method commonly called “Cross Multiplication;" or (3) by changing the number of feet and inches in the length and breadth into a number of feet and parts of feet expressed in the dnodenary scale. EXAMPLE 1 29 2 2 3 83 length = 20— ft. - ft. 4 4 29 83 2407 X sq. ft. =— 2 4 8 = 300 sq. ft. 126 sq. in. ft. pr. sec. By second method 20 9 0 .. area= sq. ft. Observations on the 2nd method. In lineal measure inches are also called prinies, and twelfths of inches seconds; in square measure, a rectangle 12 in. by 1 in. i.e. the twelfth of a square foot is called a superficial prime; the twelfth of a prime, i.e. a square inch is called a superficial second; the twelfth of a second, a third; and so on. Hence multiplying the number of feet in a length by the number of inches in the breadth we find the number of primes in the area; or feet multiplied by inches give superficial primes. So inches multiplied by inches give square inches or superficial seconds; inches multiplied by seconds give thirds. Observing these facts, the Rule for the performance of the second method is as follows: Write the multiplier under the multiplicand, placing feet under feet, &c. Beginning with the highest denomination multiply by each term of the multiplier, taking care to write quantities of the same denomination under each other. Add the products. The result will be in sq. ft., primes, seconds, &c. II. TO FIND THE SOLID CONTENT OF A RECTANGULAR PARALLELOPIPED. Rule. Multiply together the units in the length, breadth, and thickness; the result is the number of cubic units. This operation (as the last) may be conducted by each of the three methods, already mentioned. The twelfth part of a solid foot is called a solid or cubic prime; the twelfth part of a prime, a second ; and so on. Since the twelfth part of a solid foot is a solid whose base is a square foot, and height 1 inch; therefore a number of square feet multiplied into a number linear inches give a number of cubic primes. Also since the twelfth part of a cubic prime is a solid whose base is a superficial prime and height an inch, or whose base is a square inch or superficial second, and height a foot; therefore a number of superficial primes multiplied into a number of linear inches, or a number of superficial seconds multiplied into a number of linear feet, give a number of cubic seconds. Also since the twelfth part of a cubic second is a cubic inch, therefore a number of superficial seconds multiplied into a number of linear inches give a number of cubic thirds. Observing the above facts, the working of the 2nd method will be understood, the operation being performed as in the previous case. EXAMPLE. How many solid feet and inches are there in a block of stone, whose dimensions are 6 ft. 7 in., 5 ft. 8 in., 3 ft. 4 in. ? By the first method. 7 2 1 12 3 3 6715 cub. ft. = - cub. ft. 54 54 By the second method. ft. pr. sec. By the third method. The dimensions expressed in the duodenary scale are 6.7 ft., 5.8 ft., 3.4 ft. ft. 6.7 5.8 APPENDIX. I. TO ADD RECURRING DECIMALS. Rule 1. Convert the decimals into vulgar fractions, and add. Rule 2. Write down the decimals at length, under one another as in common addition, to three or four places more than are necessary to obtain two vertical columns alike; add as in common addition ; the period will readily be seen in the result. EXAMPLE. 1.03 573573573573 1.94923092645864921 Ans. 1.9492309264586. II, TO SUBTRACT RECURRING DECIMALS. Rule 1. Convert the decimals into vulgar fractions, and add. Rule 2. Write down the decimals, as in common subtraction, to four or five more places than are necessary to obtain two vertical columns alike; subtract as in common subtraction; the period will readily be seen in the result. EXAMPLE. 1.54302543025430254302 .88937177660064888937 Ans. .889371776600648. III. TO MULTIPLY RECURRING DECIMALS. Rule 1. If the exact result is required, convert the decimals into vulgar fractions, and multiply; then re-convert the product into a decimal. Rule 2. If only an approximate answer be required, multiply by the abbreviated form for multiplication of decimals. EXAMPLE. Multiply 27.3 by 4.7. 1 7 82 43 27.3 x 4.7 = 27- X 4- -X 3 9 3 9 3526 16 = 130— 27 27 = 130 592. Rule 1. Convert the decimals into vulgar fractions, and divide. Rule 2. If the divisor be not a recurring decimal, perform the division in the ordinary manner. Rule 3. If an approximate answer be required, divide by the abbreviated method for division of decimals. EXAMPLE. Divide 130.592 by 4.7. 592 7 16 9 130.592 = 4.1 = 130--4-= 130-X 999 27 43 3526 82 V. TO DETERMINE THE TOTAL NUMBER OF FIGURES WHICH THERE WILL BE IN ANY QUOTIENT. Rule. If the significant figures of the divisor represent a number not greater than the first equal number of significant figures in the dividend, the number of figures in the quotient will be equal to the difference between the numbers in the dividend and divisor, increased by 1. But if the figures of the divisor represent a larger number than the first equal number of figures in the dividend, the number of figures in the quotient will be equal to this difference. Thus in dividing 624309 by 8275, since 8275 denote a larger number than 6243, therefore there will be two figures in the quotient. Hence in division of decimals, knowing the whole number of figures in the quotient, and the number of decimals, we can easily find the number of integers or ciphers immediately after the decimal point. Thus in dividing 356.5043 by 7.253, since 7253 denote a larger number than 3565 therefore there will be 3 figures in the quotient, and since one must be a decimal, therefore 2 will be integers. Again in dividing .3565043 by 7.253, there will be 3 figures in the quotient, but there should be 4 decimals, therefore there must be one cipher after the decimal point. |