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When you have obtained two or three decimal figures in the quotient, it is useless to carry the calculation any further, as they will be too small.

We shall now endeavour to apply these principles.

PART I.

OF LINES.

PROBLEM I. To find a side of a rectangular triangle, the two others being known.

RULE. Multiply by itself each of the known sides, then add them together if you wish to find the greater side; and subtract the lesser number from the greater if you wish to find one of the lesser sides. Then you will have the same result as if you had multiplied the unknown side by itself. Of course, you have only to find what number multiplied by itself, will give this result.

Example 1. The smaller sides of a rectangular triangle, (figs. 12 and 13) are one 3, and the other 4 inches, find the larger side.

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5 multiplied by itself makes 25, and the greater side or side required, must be 5 inches.

Example 2. In a rectangle, (1st Class, fig. 14,) it is known that the base is 8 inches, 54 hundredths, the

Class I

height is unknown, but the diagonal (a right line drawn from corner to corner) is found by measurement to be 15 inches, 32 hundredths. What is the height?

Note. A diagonal cuts the oblong square or rectangle into two rectangular triangles, of which the height above required is one of the smaller sides.

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It remains to find a number, which multiplied by itself will give 161,770. A few trials will show that this is (as near as possible) 12,719 as may be found by multiplying this number by itself. The height then, is 12 inches, and 719 thousandths of an inch.

Example 3. Find the height of an isoceles triangle, (1st Class, Prob. 27.) whose base is ,52 and the equal sides,,87. The perpendicular drawn from the summit, cuts the base in halves, and is a small side of a rectangular triangle, of which

the base is half the larger one, or The great side of the new angle, which

was one of the equal sides of the isoceles, is

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,26

,26

,87

156

,87

52

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A few trials will show, that the height required is, 83 nearly, for 83 times 83 are 6889.

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4. The smaller sides of a rectangular scalene trian gle, (1st Class, fig. 13,) are 5 and 7 inches, required the larger side. Ans: 8,6 in nearly

5. One small side of a rectangular scalene triangle is, 8,3 and the other 4,8 what is the size of the third side? Ans? 9,59 in nearly.

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nearly

6. The two equal sides of a rectangular isoceles triangle, are 4,8 in length, what is the length of the base? 6,785 in 7. The base of a rectangle (Ex. 2) is 7,15; the diagonal 13,25; required the height (that is, the third side of the triangle.) Ans. 14, 14 in nearly.

8. The base of a rectangle is 4,75 and the height 7,25 required the diagonal, or longest side of the triangle. Ans. 8,67 in. nearly.

9. What is the height of an isoceles triangle, of which the base is 6 inches, and the equal sides 3in.,8 each? Ams 2,33 in nearly.

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The instructer may increase these examples at pleas

ure.

PROBLEM II. To find the circumference of a circle when lengthened out into a right line.

RULE. Multiply its diameter by 3, and add a seventh of a diameter.

Example 1. The diameter of a circle is 4,523.

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2. The width of a basin is 5i,5 how long must a string be to reach round it?

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3. The diameter of a circle is 4,45 what is the circumference? Ans. 13,98 in.

4. The diameter of a ring is 1,5 what is the circumference? Ans. 4,7 in. 5. The diameter is 4,17 what is the circumference? Ans. 13,10 in. PROBLEM III. The circumference being known, to find the radius

RULE. Multiply the circumference by 0,159 and you will have the radius.

Example 1. To find the thickness of a column, its circumference has been measured with a string and found to be 12,542-required the radius.

12,542
0,159

112878

62710

12542

1,994178 radius.

If 1,994 thousandths be the radius or half diameter, 3,988 will be the whole diameter of the column. Six figures are separated, because there are three in the multiplier, and 3 in the multiplicand. The three right hand decimals are unimportant.

2. The circumference of a column is 10,5 what is its radius? what its diameter? Radius 1.67 - Diam 3,34

3. The circumference of a ship's mast is 136,15 what is its diameter ? Ans. 43, 28 in.

4. The circumference of a wheel is 48,75 what is its radius and diameter ?

Radius - 775
Diam. 15,5

PART II.

OF SURFACES.

PROBLEM I. To find the surface of a parallelogram.

RULE. The surface of a parallelogram (fig. 16,) or rectangle (fig. 14,) is found by multiplying the base by the height. That of a square is found by multiplying one of the sides by itself.

Example 1. A rectangle has 2,24 for its base, 4,31 for its height, what is its surface?

2,24

4,31

224

672

896

9,6544 surface.

That is, 9 square inches, and 65 hundredths of a square inch.

2. A room is 154,6 inches long, and 75,3 wide, what is its surface or area?

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