9. 75 lb. 10 pwt. 123 gr. ÷ 15; 10 oz. 7 pwt. 20 gr. ÷ 16. 10. 100 T. 15 cwt. 75 lb. ÷ 18; 18 cwt. 90 lb. 10 oz. ÷ 20. 11. 170 A. 19 sq. rd. 11 sq. yd. ÷ 46; by 65; by 78; by 85; by 97; by 105; by 126; by 154; by 175; by 250. 12. If 9 loads of hay weigh 10 T. 15 cwt. 75 lb., what is the average weight? If 10 loads? 11 loads?

13. Divide 98 mi. 60 rd. 5 yd. 2 ft. 6 in. by 45; by 54; by 67; by 85; by 98; by 106; by 125; by 150; by 199.

14. If 9 hhd. of sugar weigh 5 T. 6 cwt. 253 lb., what is the average weight? If they weigh 6 T. 10 cwt. 50 lb.? 15. How much is 3 of 15 lb. 10 oz. 2 scr. 15 gr.? ?

16. Find 3 of 10 mi. 4 rd. 1 ft.; ; ; ; ; ; 10; 1. 5 7 }; 6

17. Add 10 A. 125 sq. rd. 20 sq. yd. and 25 A. 200 sq. rd. 15 sq. yd. 8 sq. ft., and divide the sum by 9; by 12; by 25.

18. From the sum of 40 gal. 2 qt. 2 gi., and 20 gal. 1 pt. take 15 gal. 1 qt. 3 gi., and divide the result by 15; by 18. 19. To the difference between 21 sq. rd. 18 sq. yd. 5 sq. ft. and 15 sq. rd. 8 sq. ft. add their sum, and divide by 15. 20. How many bottles, each holding 3 qt. 1 pt., can be filled from a cask containing 31 gal. 2 qt.?

21. How many baskets of peaches, each 2 pk. 4 qt., will make 31⁄2 bu. ? 51 bu.? 7 bu.? 71⁄2 bu.? 10 bu.?

22. If a man steps 2 ft. 4 in., how many steps will he take in walking of a mile? 2 mi.? mi.? 3 mi.?

23. If a train of cars runs 25 mi. 60 rd. in an hour, in what time can it run 151 mi. 40 rd.?

24. If 8 hogsheads of molasses, each 42 gal. 2 qt., sell for $136.717, what is the selling price per gallon?

25. The sum of three numbers is 75 mi., and two of them are 25 mi. 275 rd. and 30 mi. 5 yd. Find the third.

A.

26. Subtract the quotient of 21 75, from the product of 21 A. 35 75, and divide the remainder by 225.

35 sq. yd. 30 sq. in. sq. yd. 30 sq. in. ×

SECTION XXVI.

MEASUREMENTS.

CASE I.

Measurement of Surfaces.

432. A Rectangle is a flat surface which has four straight sides and

four square corners.

A rectangle has two dimensions: length and breadth.

433. The Area of a rectangle is the surface bounded by the sides of the rectangle.

Thus, the area of a blackboard is the surface bounded by the sides or edges of the blackboard.

Let the figure A B C D represent a surface 4 A in. long and 3 in. wide. Now, there can be

B

placed along the length of the surface as many square inches as there are inches in the length, making a row of 4 sq. in.; and there can be as

D

many rows of square inches as there are inches in the width, making 3 rows. Hence, the surface contains 3 rows of 4 sq. in. each; and 3 times 4 sq. in., or 12 sq. in., must be the area of the surface.

434. Principle.

The area of a rectangle is the product of the length and the breadth.

duct of the length, 163 ft., multiplied by the breadth, 10 ft., must

be 175 sq. ft., the area required.

436. Example 2.-A lot of ground 18 ft. wide contains 181 sq. yd. How long is it?

SOLUTION.

9 sq. ft. X 181=1629 sq. ft. Area

breadth = length.

1629 sq. ft. ÷ 18 ft. = 90 ft.

EXPLANATION.-181 sq. yd.

equal 1629 sq. ft.

Since the area is the product of the length and the breadth, the quotient of the area, 1629 sq. ft., divided by the breadth, 18 ft., must be 90 ft., the length required.

437. Rules for the Measurement of Surfaces. I. Multiply the length by the breadth. The product will be the area of the surface required.

