Note. If common divifion be used, you must bring down as many figures, as there where periods to come down when you began with divi fion. See the last Example. Numbers like thofe in Example 8, are called furds, whofe fquare root cannot be exactly found; but by annexing cyphers as above, you may come extremely near the truth, and the further you proceed, the more exact will the root be; but for common purpofes four or five places of decimals are fufficient. TO EXTRACT the SQUARE ROOT of VULGAR FRACTIONS. RULE.. Reduce the fraction or fractional parts to their lowest terms, and if it be a mixed number, to an improper fraction; then extract the fquare root of the numerator for a new numerator, and the fquare root of the denominator for a new denominator. But if the fraction be not a compleat power, then reduce it to a decimal, and proceed as taught before, EXAMPLE I. What is the fquare root of Firft E 2. in its lowest terms is = ; then √ Firft 2704 4225 E. 3. What is the fquare root of 2204 12544 First 92625 in its lowest terms; then 12544 ? the root required. the root required. the root required. LII. THE USE OF THE SQUARE ROOT. CASE 1. To find a mean proportion between any two given numbers, RULE. Multiply the two given numbers together, and extract the fquare root of the product, which root will be the mean proportional fought. EXAMPLE EXAMPLE I. What is the mean proportional between 7 and 9? First 9x7=63; Then 63(7,93 Answer 49 149)1400 1341 1583)5900 4749 1151 Therefore, as 36: 48:: 48: 64 CASE 2. To find the fide of a fquare, equal in area to any given fuperficies, RULE. Extract the fquare root of the givien fuperficies, which root will be the fide of the fquare fought. E. 3. If the area of a circle be 33124, I demand the side of a square, whofe fuperficial content shall be equal thereto ? 33124(182 Answer. I 28)231 224 362)724 E. 4. A gentleman has a piece of ground in the form of a parallelo. gram, whofe longeft fide is 134 chains, and shortest 80 chains, which he intends to change for a fquare piece of ground of the fame area, which is to be inclosed out of a large field; you are required to find the length of the fide? First 134X80=10720; then : CASE 3. 10720,00(103,5 Anfwer: I 20310720 2065)11100 775 To find the diameter of a circle, equal in area to an ellipfis, whofe tranfverfe and conjugate axes are given, RULE. Multiply the two axes of the ellipfis together; and the square root of the product is the diameter of a circle equal to the ellipfis. CASE 4. Having the area of a circle, to find the diameter, RULE. As 355: 452 or, as: 1,273239: the area to the fquare of the diameter; or, multiply the fquare root of the area by 1,12837, and the product will be the anfwer. Note. 7854, and 3,1416, are areas of circles, whofe diameters are I and 2, and,079577 is the area of a circle, whofe circumference is 1 ; likewife 452 and 1,273239, are fquares of the diameters of circles, whose areas are 355; and 1, and 1,12831, is the diameter of a circle, whofe area is equal to a square whofe fide is 1. E. 7. In the midst of a medow well flored I took just three acres to tether my horfe; round, He mayn't graze lefs or more than three acres of ground? First 4840 X 3=14520 yards, the content of three acres; then, as 355 : 452 14520 18487,4 yards fquare of the diameter. Therefore 2)135,96 the Diameter. 67,98 Yards, length (of the cord required CASE 5. Any two fides of a right-angled triangle, A, B, C, being given, to find the remaining fide. RULE. From the fquare of the hypothenufe, fubtract the fquare of the given fide, the fquare root of the remainder gives the fide required. E. 8. The top of a castle from the ground is 45 yards high, and furrounded with a ditch 60 yards broad; what length must a ladder be, to reach from the outside of the ditch to the top of the castle; In the above figure, A B = the breadth of the ditch 60 yards; 45 yards, the height of the caftle; and A C the BC ladder required. 2025 length of the 5625(75 Yards = A C, the lenght AC, First 60X60 3600 And 45X45 49. 145)725 (of the ladder. 725 E. o. At Matlock, near the Peak, in Derbyshire, where are many furprifing curiofities in nature, is a rock by the fide of the river Derwent, rifing perpendicular to a wonderful height, which being inacceffible, I endeavoured to measure, and found by a mathematical method, that the distance between the place of obfervation and the foot of the rock, to be 55 yards, and from the top of the rock to the faid place, to be 1401 yards, (nearly); required the height of this ftupendous rock? |