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56. The principal use of the arithmetical complement, is in working proportions by logarithms; where some of the terms are to be added, and one or more to be subtracted. In the Rule of Three or simple proportion, two terms are to be added, and from the sum, the first term is to be subtracted. But if, instead of the logarithm of the first term, we substitute its arithmetical complement, this may be added to the sum of the other two, or more simply, all three may be added together, by one operation. After the index is diminished by 10, the result will be the same as by the common method. For subtracting a number is the same, as adding its arithmetical complement, and then rejecting 10, 100, or 1000, from the sum. (Art. 53.)

It will generally be expedient, to place the terms in the same order, in which they are arranged in the statement of the proportion.

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57. In compound, as in single proportion, the term required may be found by logarithms, if we substitute addition for multiplication, and subtraction for division.

Ex. 1. If the interest of $365, for 3 years and 9 months, be $82.13; what will be the interest of $8940, for 2 years and 6 months?

In common arithmetic, the statement of the question is made in this manner,

365 dollars 3.75 years

: 82.13 dollars ::

8940 dollars
2.5 years

And the method of calculation is, to divide the product of the third, fourth, and fifth terms, by the product of the two first.* This, if logarithms are used, will be to subtract the sum of the logarithms of the two first terms, from the sum of the logarithms of the other three.

Two first terms S 365 log. 2.56229

3.75

0.57403

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Sum of the logs. of the 3d, 4th, and 5th 6.26378

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58. The calculation will be more simple, if, instead of subtracting the logarithms of the two first terms, we add their arithmetical complements. But it must be observed, that each arithmetical complement increases the index of the logarithm by 10. If the arithmetical complement be introduced into two of the terms, the index of the sum of the logarithms will be 20 too great; if it be in three terms, the index will be 30 too great, &c.

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The result is the same as before, except that the index of the logarithm is 20 too great.

*See Arithmetic.

Ex. 2. If the wages of 53 men for 42 days be 2200 dollars; what will be the wages of 87 men for 34 days?

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59. In the same manner, if the product of any number of quantities, is to be divided, by the product of several others; we may add together the logarithms of the quantities to be divided, and the arithmetical complements of the logarithms of the divisors.

Ex. If 29.67 x 346.2 be divided by 69.24 × 7.862 × 497 ; what will be the quotient?

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In this way, the calculations in Conjoined Proportion may be expeditiously performed.

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60. In calculating compound interest, the amount for the first year, is made the principal for the second year; the amount for the second year, the principal for the third year, &c. Now the amount at the end of each year, must be portioned to the principal at the beginning of the year. If

pro

In the right angled triangle BCP, (Trig. 134.)

R: BC::sin B:: PC

18.79.

And the solidity of the pyramid is 225.48 feet.

3. What is the solidity of a pyramid whose perpendicular height is 72, and the sides of whose base are 67, 54, and 40? Ans. 25920.

PROBLEM IV.

To find the LATERAL SURFACE of a REGULAR PYRAMID.

49. MULTIPLY HALF THE SLANT-HEIGHT INTO THE PERIMETER OF THE BASE.

Let the triangle ABC (Fig. 18.) be one of the sides of a regular pyramid. As the sides AC and BC are equal, the angles A and B are equal. Therefore a line drawn from the vertex C to the middle of AB is perpendicular to AB. The area of the triangle is equal to the product of half this perpendicular into AB. (Art. 8.) The perimeter of the base is the sum of its sides, each of which is equal to AB. And the areas of all the equal triangles which constitute the lateral surface of the pyramid, are together equal to the product of the perimeter into half the slant-height CP.

The slant-height is the hypothenuse of a right angled triangle, whose legs are the axis of the pyramid, and the distance from the center of the base to the middle of one of the sides. See Def. 10.

Ex. 1. What is the lateral surface of a regular hexagonal pyramid, whose axis is 20 feet, and the sides of whose base are each 8 feet?

The square of the distance from the center of the base to one of the sides (Art. 16.)

The slant-height (Euc. 47. 1.)

=48.

=√48+(20)=21.16. sq. feet. xx

And the lateral surface =21.16X4X6=507.84

2. What is the whole surface of a regular triangular pyra

mid whose axis is 8, and the sides of whose base are each

20.78 ?

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3. What is the lateral surface of a regular pyramid whose axis is 12 feet, and whose base is 18 feet square?

Ans. 540 square feet.

The lateral surface of an oblique pyramid may be found, by taking the sum of the areas of the unequal triangles which form its sides.

PROBLEM V.

To find the SOLIDITY of a FRUSTUM of a pyramid.

50. ADD TOGETHER THE AREAS OF THE TWO ENDS, AND THE SQUARE ROOT OF THE PRODUCT OF THESE AREAS; AND MULTIPLY THE SUM BY OF THE PERPENDICULAR HEIGHT OF THE SOLID.

Let CDGL (Fig. 17.) be a vertical section, through the middle of a frustum of a right pyramid CDV whose base is a square.

Let CD=a,

LG=b,

RN=h.

By similar triangles, LG: CD::RV : NV.

Subtracting the antecedents, (Alg. 389.)
LG: CD-LG::RV: NV-RV=RN,

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The square of CD is the base of the pyramid CDV;
And the square of LG is the base of the small pyramid LGV.
Therefore, the solidity of the larger pyramid (Art. 48.) is

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And the solidity of the smaller pyramid is equal to

hb

hb3

LG ×RV=ba ×· 3a-3b3a-36°

If the smaller pyramid be taken from the larger, there will remain the frustum CDLG, whose solidity is equal to

ha3-hb3
Sa-3b=3hx

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a-b=h×(a2+ab+b2)

Or, because a2b2 =ab, (Alg. 259.)

}h×(a2+b2+√a2b2)

(Alg. 466.)

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