II. Divide the area of the surface by the length. The quotient will be the breadth required.

III. Divide the area of the surface by the breadth. The quotient will be the length required.

Note 1. - When the length and the breadth are given to find the area, two factors are given to find their product.

2. When the area and one dimension are given to find the other dimension, the product of two factors and one factor are given to find the other factor.

Problems.

Find the areas of the surfaces having the following dimensions, and change each result to higher denominations:— 1. 7 in. by 6 in.; 8 in. by 7 in. 4. 18 ft. 9 in. by 15 ft. 2. 8 ft. by 5 ft.; 9 ft. by 8 ft. 5. 30 yd. by 20 yd. 2 ft. 3. 10 ft. by 6 ft.; 151 ft. by 8 ft. 7. How many sq. in. in a window-pane 181⁄2 in. by 121 in.? 8. How many sq. ft. of surface in a board 10 ft. long, 8 in. wide? In a board 12 ft. 9 in. long, 101⁄2 in. wide?

6. 45.5 rd. by 10 rd.

9. Find the area of a cellar of a square building, each side of which is 30 ft. 6 in. What is the area in sq. yd.?

10. In a village lot 90 ft. by 200 ft. are how many sq. yd.? How much is it worth at $1.25 a sq. rod?

Find the required dimension of the following surfaces:11. Area 437 sq. in., breadth 171⁄2 in.; breadth 18 in. 12. Area 762.5 sq. ft., length 30.5 ft.; length 24.4 ft. 13. Area 2318 sq. yd., length 303 yd.; length 25 yd. 9 in. 14. Area 2 acres, breadth 30.25 rd.; breadth 20 yd. 2 ft. 15. Of a blackboard 7 ft. long, the area being 24 sq. ft.? 16. How long is a window-curtain 3 ft. wide, if its surface is 2 sq. yd. 3 sq. ft.? If its surface is 34 sq. yd.?

17. I have a lot which is 6 rods wide, and contains 77 sq. rd. How many rods of fence are needed to inclose it? How many yd.? Feet?

CASE II.

Measurement of Solids.

438. A Rectangular Solid is a body which is bounded by six flat surfaces having

square corners.

A rectangular solid has three dimensions: length, breadth, and thickness.

439. The Solidity, or Capacity, of a solid is the space bounded by the surfaces of the solid.

Thus, the solidity of a block, or the capacity of a box, is the space bounded by the surfaces of the block or the box.

Let the figure represent a solid 4 in. long, 2 in. wide, and 3 in. high.

Now, there can be placed along the length of the solid as many cubic inches as there are inches in the length, making a row of 4 cu. in.; and there can be as many

rows of cubic inches on the lower surface as there are inches in the width, making 2 rows of 4 cu. in. times 4 cubic inches, or 8 cu. in.

each, which form one layer of 2

Also, there can be as many layers

as there are inches in the height, making 3 layers.

Hence, the solid contains 3 layers of 8 cu. in. each; and 3 times 2

times 4 cu. in., or 24 cu. in., must be the solidity of the solid.

440. Principle.

The solidity of a rectangular solid is the continued product of the length, the breadth, and the thickness.

Written Exercises.

441. Example 1.-Find the solidity of a block of granite 73 ft. long, 21 ft. wide, and 2 ft. thick.

SOLUTION.

EXPLANATION. Since

Length breadth thickness solidity. the solidity of a solid is the ጞ ft. ft. × 2 ft. × 2 ft. = 33 cu. ft.

7 2ft.

continued product of the length, the breadth, and the thickness,the continued product of the length, 7 ft.,

by the breadth, 24 ft., by

the thickness, 2 ft., must give 33 cu. ft., the solidity required.

442. Example 2.- A pile of stone 8 ft. long and 63 ft. wide contains 225 cu. ft. How high is it?

product of the length, 83 ft., by the width, 62 ft., must give 4 ft., the height required.

443. Rules for the Measurement of Solids.

I. Multiply the length, the breadth, and the thickness together. The product will be the solidity or capacity required.

II. Divide the solidity by the product of the length and the breadth. The quotient will be the thickness required.

III. Divide the solidity by the product of the length and the thickness. The quotient will be the breadth required.

IV. Divide the solidity by the product of the breadth and the thickness. The quotient will be the length required.

Note 1. When the length, the breadth, and the thickness are given to find the solidity, three factors are given to find their product.