Suppose the square A, Fig. 11, has an area of 5,184. Let us find the length of the side. Now, 5,184 contains two periods of two figures each, and the length of the side. will accordingly be represented by a number of two figures. The greatest number whose square is contained in 51, the left-hand period, is 7. We then form the square B, whose side is 70. This square is 702, for in the complete root 7 is to occupy the tens place. This is true because 5,184 consists of two periods, and 7 is the left-hand one of the two corresponding figures in the root.

It is clear that we must now add area to the square B in three pieces if the square A is to be duplicated. Two of these must be the length of one side of B, or 70, with an unknown width, while the third piece must be a square whose side is equal to the unknown dimension of the two rectangles. Let G represent the remaining area, or the difference between A and B. Let C and E be the two rectangles, and D the square. In F the clear portion represents that part of the square already found, while the shaded portion represents C and E.

Now, in order to complete all the given area, the width of the two rectangles and the square, or the width of G, must be nearly one-half of the difference between A and B divided by 70, the side of B. It will be not quite one-half, for we still have the square D to consider, as shown in H. In other words the area of A = that of (BG), for GC+D+ E. Then to find the unknown dimension we annex a cipher to that part of the root already found (7), double the result (70), and see how many times this result (140) is contained in G, the difference between A and B. We find that it is contained twice; so we place 2 as the other figure of the root, add it to our trial divisor (140) to form the true divisor, which is the actual length of G (142). We multiply the true divisor by 2, the second figure of our root, which represents the width of G, and we have the area of G, which is equal to the remaining area, the difference between A and B.

We have now used up all our area and have found a number (72) which, when multiplied by itself, will produce the original number, 5,184.

APPLICATION OF SQUARE ROOT.

310. A triangle is a figure bounded by three straight lines. 311. A right-angled triangle has one of its angles a right angle. The side opposite the right angle is called the "hypothenuse." The other two sides are the "base" and the "perpendicular."

312. The square of the hypothenuse of a right-angled triangle equals the sum of the squares of the other two sides. Conversely,

4

3

Fig. 12.

the square of either side equals the difference between the square of the hypothenuse and the square of the other side.

This is shown in Fig. 12, in which the sides are respectively 3 and and the hypothenuse is 5. The square of 3 is 9, the square of 4 is 16, and the square of 5 is 25; also, 16 +9 = 25.

313. To find the hypothenuse when the base and perpendicular are given, add the square of the base to the square of the perpendicular and extract the square root of the sum. To find either side when the hypothenuse and the other side are given, find the square of the hypothenuse, subtract the square of the given side, and extract the square root of the remainder.

What is

8. 769 to four decimal places..

10. A square piece of land contains 734,449 sq. ft. the length of each of its sides?

Ans. 857 ft.

11. A square field contains 1,016,064 sq. ft. What is the+ length of each side? Ans. 1,008 ft.

12. A square piece of pavement contains 48,841 stones each + with a face six inches square. What is the length of one side of the pavement? Ans. 110 ft. 6 in.

13. Of two pulleys one is placed on a ceiling with its center 3 ft. from the wall, and the other on the wall with its center four feet from the ceiling. What is the length of a belt connecting the two, making no allowance for the size of the pulleys?

MISCELLANEOUS PROBLEMS FOR PRACTICE.

Ans. 5 ft.

1. After a certain bridge had been built it was decided to add a promenade for pedestrians on one side. The promenade was to be 12 feet wide. The girder supporting the main bridge structure was 3 feet deep. How long were the braces connecting the lower side of the girder with the outer side of the promenade?

Ans. 12.36 ft.

2. In a Pratt truss bridge across Saucon Creek, Pa., the truss was 21 ft. 6 in. high, and the diagonal rods connecting opposite corners of the panels were 28 feet long. What was the width of each panel? How many panels were there if the bridge was 107 ft. 4.8 in. long?

6 panels. Ans. 17.9 ft. 81 feet across X

3. The roof of a shop in East Berlin, Ct., is at the eaves, and rises 20 feet 3 in. to the ridge. distance from ridge to eaves?

What is the

Ans.

45.28 + ft.

CUBE ROOT.

314. The Cube Root of any number is the third root or one of the three equal factors of that number. To find the cube root of any number is to resolve it into three equal factors.

First point off the number into periods of three figures, beginning (as in square root) at the decimal point, thus determining the number of figures in the root. For example, there are two figures in the cube root of 74,088 because there are two periods, thus, 74 088.

After pointing off the number into periods of three figures we find the greatest number whose cube is less than 50. This we ought to know at a glance, or we can find it by trial, as in square root. We find it to be 3 and its cube is 27. We place the 3 as the first figure of the root, and the cube (27) under the first period. Subtracting 27 from 50 and bringing down the next period gives us 23,653 for a new dividend.

To find our trial divisor we square the part of the root already found (3), and multiply by 300. Thus, 32 x 300 2,700. This is our trial divisor. We next ascertain how many times 2,700 is contained in 23,653. It is contained 7, 8, or 9 times, but as either 8 or 9 would produce too great a number we use the 7. We place 7 as the second figure of the root. Multiply the part of the root already found by 7 and by 30 and add this product (630), together with the square of the second figure of the root, to the trial divisor (2,700). Now multiplying our new divisor (3,379) by 7 we find the product equal to our dividend (23,653). We have thus as the cube root 37.