SOLUTION. H2=6282=36+64-100. H=√100=10.

2. The base is 12, the perpendicular is 16. Find the

hypothenuse.

3. The base is 15, the perpendicular is 20.

hypothenuse.

4. The hypothenuse is 10, the base is 6.

perpendicular?

Find the

What is the

SOLUTION. Since H2=B2+P2, P2=H2 - B2. That is, P2=102-62 = 100-36-64. Since P2-64, P=√/64=8.

5. The hypothenuse is 10, the perpendicular is 8. What is the base?

SOLUTION. Since H2-B2+P2, B2=H2-P2. That is, B2=10a —82— 100-64-36. Since B2=36, B=√36=6. In general,

551. I. The hypothenuse of any right-angled triangle is equal to the square root of the sum of the squares of the other two sides.

H=√B2+P2.

II. Either side about the right angle of a right-angled triangle is equal to the square root of the difference of the other two sides.

B=VH-P2; and P=VH2-B2.

1. What is the base, when the hypothenuse is 50, and the perpendicular 40 ?

2 What is the perpendicular when the hypothenuse is 30 and the base 18 ?

3. Base is 7, hypothenuse 9. What is the perpendicular?
4. Hypothenuse is 11, perpendicular is 5. What is the
base?

5. How far from the base of a perpendicular telegraph
pole 60 ft. high could a kite string 100 ft. long reach,
if the kite were lodged on the top of the pole?
6. A ladder 20 ft. long leans against the side of a house.

How far from the ground will the upper end of the ladder be if the foot is 3 feet from the wall?

7. What is the diagonal (Art. 585.) of a blackboard 10 ft. long, 4 ft. wide?

8. What is the diagonal of a slate 6 inches by 10 inches? 9. What is the diagonal of a square whose side is 1? SOLUTION. Diagonal=√12+12=√. Therefore,

The ratio of the side of a square to its diagonal is 1 : √T.

10. What is the diagonal of a square 5 inches on a side? 11. What is the largest square that can be cut from a circular piece of paper 4 inches in diameter ?

12. What is the largest square beam that can be cut from a log 20 inches in diameter?

13. Find the distance from the upper N. E. corner to the lower S. W. corner of a room 16 ft. by 12 ft. by 10 ft. This distance is called the diagonal of the room. SOLUTION. D=V16+12+10=500-22.3+ft.

The diagonal of any rectangular solid is equal to the square root of the sum of the squares of its three dimensions.

14. Find the diagonal of a room 20 ft. by 16 ft. by 12 ft. 15. Find the diagonal of a box 17 in. by 9 in. by 6 in., inside measurement.

16. Find the largest cube that can be cut from a globe 18 in. in diameter.

17. Find diagonal, in rods, of 18. A steamer goes east 18 north 4 miles an hour. hours?

a square acre.

miles an hour; a sloop goes How far apart are they in 5

19. A cistern holding 2400 gallons is 10 ft. deep. The length and width are equal; what are they?

20. Find the side, in rods, of a square field equal in area to three fields containing respectively 10 acres, 4 acres, and 2 acres 144 sq. rd.

CUBE ROOT.

552. The cube of any number represented by two or more figures, as 65, may be thus obtained:

65-60+5, that is, 6 tens plus 5 units.

= 274625

60+3(60×5)-'-3(60×5°)+53

t3 +3(txu)+3( t xu2)+u3

Every perfect cube consists of four parts, as shown above, viz:

1. The cube of the tens.

2. Three times the product of the square of the tens by the units.

3. Three times the product of the tens by the square of the units.
4. The cube of the units.

Or, more briefly, using t for tens and u for units,

(t+u)3= t3+3(t2×u)+3(txu) + u3.

This is illustrated by the following figures, in which the entire cube (tho cube of 65) is composed of

ANALYSIS. 1. Separate the number into periods of three figures each. Since there are two periods, the root will contain two figures. 2. The first period considered as integral thousands is 274, the greatest cube in which is 216 thousands, the cube root of which is 6 tens, or 60. 3. From the entire cube deduct the cube of the 6 tens = 216000, and there remains 58625, equal to 3 t u + 3 t u2 + u3. 4. 3 12 u is hence contained in 58635, and is a product of two factors, one of which is 3 t2; and 58625 divided by 3 t2 or 3 × 60o € 10800 will give 5 for the probable next figure of the root. 5. 3 t u or 10800 X5=54000, which deducted from 58624 leaves 4625, equal to 3 t u2. 3 tu or 3 (60 × 32)=4500, deducted from 4625 leaves 125, the cube of the 5 units.

NOTE. To find the cube root of a number of more than two periods, find the root of the greatest cube in the first two periods as already analyzed, and proceed as before with the next period to find the next order of the root, and in like manner until no period remains. 2. What is the cube root of 34012224 ?

probable second figure of the root, considered as units. As in the first step we considered the first period without regard to the others, so now we shall consider the first two periods without regard to the third. From the trial dividend, or that which remains after the cube