progression is a series of numbers which increase or decrease by a Common Difference.

Thus, 1, 3, 5, 7, 9 is an increasing arithmetical progression in which the common difference is 2.

And 10, 8, 6, 4, 2 is a decreasing arithmetical progression in which the common difference is 2.

The numbers composing the series are called the Terms.

268. To find the last term of an arithmetical progression.

WRITTEN EXERCISES.

1. Find the 11th term of the series 3, 5, 7, 9, 11, etc.

PROCESS.

The first term is 3. 1st 3

The second term equals 3 plus 2d 3 + 1 x 2 5

once the common difference. 3d 3 + 2 x 2 7

The third term equals 3 plus 4th 3 + 3 x 2

twice the common difference. Hence 11th 3 + 10 x 2 23. Hence the eleventh term equals

the first term plus ten times the common difference, which equals 3 + 10 x 2

23. From this problem we derive the

RULE.

The last term equals the first tern increased by the common difference multiplied by the number of terms less one.

2. Find the 12th term of the series 1, 3, 5, etc. 3. Find the 31st term of the series 2, 4, 6, etc. 4. Find the 50th term of the series 1, 4, 7, etc. 5. Find the 80th term of the series 10, 15, 20, etc. 6. Find the 103d term of the series 6, 73, 83, etc. 7. Find the 21st term of the series 8, 12, 16, etc. 8. Find the 100th term of the series 15, 25, 35, etc. 9. Find the 60th term of the series 216, 214, 212, etc. 10. Find the 12th term of the series , 1, 13, etc.

11. If a body falls 16 1 ft. the first second, and in each succeeding second 32 feet more than in the preceding one, how far will it fall in the 25th second ?

269. To find the sum of an arithmetical progression.

WRITTEN EXERCISES.

1. Find the sum of 8 terms of the series 1, 5, 9, etc.

Sum 1 + 5 + 9 + 13 + 17 + 21 + 25 + 29

Sum = 29 + 25 + 21 + 17 + 13 + 9 + 5 + 1 2 sum

30 + 30 + 30 + 30 + 30 + 30 + 30 + 30 2 sum 30 X 8 (1 + 29) 8

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The sum of the series is 1 + 5 + 9 + 13 + 17 + 21 + 25 + 29.

Writing the sum of the series in a reverse order—that is, 29 + 25 + 21 + 17 + 13 + 9 + 5 + 1-and adding, we have twice the sum of the series equal to 30 taken 8 times, or (1 + 29) 8. Hence the sum of the series equals

1 + 29

x 8, or 120.

2 From this problein we derive the

RULE. The sum of an arithmetical progression equals half the sum of the extremes multiplied by the number of terms.

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