time.

solve a practical problem at the same We want a square reamer to ream a hole 3 inches in diameter; what will be the dimension of the flat sides? The diagonal is what reams the holes and this must be 3 inches, so we have c= √ a2+b2 and c=3 inches.

To find a we have a=√2—62 or knowing that a and b are the same and that c2=a2+b2 we can say a2+b2=c2 or a2 = 1⁄2c2 or a= √ 1⁄2c2.

As c=3, c2=9, and 1⁄2 of 9=4.5, then the square root of 4.5=2.121 inches which gives the side of reamer.

Calling the square in figure 2 one inch each way, we know that its area must be one square inch. Dividing this into quarters as shown, will aid in dispelling a little fog which too often arises in the matter and which leads some to claim that because a square whose sides are one inch contains one square inch, a square whose sides are 1⁄2 inch contains 1⁄2 a square inch.

A glance at the figure shows this area to be 4 of a square inch and figures back

it up by saying 1⁄2 x 1⁄2

or.5x.5=.25.

A rectangle is subject to the same laws as a square and triangle, the area and diagonal being found as with square.

The rules for circles were given under the chapter on "formula" and need not be

repeated, although a few examples will be given later.

The square is also a rectangular figure, that is all its angles are right angles; its sides are also equal. The figure A B H G in figure 2 is also a rectangle although not a square.

Figure 3 is a hexagon (meaning six sided figure), and the area can be found

by dividing it into six triangles as A O B, BOC, CO D, etc., and after finding the area of one triangle, the whole area can

be easily calculated. A handy table has been worked out as follows.

Area of Regular Figures.

To use this table, square (multiply

by itself) the length of one side and multiply by the constant.

Taking the octagon in figure 4 we can either calculate its area by dividing it into the numerous parts shown; into triangles as B O C, FOG, etc, or into rectangles as H C D G, A BJ I, and LKE F, with the four triangles H I A, BJ C, etc., and it is best to know how to do this as the table may be mislaid. But taking the table and calling one side, as B C, 2 inches, we square this (2× 2), and have 4 × 4.828427=19.313708, as the area of the octagon. The area of any other regular figure (having sides and angles equal) up to one with twelve sides, can be found by using the table as shown, still it is well to know how to work them out without the table.

The formula for the area of a ring was given in chapter 4, but in figure 5 we have a case which might readily puzzle one at first glance.

Call the outer diameter of the shaft 10 inches and the whole which is eccentric to the shaft, 5 inches in diameter; what is the area of the one sided ring?

Of course the area is exactly the same

as though the hole was in the centre of the shaft, and we use the formula:

Area .7854 (D2-d2) as before. A little study will set these problems straight.

FIG. 5.

Another problem in calculating weight will give a very good illustration.

What will be the weight of a turret as shown in figure 6, with an outside diameter of 20 inches, depth of 6 inches, center hole for stud 3 inches, and having 6 holes for tools, each 2 inches in diameter. First