Therefore the sides of the triangle are expressed in the integers 3,

4, and 5.

Again, 541. Therefore the hypothenuse exceeds the base by unity.

Now 3 X 4

12 double the area of the given triangle.

2.4

the sine of the given arc.

Then 12 5
And 4 × 4—16 the square of the radius or base of the triangle.
Then 16÷ 5 =3.2 : cosine of the given arc. See Plate 9, Fig-

ure 1.

Subtracting the cosine from the radius will give the versed sine of the arc.

PROBLEM 2.

Let 5 be the perpendicular of the given triangle; then 52 25 = the sum of the other two sides.

Therefore the sides of the given triangle are expressed by the integers 5, 12, and 13. See Plate 9, Figure 2.

PROBLEM 3.

Let 7 be the perpendicular of the given triangle; then 72 49 = the sum of the other two sides.

=

For 49 + 1 ÷ 2 = 25 the hypothenuse of the given triangle.

Again, 491224 the base of the given triangle.

Therefore the sides of the given triangle are expressed by the integers 7, 24, and 25. See Plate 9, Figure 3.

PROBLEM 4.

Let 9 be the perpendicular of the given triangle; then 92 81 = the sum of the other two sides.

For 81+1 2 41

the hypothenuse of the given triangle.

And 81 - 1 ÷ 240 the base of the given triangle. Therefore the sides of the given triangle are expressed by the integers 9, 40, and 41, See Plate 9, Figure 4.

The other parts to the triangle are found exactly in the same manner as shown in Problem 1, and the student is recommended to make

the calculation for himself. The triangles, as may readily be seen, can be continued ad infinitum.

The following table will give the first 14 of the series, which the student can continue as far as he may desire.

APPENDIX.

A CHAPTER ON CONSTRUCTION

FOR THE USE AND BENEFIT OF

DRAUGHTSMEN, BUILDERS,

AND

MECHANICS.