in the last column bringing down tho next period, 677, obtaining 1834677, the work is completed preparatory to finding the third root figure. To determine that figure, as before, we divide the last term of the third column by the last term of the second. We thus obtain 3, which, added to the last term of the first column, gives 1353, which sum, multiplied by the 3, gives 4059; and that product, being added to the last term of the second column, gives 611559; and that sum, multiplied by the 3, gives 1834677, which being exactly as large as the last term of the third column, on being written under it and subtracted there is no remainder. The given number is therefore a perfect power, and the cube root sought is 453. In practice, the work may be performed with less figures, by writing down in the several columns only the results. Rule.— Commence as many columns as there are units in tlte exponent of the root to be extracted, by writing the given number as the head of the right-hand column, and a cipher as the head of each of the others. Separate the given number into as many periods as possible of as many figures each as the exponent of the root requires; and having found the nearest root of the left-hand period, write it as the first figure of the required root. Write this figure in the first column, and, having added it to what stands above it, multiply the sum by the same figure, and write the product in the second column; add, in like manner, in the second column, and multiply the sum by the same figure, writing the product in the third column; and so proceed, writing the last product in the last column, and subtracting it from what stands above it. Then add the same figure to the last term of the first column, multiply the sum by the same figure, and add the product to the last term of the second column; and so on, writing the last product in the last column but one. Repeat the process, stopping each time with one column farther to the left, till the last product shall fall in the second column. Add the figure found for the root to the last term of the first column; annex one cipher to the last number in the first column, two ciphers to the last number in the second column, and so on; and to the last number in the last column bring down the next period for a dividend. Take the last term of the column next to the last for a trial divisor, and see how of en it is contained in the dividend, and write the result as the next figure of the root. Add this figure to the last term of the first column, multiplying the sum by the same figure, add the product to the second column, and so on; proceed as before, till all the periods have been brought down, or an answer sufficiently exact has been obtained. Note 1. — When any dividend will not contain its corresponding trial divisor, write a cipher in the root. bring down to the dividend another period. annex an additional cipher to the last terra of the first column, two additional ciphers to the last term of the second column, and so on; and use the same trial divisor as before, increased however by the additional ciphers. Note 2. — When the given number does not have an exact root, periods of ciphers may be annexed. Note 3. — When the root is required to many places of decimals, the work may be contracted by rejecting one figure at the right from the number in the column next to the last, two from the number in the column next farther to the left, and so on, and otherwise proceeding as directed in the rule, except that the new figure of the root is not added to the first column. As soon as all the figures are rejected in the number of the first column, the remainder of the work may be performed as in contracted division of decimals (Art. 276). Examples. 2. Required the fourth root of 1.016397 to eight places of decimals. Ans. 1.00407427. OPERATION. ,3. What is the cube root of 41673648563? Ans. 3467. 4. What is the cube root of 43614208? 5. What is the cube root of 1.05 to six places of decimals? Ans. 1.016397. 6. What is the fifth root of 184528125? Ans. 45. 7. Required the fourth root of 100 to six places of deci mals? Ans. 3.102278. 8. Required the fifth root of the fourth power of 9 to seven places of decimals? Ans. 5.7995466. APPLICATIONS OF POWERS AND ROOTS. 534i A Triangle is a figure having three sides and three angles. When one of the sides of a triangle is perpendicular to another side, the opening between them is called a right angle, and the triangle is called a right-angled triangle. The lowest side, A B, is called the base of the triangle ABC, the side B C the perpendicular, the longest side, A C, the hypothenuse, and the angle at B is a right angle. Also, the line B C, being perpendicular to the base, is the altitude.' 535. The square described upon the hypothenuse of a rightangled triangle is equivalent to the sum of the squares described upon the other two sides. Thus, if the hypothenuse A C be 5 feet, the base A B 4 feet, and the perpendicular B C 3 feet, then 52 = 43 -4- 32, or 25 = 16 -f- 9. 536. To find the hypothenuse, the base and perpendicular being given. Add together the square of the base and the square of the perpendicular, and extract the square root of the sum. Thus, if the base be 4 and the perpendicular 3, the hypothenuse will equal V^-(-~3* = \^25 = 5. 537. To find the perpendicular, the base and hypothenuse being given. Subtract the square of the bane from the square of the hypothenuse, and extract the square root of the remainder. Thus, if the base be 4 and the hypothenuse 5. the perpendicular will equal 5s _ 4* = 9~ = 3. 538. To find the base, the hypothenuse and perpendicular being given. Subtract the square of the perpendicular from the square of the hypothenuse, and extract the square root of the remainder. Thus, if the perpendicular be 3 and the hypothenuse 5, the base will equal ^ o2 — 3' = is/ 16 = 4. 539. All triangles having the same base are to each other as their altitudes. All similar triangles, and other similar rectilineal figures, are to each other as the squares of their homolo- E gous or corresponding sides. /1 Thus, the triangles ACE and A C D, I / , having the same base, A C, are to each other *~~%--~~~^,. D as the altitude E C of the one is to the alti-' tude D C of the other. ^_J/__j Also, the triangles ACE and BCD,' having their corresponding angles the same, and their sides in direct proportion, are said to be similar, and are to each other as the squares of their corresponding sides, or as (A E)2 is to (B D)2, (A C)a is to (B C)2, and (C E)s is to (C D)2 Likewise the larger square, of which A C is one of the equal sides, is to the smaller square, of which B C is one of the equal sides, as (A C)2 is to (B C)2. 54*). All circles (Art. 143) are to each other as the squares of their diameters, semidiameters, or circumferences. The circumference of a circle is the line which bounds it; and the diameter is a line drawn through the center, and terminated by the circumference; as A B and C D. Then, the larger circle, of which A B is the diameter, is to the smaller, of which C D is the diameter, as (A B)2 is to (C D)2, &c. 541. To find the side, diameter, or circumference of any surface, which is similar to a given surface. State the question as in Proportion, and square the qiven sides, diameters, or circumferences, and the square root of the fourth term of the proportion wilt be the answer required. Thus, if 12 feet be the length of a side of a triangle whose area is 72 square feet, the length of the corresponding side of a similar triangle whose area is 32 square feet would be found as follows: 72 : 32 : : 12s = 144 : 64; x/U = 8 feet, "length required. 542^ To find the area of any surface which is similar to a given surface. State the question as in Proportion, and square the given sides, diameters, or circum ferences, and the fourth term of the proportion will be the answer required. Thus, if 72 square feet be the area of a triangle of which 12 feet is one of the sides, the area of a similar triangle of which the corresponding side is 8 feet would be found as follows: 12* = 144 : 82 = 64 : : 72 sq. ft. : 32 sq. ft., area required. 54i5i To find the side of a square equal in area to any given surface. Find the square root of the given area, and that root will be the side of the area required. 544i A sphere \% a solid bounded by a continued convex surface, every part of which is equally distant from the point within called the centre. A The diameter of a sphere is a straight line passing through the centre, and terminated by the surface; as A B. B 545^ A Cone is a solid having a circle for its base, and tapering uniformly to a point, called the vertex. The altitude of a cone is its perpendicular height, or a line drawn from the vertex perpendicular to the plane of the base, as B C. The diameter of its base is a straight line drawn through the centre of the plane of the base from one side of the circle to the other; as A D. 546. Spheres are to each other as the cubes of their diameters, or of their circumferences. Similar cones are to each other as the cubes of their altitudes, or of the diameters of their bases. All similar solids are to each other as the cubes of theii homologous or corresponding sides, or of their diameters. |