52. What is the radius of a circular acre?

Operation.-Side of a square (p. 81)×1.128 diameter of an equal

circle.

The side of a square acre (p. 13) is 208.710321, which, × 1.128= 235.5, and 235.5÷÷2 to obtain radius=117.75 feet, the Ans.

53. The time of the day is between 4 and 5, and the hour and minute hands are exactly together; what is the time?

Operation. The speed of the hands is as 1 to 11.

4 hours × 60=240, and 240÷11=21 min. and 49 sec., which, added to 4, =4 hours 21 min. and 491 sec., the Àns.

54. A person being asked what o'clock it was, replied that it was between 5 and 6; but, to be more particular, the minute-hand was as far beyond the 6 as the hour-hand wanted of being to the 6; that is, that the hour and minute hands made equal acute angles with a line passing from the 12 through the 6; required the time.

Operation.-5 hours=300 minutes, and 6 hours=360 minutes.

Then, 300+the time by the hour-hand past 5=360-the time by the minute-hand past 6.

As the relative speed of the hour and minute hands is as 1 to 12, 300+1, 360-12.

Consequently,

360-300

1+12

=fg=4, 36}}=the space between the hour

and minute hands, which, ÷2 to obtain the half space (each side of the 6), gives 2 min. 18 sec., which, added to 5 hours and 30 min., =5 hours 32 min. and 18 sec., the time required.

55. Two persons, A and B, start at the same time to meet each other when apart 100 miles; after 7 hours they meet, when it appears that A had ridden 11⁄2 miles per hour faster than B; at what rate per hour did each ride?

Ans. A 7.893, B 6.392 miles per hour. 56. Swift can travel 7 miles in of an hour, but Slow can travel only 5 miles in of an hour; both started from one point at the same time to walk a distance of 12 miles; how much sooner will Swift arrive than Slow?

Ans. 12.467 seconds.

57. At a certain time between two and three o'clock, the minute-hand of a clock was between three and four; within an hour after, the hour and minute hands had exactly changed places with each other; what was the precise time when the hands were in the first position?

58. If a traveler were to leave New Haven at 8 o'clock on a morning, and walk toward Albany at the rate of 3 miles an hour, and another traveler were to set out from Albany at 4 o'clock in the evening, and walk toward New Haven at the rate of 4 miles an hour, whereabout on the road would they meet, supposing the distance to be 130 miles?

Ans. 69.4286 miles from New Haven. 59. A thief, escaping from an officer, has 40 miles the start, and travels at the rate of 5 miles an hour; the officer in pursuit travels at the rate of 7 miles an hour; how far must he travel before he overtakes the thief?

Ans. 20 hours, and 140 miles.

60. If 12 oxen graze 3 acres of grass in 4 weeks, and 21 oxen 10 acres in 9 weeks, how many oxen would it require to graze 24 acres in 18 weeks, the grass to be growing?

Operation.-Each ox grazes a certain quantity in each week, which we suppose to be 100 pounds, and of the whole quantity grazed in each case, a part must have grown during the time of grazing.

Then, by the first condition,

12×4 × 100=4800 lbs.=whole quantity on 34 acres for 4 weeks.
4800÷33=1440 lbs.=whole quantity on 1 acre for 4 weeks.
By the second condition,

21 × 9 × 100=18900 lbs.=whole quantity on 10 acres for 9 weeks.
18900÷10=1890 lbs.=whole quantity on 1 acre for 9 weeks.

1890-1440=450 lbs.=the quantity grown on an acre for 9-4-5 weeks. 450÷9¬4=90 lbs.=the quantity which grows on each acre for 1 week. 90×3×4=1200 lbs. quantity grown on 34 acres for 4 weeks. 4800-1200=3600 lbs.=original quantity of grass on 34 acres. 3600÷÷31=1080 lbs.=original quantity on 1 acre.

And by the last condition,

24 × 1080=25920 lbs.=original quantity on 24 acres.

