he gave orders to form into a square body 4 feet CASE II. by having the area given. Example. First circle will contain 7854 sq. ft. Fourth circle will contain: 31416 sq. ft. diam. ft. 7854-7854710000=100 150 first rope. 15708-785457 20000=1411==70·5 sec. rope, 23562--78543730000=1737-1786.5 third rope. 31416+7854 V 40000=200 3100 fourth rope. Note. — The following figure will serve to illustrate the idea ; each ring in the figure contains an equal area, viz. 2 2 2 2 •7854 sq. ft. Note.--The diameter of a cirele 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. 134; C. 16; B. 20); 4.50 inches. Note. The preceding figure represents the grindstone, and the four rings each mans' share in the same. CASE III, THEOREM FIRST, hypothenuse. RULE.-Square the base and perpendicular, add the two squares together, extract the root of their süm ; 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 caves of the house are 20 feet from the ground; I demand the length of the ladder. B A and A C given to find B C. 20 X 20.400 2 ✓ 544238+ B C Ans. 20 с 12 THEOREM SECOND. The base and hypothenuse given to find the perpendicular. :d ch nd RULE.-Square the base and hypothenuse; subtract the least from the greatest ; extract the root of the remainder, the root is the answer, or perpendicular, Examples 1. What is the height of the eaves of a house that requires a ladder 23.37 feet long to reach from the ground to the same; allowing the bottom of the ladder to stand 1.2 feet from the bottom of the house? ne, B A B and A C given to find C B, =144• ft. ft. 398.89=19.97, or 20 Ans. 2 the the 23.37 A 12 € THEOREM THIRD. The hypothenuse and perpendicular given, to find the base. Rule.-Square the hypothenuse and perpendicular subtract the less from the greater, extract the root of the remainder ; the root is the answer, nk base. Example 1. Hypothenuse, 23:37; perpendicular 20 ; 'required the base. Ans. 11-91, or 12 ft. CUBE ROOT. DEFINITION. A cube is the third power of any number; that is, any number multiplied by itself, and that product again by the same number, produces a Cube; thus 9X9%81%9729 is a Cube ; and 9 is the root. EXTRACTION OF THE CUBE ROOT, The extraction of the cube root is the finding of a number which has been multiplied three times by itself. RULE.-Separate the number into periods of three figures each, beginning in units place ; thus,943213427; find the greatest cube in the first period, place the root in the quotient, and its cube under the first pe riod; subtract it therefrom, and to the remainder bring down the next period, which call the dividend, or resolvend ; find a divisor by multiplying the square of the quotient by 300, seek how many times the divisor is contained in the resolvend, and put the answer in the quotient. 1 To find the Subtrahend. Multiply the divisor by the last quotient figure, and place the product under the last dividend (units under units, &c.): then square the last quotient figure, and multiply the square by the preceding figures in the quotient, and the last product multiply by 30, and place this product also under the dividend (units under units, &c.): then cube the last quotient figure, and place its cube under the dividend also (units under units, &c.) add the three numbers (which have placed under the dividend) together, and subtract their sum from the last dividend; and to the remainder bring down the next period of figures, and proceed as before, until all have been brought down, and the quotient is the root. PROOF.-Cube the quotient, or root, add in the remainder (if there is any) and the sum will be like the given cube if the work is right. 4 Examples. 3X3X3.-27)33076161(321 root Ans. 27 1st Divis. 3X3X300 2700)6076 1st dividend or resolvend. Nore.-In the above example I first found the greatest cube that was in the first period, viz. 27, and its root was 3, which I put in the quotient, and the cube I put under the first period, and subtracted it therefrom; and to the remainder (6) I brought down the next period of figures, viz. 076; which completed my first dividend or resolvend. Then for the first divisor, I sqnared the quotient or root, and multiplied the square by 300, which gave me 2700 for the first divisor; I found my divisor was contained in the dividend but twice, which I put in the quotient.--Then for the subtrahend 1 multiplied my last divisor, viz. 2700, by the last figure in the quo; tient, viz . 2, which product I put under the last dividend; and plied its square by the preceding figures in the quotient, viz. 3, and that product again by 30 and put this last product under the dividend ; I then cubed the last figure in the quotient, viz. 2, and put its cube also under the dividend ; I then found the sum of the three last numbers (standing under the dividend) and subtracted their sum from the dividend ; and to the remainder, viz. 308, I brought down the next and last period, viz, 161, and proceeded as before. 2. The cube root of 1728000 is required. Ans. 120. 3. The cube root of 997002999 is required. Ans. 999. |