at the right of it, to complete the divisor. We double the root for a divisor, because the additions must be made on two sides of our Figure rstu, to preserve it in a square form, each side being now 20 rods in length, the additions on the two sides must be 40 rods. And after doubling our root for a divisor, it expresses 40 in the divisor, because to complete our divisor another figure must be placed at the right of it; therefore, double the root is just equal to the additions which must be made on the two sides of our Figure rstu. We next inquire how often our divisor 4, is contained in our dividend exclusive of the right hand figure; (we except the right hand figure of the dividend, because we have another figure to place in the divisor,) and we find that 4 is contained in 22 five times; we then write five in the quotient for the second figure of the root, and also place the 5 at the right hand of the 4, in the divisor; we then have 45 for a divisor, which is the whole length of the additions which must be made around the Figure rstu, to preserve it in a perfect square form; because when we made the two first additions, that is, the long squares A and C, which equal in length the two sides of our first square r s t u, there was left a vacancy which is supplied by the little square, B; our divisor, 45, is then the whole length of A, B and C, and dividing by the length, our second quotient figure must express the width of the additions. We next multiply 45, our divisor, by 5, the last figure in the root, placing the product under 225, the remaining part of the dividend, which we find it equals, consequently we have no remainder. Our work is now finished, and it is plain, on inspecting the Figure, that the first figure of the root shows the length of one of the sides of the square r s t u, and the second figure of the root shows the width of the additions made to our first square, consequently 25, the root, shows the length of one of the sides of the square after it is completed. The work may be proved by multiplying the root by itself, that is, one of the sides of the square by itself; or it may be proved by adding in one sum, the area of the different parts of the Figure, and if it be equal to the given sum, the work is right. rds. Proof by adding the parts. 20 20 400-Area of the square r s t u. 20×5=100=Area of the long square A. 20X5=100=Area of the long square C. | 5X5= 25 Area of the little square B. 625 Rods, area of the Fig. completed. Proof by raising the root to the second power or square. 25 rds. 25 125 50 625 Rods, a number equal to the given sum. 3. A man has 26896 square rods of meadow grourd; he wishes to know the side of a square equal in area to it. 160, 640, do. B. C. n. D. E. 16, do. e. : : NOTE. We double the right hand figure of our last divisor in forming a new divisor, in order to get the length of the two sides of the little square placed in the corner, because the two additions to be made, must extend on two sides of the little square placed The attentive stu in the corner. dent must now discover, on inspecting this and the two first Fi 26896 The area of: gures under this rule, that for eve whole Figure. ry figure in the root, after the first, there must be an addition made around the square, and that the area of the Figure may always be calculated, by adding the area of the different parts of the Figure, giving to the figures representing the different parts their proper local value. 4. What is the square root of 81? Ans. 9. Ans. 36. Ans. 256. Ans. 654: Ans. 4802. NOTE.--If there be a remainder after all the periods are brought down, the operation may be continued, at pleasure, by annexing periods of ciphers. 9. What is the square root of 42 ? 625,8950(25,017 4 45)225 225 5001) 8950 5001 50027) 39490C 350189 44711 rem. Ans. 6,48. Ans. 5,477. Ans. 25,017. NOTE.-In placing periods in decimals and whole numbers, a period is placed over units, in whole numbers, and so on, to the left over every second figure, and the first period in decimals must be placed over tenths, and so on, to the right, placing a period over every second figure. And if the decimals be odd, a cipher must be joined to the right hand of the last period to complete it; the student will recollect that joining a cipher to the right of a decimal does not alter its value. 12. What is the square root of 262,44? 13. What is the square root of 24,7009 ? Ans. 16,2. Ans. 4,97. To extract the square root of a Vulgar Fraction. RULE.