APPLICATIONS OF THE SQUARE ROOT. 353. The principles of the square root may be 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 Hypothenuse. Perpendicular. 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 1/25=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 19=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. Perpendicular (32)2=32×32=1024 Base (24)2=24×24= 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? 30. Two men start from the same place, one goes exactly south 40 miles a day, the other goes exactly west 30 miles a day: how far apart will they be at the close of the first day? 31. How far apart will the same travelers be at the end of 4 days? 32. A line 75 feet long fastened to the top of a flag staff reaches the ground 45 feet from its base: what is the height of the flag staff? 33. Suppose a house is 40 feet wide, and the length of the rafters is 32 feet: what is the distance from the beam to the ridge pole? 34. The side of a square field is 30 rods: how far is it between its opposite corners ? 35. If a square field contains 10 acres, what is the length of its side, and how far apart are its opposite corners? EXTRACTION OF THE CUBE ROOT. 360. To extract the cube root, is to resolve a given number into three equal factors; or, to find a number which being multiplied into itself twice, will produce the given number. (Art. 345.) 1. What is the side of a cubical block containing 27 solid feet? Solution. Let the given block be represented by the adjoining cubical figure, each side of which is divided into 9 equal squares which we will call square feet. Now since the length of a side is 3 feet, if we multiply 3 into 3 into 3, the product 27 will be the solid con tents of the cube. (Art. 164.) 3X3X3=27. Hence, if we reverse the process, i. e. if we resolve 27 into three equal factors, one of these factors will be the side of the cube. (Art. 344. Obs.) Ans. 3 ft. 2. A man wishes to form a cubical mound containing 15625 solid feet of earth: what is the length of its side? QUEST.-360. What is it to extract the cube root? 8 Operation. 1. We first separate the given num15625(25 ber into periods of three figures each, by placing a point over the units' figure, then over thousands. This shows us that the root must have two figures, (Art. 342. a. N. 3,) and thus enables us to find part of it at a time. Divisor. Dividend. 1200 7625 1525 7625 2. Beginning with the left hand period, we find the greatest cube of 15 is 8, the root of which is 2. Placing the 2 on the right of the given number for the first figure in the root, we subtract its cube from the period, and to the remainder bring down the next period for a dividend. This shows that we have 7625 solid feet to be added to the cubical mound already found. 3. We square the root already found, which in reality is 20; for since there is to be another figure annexed to it, the 2 is tens, and multiplying its square 400 by 3, we write the product on the right of the dividend for a divisor; and finding it is contained in the dividend 5 times, we place the 5 in the root. 4. We next multiply 20, the root already found, by 5, the last figure placed in the root; then multiply this product by 3 and place it under the divisor. We also place the square of 5, the last figure placed in the root, under the divisor, and adding these three results together, multiply their sum 1525 by 5, and subtract the product from the dividend. The answer is 25. PROOF. (25)3=25×25×25—15625, the given number. DEMONSTRATION BY CUBICAL BLOCKS. 361. The simplest method of illustrating the process of extracting the cube root to those unacquainted with algebra and geometry, is by means of cubical blocks.* 1. Place the large cube upon a table or stand. Let this represent * A set of these blocks contains 1st, a cube, the side of which is usually about 1 in. square; 2d, three side pieces about in. thick, the upper and lower base of which is just the size of a side of the cube; 3d, three corner pieces, whose ends arein. square, and whose length is the same as that of the side pieces: 4th, a small cube, the side of which is equal to the end of the corner pieces. It is desirable for every teacher and pupil to have a set. If not conveniently procured at the shops, any one can easily make them for himself. the greatest cube in the left hand period, which in the example above is 8, the root of which is 2. By subtracting 8 from the period, we find we have 7625 feet of earth to be added to this cube. In making this addition, it is plain the cube must be equally increased on three sides; otherwise its sides will become unequal, and it will then cease to be a cube. (Art. 154. Obs. 2.) 2. The object of squaring the root already found is to find the area of one side of this cube; (Art. 163 ;) we then multiply its square by 3, because the additions are to be made to three of its sides; and divide the dividend by this product to find the thickness of these additions, which is 5 feet. Now placing one of the side pieces on the top, and the other two on two adjacent sides of the cube, they will represent these additions. 3. But we perceive there is a vacancy at three corners, each of which is of the same length as the side of the cube, viz: 20 ft., and the breadth and thickness of each is 5 ft., the thickness of the side additions. Placing the corner pieces in these vacancies, they will represent the additions necessary to fill them. The object of multiplying 20 ft. their common length, by 5, their common width, is to obtain the area of a side of one of them; we then multiply this area by 3, to find the area of a side of each of them. 4. We find also another vacancy at one corner, whose length, breadth, and thickness are each 5 ft., the same as the thickness of the side additions. This vacancy therefore is cubical. It is represented by the small cube, which being placed in it, will render the mound an exact cube again. The object of squaring 5, the figure last placed in the root, is to find the area of a side of this cubical vacancy. We now have the area of one side of each of the side additions, viz: 20X20X3=1200, the divisor; also the area of one side of each of the corner additions, viz: 20×5X3=300; and the area of one side of the cubical vacancy, viz: 5×5-25. We next multiply the sum of these areas, 1525, by 5, the thickness of each, in order to find the cubical contents of the several additions. (Art. 164.) These areas are added together, and their sum multiplied by the last figure placed in the root, for the sake of finding the solidity of all the additions at once. The result would obviously be the same, if we multiplied them separately, and then subtracted the sum of their products from the dividend. 362. From the preceding illustrations we derive the following general QUEST.-362. What is the first step in extracting the cube root? The second? Third? Fourth? Fifth? How is the cube root proved? Why point the given number into periods of three figures each? Why subtract the greatest cube from the left hand period? In finding the divisor, what does the square of the root show? Why multiply its square by 3? What does the quotient figure obtained by the divisor show? Why multiply the root previously found by the last figure placed in it? Why multiply this product by 3? What does the square of this last figure in the root show? What does the sum of these results show? Why multiply this sum by the last figure placed in the root? |