The square square. For this reason the root is doubled for a divisor in the operation. The parallelograms AEFH and GFIC will therefore represent the additions made to the two sides, each of which is 4 ft. wide; consequently the area of each is 20 X:4=80 ft., and the area of both is 40 X4=160 ft. But having made these additions to two sides of the square, there is a vacancy at the corner. BIFH represents this vacancy, the side of which is 4ft., or the same as the width of the additions; and its area is 4X4=16 ft. For convenience of finding the area of this vacancy, it is customary in the operation to place the last figure of the root on the right of the divisor, and thus it is multiplied into itself. The figure is now a perfect square, the length of whose side is 20+4-24 ft. 351. From these principles and illustrations we derive the following general RULE FOR EXTRACTING THE SQUARE ROOT. I. Separate the given number into periods of two figures each, by placing a point over the units' figure, another over the hundreds, and so on over each alternate figure. II. Find the greatest square number in the first or left hand period, and place its root on the right of the number for the first figure in the root. Subtract the square of this figure of the root from the period under consideration ; and to the right of the remainder bring down the next period for a dividend. III. Double the root just found and place it on the left of the dividend for a partial divisor, find how many times it is contained in the dividend, omitting its right hand figure ; place the quotient on the right of the root, also on the right of the partial divisor; multiply the divisor thus completed by the last figure of the root ; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend, as before. IV. Double the root already found for a new partial diQuest.–351. What is the first step in extracting the square root? The second? Third ? Fourth ? visor, divide, fc. as before, and thus continue the operation till the root of all the periods is extracted. Proof.-Multiply the root into itself; and if the product is equal to the given number, the work is right. (Art. 344.) Obs. The reason for the several steps in the rule may easily be inferred from the preceding illustrations. The following is a summary of them : 1. Separating the given number into periods of two figures each shows how many figures the root is to contain, (Art. 348. Obs. 3,) and thus enables us to find part of the root at a time. 2. The square of the first figure of the root is the number of feet, yards, &c. disposed of by the first figure of the root; it is subtracted from the period to find how many feet, yards, &c., remain to be added. 3. The root thus found is doubled for a partial divisor, because the addition must be made on two sides of the square already found, or it will cease to be a square. 4. In dividing, the right hand figure of the dividend is omitted, because the cipher on the right of the divisor is omitted ; otherwise the quotient would be 10 times too large for the next figure in the root. 5. The last figure of the root is placed on the right of the divisor for convenience of multiplying. The divisor is then multiplied by the last figure of the root to find the area of several additions thus made. 3. What is the square root of 625 ? 4. What is the square root of 900 ? 5. What is the square root of 1225 ? 6. What is the square root of 1764 ? 7. What is the square root of 2916 ? 8. What is the square root of 4761 ? 9. What is the square root of 8649 ? 10. What is the square root of 12321 ? 11. What is the square root of 53824 ? 12. What is the square root of 531441 ? Quest.-How is the square root proved? Obs. Why do we separate the given number into periods of two figures each? Why subtract the square of the first figure in the root from the first period Why double thie root thus found for a divisor? Why omit the right hand figure of the dividend? Why place the last figure of the root on the right of the divisor? Why multiply the divisor by the last figure in the root ? 352. When there are decimals in the given number, how are they pointed off? When there is a remainder, how proceed ? 352. If there are decimals in the given sum, they must be separated into periods like whole numbers, by placing a point over units, then over hundredths, and so on, over every alternate figure towards the right. If there is a remainder after all the periods are brought down, the operation may be continued by annexing periods of ciphers. Obs. 1. There will always be as many decimal figures in the root, as there are periods of decimals in the given number. 2. The square root of a common fraction is found by extracting the root of the numerator and denominator. 3. A mixed number should be reduced to an improper fraction. When either the numerator or denominator of a common fraction is not a square number, the fraction may be reduced to a decimal and the approximate root be found as above. 13. What is the square root of 6.25 ? Ans. 2.5. QUEST.–Obs. How do you determine how many decimal figures there should be in the root? How is the square root of a cominon fraction found? Of a mixed number? APPLICATIONS OF THE SQUARE ROOT. be Hypothenuse. Perpendicular. 353. The principles of the square root may applied to the solutions of questions in which two sides of a right-angled triangle are given, and it is required to find the other side. 354. A triangle is a figure which has three sides and three angles, as in the adjoining diagram. When one of the sides of a triangle is perpendicular to another side, the angle between them is called a right-angle. (Legendre, B. I. Def. 12.) A Base. B 355. A right-angled triangle is a triangle which has a right-angle. (Leg. B. I. Def. 17.) The side opposite the right-angle is called the hypothenuse, and the other two sides, the base and perpendicular. The triangle ABC is right-angled at B, and the side AC is the hypothenuse. 356. It is an established principle in geometry, that the square described on the hypothenuse of a right-angled triangle, is equal to the sum of the squares described on the other two sides. (Leg. IV. 11. Euc. I. 47.) Thus if the base of the triangle ABC is 4 feet, and the perpendicular 3 feet; then the square of 4 added to the square of 3 is equal to the square of the hypothenuse BC; that is, (4)2+(3)2, or 16+9=25 the square of the hypothenuse; therefore the square root of 25, which is 5, must be the hypothenuse itself. Hence, when any two sides of a right-angled triangle are given, the third side may be easily found. QUEST.–354. What is a triangle? What is a right-angle? 355. What is a-right-angled triangle ? Draw a right-angled triangle upon the black board. What is the side opposite the right-angle called? What are the other two sides called ? 356. What is the square described on the hypothenuse equal to ? Draw a right-angled triangle, and describe a square on each of its sides. 357. When the base and perpendicular are given, to find the hypothenuse. Add the square of the base to the square of the perpendicular, and the square root of the sum will be the hypoth enuse. Thus, in the right-angled triangle ABC, if the base is 4 and the perpendicular is 3, then (4)2+(3)2 =25, and V25=5, the hypothenuse. 358. When the hypothenuse and base are given, to find the perpendicular. From the square of the hypothenuse subtract the square of the base, and the square root of the remainder will be the perpendicular. Thus if the hypothenuse is 5 and the base 4, then (5)2 -(4)2 =9, and V9=3, the perpendicular. 359. When the hypothenuse and the perpendicular are given, to find the base. From the square of the hypothenuse subtract the square of the perpendicular, and the square root of the remainder will be the base. Thus if the hypothenuse is 5, and the perpendicular 3, then (5)2 –(3)2 =16, and V16=4, the base. 28. What is the length of a ladder which will just reach to the top of a house 32 feet high, when its foot is placed 24 feet from the house ? Operation. (24)2 =24 X24= 576 The square root of their sum 1600 (40. Ans. 29. The side of a certain school-room having square corners, is 8 yards, and its width 6 yards : what is the distance between two of its opposite corners ? Quest.-357. When the base and perpendicular are given, how is the hypothenuse found? 358. When the hypothenuse and base are given, how is the perpendicular found? 359. When the hypothenuse and perpendicular are given, how is the base found? |