120 760 A SQUARE ROOTDI ོ་ ། quotient or root, and use it for a divisor; divide the resolvend, omitting the right hand figure, and place the answer in the quotient, and also at the right of the last divisor; multiply the divisor by the figure last put on its right,(and in the quotient ;) place the product under the resolvend, and subtract it therefrom, and to the remainder bring down the next period; double the right hand figure of the last divisor, and use it for a new one; divide the resolvend as before, omitting its right hand figure; thus continue, until the periods are all brought down, the quotient is the root sought. Examples. 1. The square root of 37491129 is required. 6X636 greatest sqr.)37491129(6123 root or Ans. NOTE.-If a remainder is left after the periods are all brought down, annex periods of cyphers, and continue the operation to any exactness: the root thus found, must be expressed decimally. Examples. 1. The square root of 234321 is required. Ans. 484-0671 Ans. 8211 7t. 7. The square root of 40000000 is required. Ans. 6324.5t. CASE II. To extract the square root of whole numbers and decimals. RULE-Prepare the decimals by annexing cyphers (if occasion require) so that a dot may fall on units place of the whole numbers, then proceed as in case first. 2. The square root of 12-123 is required. The decimal prepared 12.1230. Root 3-48+ Ans. 3. The square root of 9'181 is required. Ans. 3.03t. L 4. The square root of 20.3331 is required. Ans. 4.509†. 5. The square root of 111111 is required. CASE III. Ans. 3.3333t. To extract the square root of Vulgar Fractions. RULE. Reduce compound fractions to simple ones, mixed numbers to improper fractions, and all to a common denominator, and also to its lowest terms: then extract the root of the numerator for a new numerator, and the root of the denominator for a new denominator. NOTE: If the fraction be a surd, that is such an one whose root can never exactly be found, reduce it to a decimal, and extract the root. Examples. 1. The square root of 304 is required. 2 8849 394 3 0 4 4 / 4 = Ans. 2. The square root of 49 is required. 726 SQUARE ROOT APPLIED. To form a square from any number, and to know how many can be upon a side. RULE. The square root of the number given, will be the number upon the side of the square. Examples. 1. It is required to lay out 25600 square rods of land in a square: I demand the side of the square that will contain the land. Σ 25600 160 rods on a side. 2. A gentleman purchased 3025 tiles, for the purpose of paving a square yard: I demand the number that can be upon a side. 2 ✓ 302555 upon a side. 3. A certain General commanded an army of 49284 men; and the better to secure his standard, 20 ། he gave orders to form into a square body 4 feet CASE II. Square root applied in finding the diameters of circles by having the area given. RULE.-Divide the area by 7854, the square root of the quotient is the diameter. Example. 1. I demand the length of rope to be tied to a horse's neck, that he may graze upon 7854 square feet of new feed every day, for 4 days: one end of the rope being each day fastened to the same stake First circle will contain 7854 sq. ft. Second circle will contain 15708 sq. ft. Third circle will contain 23562 sq. ft. Fourth circle will contain 31416 sq. ft. 15708785420000=141†÷1=70·5 sec. rope, 2 23562÷7854√30000=173†÷86.5 third rope. 2 31416÷785440000=200 100 fourth rope. NOTE.-The following figure will serve to illustrate the idea; each ring in the figure contains an equal area, viz. 7854 sq. ft. 1 NOTE. The diameter of a circle being given, the area is found by squareing the diameter, and multiplying its square by -7854, 2. A. B. C. and D. purchased a large grindstone, the diameter of which was 200 inches; they agreed that D. should wear off his share first, and that each man should have it alternately till they had worn off their shares; how much must each man wear off round the stone? Ans. D. 131; C. 16; B. 201; A. 50 inches. NOTE. The preceding figure represents the grindstone, and the four rings each mans' share in the same. CASE III. Square root applied in finding sides to triangles. THEOREM FIRST. The base and perpendicular given to find the hypothenuse. RULE.-Square the base and perpendicular, add the two squares together, extract the root of their sum; the root is the hypothenuse Examples. 1. A man wishes to make a ladder, that will reach from the eaves of his house, to the ground, 12 fect from the house; the eaves of the house are 20 feet from the ground; I demand the length of the ladder. |