Let abc (Fig. 392) be the given triangle. From a draw a d at right angles to b c; bisect a d in e; through e draw fg parallel to bc, from b and c draw b f and cg parallel to de; then bfgc will be a parallelogram containing a surface exactly equal to that of the triangle abc.
Unless the parallelogram is required to be a rectangle, the lines bf and cg need not be drawn parallel to de. If a rhomboid is desired they may be drawn at an oblique angle, provided they be parallel to one another. To ascertain the area of a triangle, multiply the base be by half the perpendicular height da. In doing this it matters not which side is taken for base.
Let A (Fig. 393) be the given parallelogram, and B the given line. Produce the sides of the parallelogram, as at a, b, c, and d; make ed equal to B; through d draw cf parallel to gb; through e draw the diagonal ca; from a draw af parallel to ed; then C will be equal to A. (See Art. 340.)
Let A and B (Fig. 304) be two given squares.
Place them so as to form a right angle, as at a; Join b and c; then the square C, formed upon the line bc, will be equal in extent to the squares A and B added together. Again: if a b (Fig. 395) be equal to the side of a given square, ca, placed at right angles to a b, be the side of another given square, and cd, placed at right angles to cb, be the side of a third given square, then the square A, formed upon the line db, will be equal to the three given squares. (See Art. 353.)
The usefulness and importance of this problem are proverbial. To ascertain the length of braces and of rafters in framing, the length of stair-strings, etc., are some of the purposes to which it may be applied in carpentry. (See note to Art. 503.) If the lengths of any two sides of a right-angled triangle are known, that of the third can be ascertained. Because the square of the hypothenuse is equal to the united squares of the two sides that contain the right angle.
(1.) - The two sides containing the right angle being known, to find the hypothenuse.
Rule. - Square each given side, add the squares together, and from the product extract the square root; this will be the answer.
For instance, suppose it were required to find the length of a rafter for a house, 34 feet wide - the ridge of the roof to be 9 feet high, above the level of the wall-plates. Then 17 feet, half of the span, is one, and 9 feet, the height, is the other of the sides that contain the right angle. Proceed as directed by the rule:
119 81 = square of 9.
17 289 = square of 17.
289 = square of 17. 370 Product.
( 1 ) 370 ( 19.235+ = square root of 370; equal 19 feet 2 7/8 in., 1 1 nearly; which would be the required
( 29 )270 length of the rafter.
( 382)..900 2 764
( 3843) 13600
( 38465).207100 192325
(By reference to the table of square roots in the Appendix, the root of almost any number may be found ready calculated; also, to change the decimals of a foot to inches and parts, see Rules for the Reduction of Decimals in the Appendix.)
Again: suppose it be required, in a frame building, to find the length of a brace having a run of three feet each way from the point of the right angle. The length of the sides containing the right angle will be each 3 feet; then, as before -
9 = square of one side. 3 times 3 = 9 = square of the other side.
18 Product: the square root of which is 4.2426+ ft., or 4 feet 2 inches and 7/8 full.
(2.) - The hypothenuse and one side being known, to find the other side.
Rule. - Subtract the square of the given side from the square of the hypothenuse, and the square root of the product will be the answer.
Suppose it were required to ascertain the greatest perpendicular height a roof of a given span may have, when pieces of timber of a given length are to be used as rafters. Let the span be 20 feet, and the rafters of 3 x 4 hemlock joist. These come about 13 feet long. The known hypothenuse, then, is 13 feet, and the known side, 10 feet - that being half the span of the building.
169 = square of hypothenuse. o times 10 = 100 = square of the given side.
69 Product: the square root of which is 8.3066+ feet, or 8 feet 3 inches and 5/8 full. This will be the greatest perpendicular height, as required. Again: suppose that in a story of 8 feet, from floor to floor, a step-ladder is required, the strings of which are to be of plank 12 feet long, and it is desirable to know the greatest run such a length of string will afford. In this case, the two given sides are - hypothenuse 12, perpendicular 8 feet.
12 times 12 = 144 = square of hypothenuse, 8 times 8 = 64 = square of perpendicular.
80 Product: the square root of which is 8.9442+ feet, or 8 feet 11 inches and 5/16 - the answer, as required.
Many other cases might be adduced to show the utility of this problem. A practical and ready method of ascertaining the length of braces, rafters, etc., when not of a great length, is to apply a rule across the carpenters'-square. Suppose, for the length of a rafter, the base be 12 feet and the height 7. Apply the rule diagonally on the square, so that it touches 12 inches from the corner on one side, and 7 inches from the corner on the other. The number of inches on the rule which are intercepted by the sides of the square, 13 7/8, nearly, will be the length of the rafter in feet; viz., 13 feet and 7/8 of a foot. If the dimensions are large, as 30 feet and 20, take the half of each on the sides of the square, viz., 15 and 10 inches; then the length in inches across will be one half the number of feet the rafter is long. This method is just as accurate as the preceding; but when the length of a very long rafter is sought, it requires great care and precision to ascertain the fractions. For the least variation on the square, or in the length taken on the rule, would make perhaps several inches difference in the length of the rafter. For shorter dimensions, however, the result will be true enough.