9. (487) (400+80+7)=400+2×400.80+802+ 2× (400+80).7+72. The above method of squaring a number consisting of the sum of two or more numbers, is elegantly illustrated geometrically as follows: The square ABCD may be enlarged to the square AEKF, by the addition of the two equal rectangles M P F (400+80) 7 H K N 400-80 802 D C G A 4002 B (400-+-80).7 400-80 BG and DH, whose lengths are each equal to the side AB of the original square, and whose widths are equal to BE, the quantity by which the side of the square has been augmented, also a little square, CGKH, whose side is the same as the width of one of the equal rectangles. Again, the square AEKF, having its side increased by EL, becomes augmented by the two rectangles EN, FP, and the little square KR. Thus we might continue to augment the square last found by the addition of two equal rectangles, and a little square; the length of each rectangle being equal to the side of the square which is to be augmented, and the width equal to the quantity by which the side of the square is increased; and the side of the little square being the same as the width of one of the rectangles. The diagram is adapted to the case of squaring 400+80+7=487. 132. Let us now, by reversing the above process, deduce a rule for extracting the square root. Let it be required to extract the square root of 527076. For the sake of simplicity, we will suppose we are required to find the number of feet in the side of a square whose area shall contain 527076 square feet. The smallest number, consisting of two figures, which is 10, becomes, when squared, 100; having more than two figures. Again, the largest number of two figures, 99, becomes, when squared, 9801, having four figures. Hence, when a number consists of more than two figures, and of not more than four, its square root will consist of two figures. By a similar method it may be shown, that when a number consists of more than four, and of not more than six figures, its square root will consist of three figures. Therefore, if we separate a number into groups of two figures each, the number of groups will denote the number of figures in the square root of that number. In this example, we see that there must be three figures in the root. We know that the side of the square sought must exceed 700 linear feet, since the square of 700 is 490000, which is less than 527076; we also know that the side of this square must be less than 800 linear feet, since the square of 800 is 640000, which is greater than 527076. Hence the first, or hundreds' figure of the root, is 7; which is the greatest number whose square can be contained in 52, the first or left-hand period. two equal rectangles, and a little square. The surface of the two rectangles will evidently make by far the largest portion of the whole increase. The length of one of these rectangles is the same as the length of a side of the square ABCD, which has already been shown to be 700 linear feet. The length of the two rectangles taken together will be twice 700 linear feet, or what would be the same thing, 700 linear feet added to 700 linear feet. If to BC, which is 700 linear feet, we add CD, which is also 700 linear feet, we shall have BC+CD equal to 1400 linear feet, for the length of the Iwo rectangles. Were we to multiply 1400 by the width of a rectangle, we should obtain the number of square feet in these rectangles, or nearly the 37076 square feet which require to be added. Consequently, if we divide 37076 by 1400, the quotient will give the approximate width of the rectangles. Using 1400 as a trial divisor, we find it to be contained between 20 and 30 times in 37076; hence the second or ten's figure of the root is 2. But besides the rectangles, there must be added the little square CGKH, each side of which is 20 linear feet, we will therefore add 20 to 1400, and thus obtain 1420 for the total length of the two rectangles and the side of the little square. Now, multiplying 1420 by 20, we obtain 28400 square feet for the total additions, which subtracted from 37076, leaves 8676 square feet. The square AE KF thus completed is 720 feet on a side. Again, a side of this square is to be further increased so that the added surface will amount to 8676 square feet. And, as before, the parts added will consist of two equal rectangles and a little square. The trial divisor, which is the sum of the length of the two new rectangles, is the same as the sum of two sides of the square AEKF. If, now, to 1420 already found, we add 20, we shall have 1440, which is the sum of EK and KF, and which is our second trial divisor. We find this divisor contained between 6 and 7 times in 8676; hence our third or units' figure of the root is 6. Therefore 6 is the width of the second set of rectangles. The second little square KNRP, of the same width as the rectangles, must be 6 linear feet on a side; therefore, adding 6 to 1440, we find 1446 for the whole length of the new rectangles and a side of the second little square. Multiplying 1446 by 6, we obtain 8676 square feet as the sum of the series of additions to the square AEKF, thus forming the square ALRM, which is the square sought; each side being 726 feet. The above work may be arranged as follows: If we omit the ciphers on the right, the work will take the following condensed form: From the above process, we deduce the following rule for the extraction of the square root of a whole number. RULE. 1. Separate the given number into periods of two figures each, counting from the right towards the left. When the number of figures is odd, it is evident that the left-hand, or first period, will consist of but one figure. II. Find the greatest square in the first period, and place its root at the right of the number, in the form of a quotient in division, also place it at the left of the number. Subtract the square of this root from the first period, and to the remainder annex the second period; the result will be the FIRST DIVIDEND. III. To the figure of the root, as placed at the left of the number, add the figure itself, and the sum, with a cipher annexed, will be the FIRST TRIAL DIVISOR. See how many times this trial divisor is contained in the dividend; the quotient will be the next figure of the root; this must be added to the TRIAL DIVISOR; the result will be the TRUE DIVISOR. Multiply the true divisor by this last figure of the root, and subtract the product from the dividend, and to the remainder annex the next period, for a NEW DIVIDEND. IV. To the last TRUE DIVISOR add the last figure of the root; the sum, with a cipher annexed, will be a new TRIAL DIVISOR. Continue to operate as before, until all the periods have been brought down. NOTE.-In case any dividend is not so great as its trial divisor, we must write 0 as the next figure of the root; we must also place O at the right of the divisor, and form a new dividend by annexing a new period. Ans. 526112. Ans. 111. Ans. 232. 4. What is the square root of 12321 ? |