-Reduce the fraction to its lowest terms for this, and all other roots; then extract the root of the numerator for a new numerator, and the root of the denominator for a new denominator. If the fraction be a surd, reduce it to a decimal, and extract its root; if a mixed number, reduce it to an improper fraction, and then proceed the same as with a simple fraction, or reduce the fraction to a decimal, and proceed as in whole numbers and decimals. EXAMPLES. 1. What is the square root of.35? Ans.. Ans: 7. Ans. . SURDS.-4. What is the square root of 11? Ans. ,9574. 5. What is the square root of ? Ans. ,64549. 6. What is the square root of?? Ans. ,88191. MIXED NUMBERS.-7. What is the square root of 301 ? Ans. 52. 8. What is the square root of 201? Ans. 42. Ans. 16,5, 9. What is the square root of 272 ? Application and Use of the Square Root, PROBLEM I-To find the mean proportional between two numbers. RULE.-Multiply the given numbers together, and extract the square root of the product: the root will be the mean proportional sought. EXAMPLES. 1. What is the mean proportional between 24 and 96? 24×96=2304= Ans. 48. NOTE. When the first is to the second, as the second is to the third, the second is called a mean proportional between the other two, thus, 24:48:: 96, 2. A cheese weighed in one scale of a balance with an unequal beam, 16lb., but weighing it in the other scale of the balance, it weighed 64lb.; what was the true weight of the cheese? Ans. 32lb. PROBLEM II.-To find the side of a square equal in area to any given surface or superficies. RULE.-First find the area, (if not given,) and the square root of the area is the side of the square sought. EXAMPLES. I. If the area of a circle be 1521; what is the side of a square equal in area thereto ? Ans. 39. 2. A man has 2 lots, the first contains 2 acres, 1 rood, and 25 poles; the second contains 1 acre and 2 roods; he wishes to exchange the two for one that is exactly square; the length of one of its sides is required? Ans. 25 poles. 3. A certain general has an army of 7056 men; how many must he place in rank and file, to form them into a square? Ans. 84. PROBLEM III.-To form a body of soldiers, so that the number in rank may be double, triple, &c. as many as in file. RULE.-Extract the square root of 1,,, &c. of the given number of men, and that will be the number of men in file, which double, :riple, quadruple, &c. and the product will be the number in rank. EXAMPLES. 1. Let 8192 soldiers be so formed, that the number in rank may be double the file. Ans. 64 in file, and 128 in rank. 2. Suppose a gentleman would set out an orchard of 1200 trees, so that the length shall be to the breadth as 4 to 3; how many trees must there be in length, and how many in breadth? Ans. 40 in length, and 30 in breadth. PROBLEM IV-Any two sides of a right angled triangle being given, to find the third-side. NOTE.-The square of the hypotenuse is exactly equal to the sum of the squares of the base and perpendicular. The base and perpendicular are sometimes called iegs. When the base and perpendicular are given, to find the hypotenuse. RULE 1.--Add the square of the base to the square of the perpendicular, and the square root of the sum will be the length of the hypo tenuse. When the hypotenuse and one of the legs are given, to find the other leg. RULE 2.-From the square of the hypotenuse take the square of the given leg, and the square root of the remainder will be the length of the other leg. EXAMPLES. 1. The base of the right angled triangle, A, B, C, is 40 yards, and the perpendicular 30 yards; what is the length of the hypotenuse? Ans. 50 yards. H A Base. B Perpen. NOTE. A Triangle is a Figure bounded by three sides. A right angled triangle is that which has one right angle, like the angle at B in the annexed Figure. An angle is the opening between two lines meeting in a point. A right angle is that which is made by one line perpendicular to the other, thus, B, C, falling perpendicular to A, B, forms a right angle at B. The line H, is called the Hypotenuse. 2. The length of the hypotenuse is 45 feet, and one leg 36 feet; what is the length of the other leg or side? Ans. 27 feet. 3. The wall of a building is 30 feet high, and the street in front is 40 feet wide; what is the length of a ladder that will reach from the top of the wall to the opposite side of the street? Ans. 50 feet. NOTE. A knowledge of this problem will be found very useful in mechanical business, in finding the length of braces, rafters, &c. PROBLEM. V.--Having the diameter of a tube or pipe given, which will discharge a given quantity of water in a certain time, to find the diameter of another tube which shall discharge the same, or a different quantity, in a given time. |