24 × 90 × 18=38880 lbs. quantity which grows on 24 acres in 18 weeks.

25920+38880=64800 lbs.=whole quantity on 24 acres for 18 weeks. 64800÷18=3600 lbs.=quantity to be grazed from 24 acres each week. 3600÷100=36=number of oxen required to graze the whole.

61. A tract of land, exactly square, is inclosed by a threerailed fence; the length of each rail is 15 feet, and the number of rails in the fence is equal to the number of acres inclosed; required the area of this tract in acres, and the length of its side in feet.

Operation.--If the tract of land was inclosed by one rail, then, 15÷ (4×3)=1.25 feet, the length of its side.

Then, if 43,560 square feet make an acre, as 1.252: 43,560::1 rail : 27878.4, the number of rails in the fence, or the number of acres in the tract; and (27878.4×15)÷(4×3) =34,848, the length of the side in feet, Ans.

62. What is the radius of a circular acre?

OPERATION.-Side of a square x1.128 diameter of an equal circle. By table, p. 13, 208.710321=the side of a square acre.

Then, 208.710321 × 1.128=235.50, which,÷2 (for radius), =117.75 feet, Ans.

63. There is an island 20 miles in circumference, and three men start together to travel the same way about it; A goes 2 miles per hour, B 4 miles per hour, and C 6 miles per hour in what time will they come together again?

Ans. 10 hours.

r;

64. A hare starts 12 rods before a hound, but is not perceived by him till she has been off 14 minutes; she runs at the rate of 36 rods a minute, and the dog, on view of her, makes after her at the rate of 40 rods a minute; how long will the course hold, and what distance will the dog run?

Ans. 14 minutes, and he will run 570 rods.

MENSURATION OF SOLIDS.

OF CUBES AND PARALLELOPIPEDONS.

Cube.

Definition. A solid contained by six equal square sides.

To ascertain the Contents of a Cube, Fig. 72

RULE.-Multiply a side of the cube by itself, and that product again by a side, and this last product will give the contents required.

Or, s3S, 8 representing the length of a side, and S the solidity.

Fig. 72.

a

EXAMPLE.-The side a b of the cube, Fig. 72, is 12 inches; what are the contents of it?

12×12×12=1728 inches, Ans.

Ex. 2. The side of a cube is 15 inches; what are its contents in feet and inches?

10000

Ans. 23.4375 feet, or 1 foot and 11-4375 inches. Ex. 3. The sides of a cube are 12.5 feet; what are its contents in cubic feet and yards?

(1953.125 cubic feet. Ans. 72.338 cubic yards.

Centre of Gravity. Is in its geometrical centre.

Parallelopipedon.

Definition. A solid contained by six quadrilateral sides, every opposite two of which are equal and parallel.

To ascertain the Contents of a Parallelopipedon, Fig. 73. RULE.-Multiply the length by the breadth, and that product again by the depth, and this last product will give the contents required.

Or, lxbxd=S.

Fig. 73.

с

a

EXAMPLE.-The length a b, Fig. 73, is 15, the breadth c d 12, and the depth c b is 11 inches; what are the contents? 15 × 12×11=1980 inches, Ans.

Ex. 2. The length of a parallelopipedon is 15 feet, and each side of it is 21 inches; what are its contents?

Ans. 45.9375 feet. Ex. 3. The dimensions of a parallelopipedon are 20 feet in length, 11.5 in breadth, and 7 in depth; what are its contents in feet? Ans. 1610 feet.

Centre of Gravity. Is in its geometrical centre.

PRISMS, PRISMOIDS, AND WEDGES.

Prisms.

Definition. Solids, the ends of which are equal, similar, and parallel planes, and the sides of which are parallelograms.

NOTE. When the ends of a prism are triangles, it is called a triangular prism; when rhomboids, a rhomboidal prism; when squares, a square prism; when rectangles, a rectangular prism, &c.