22. To trace the parallel, CD, on the ground. Fix upon any convenient point, G in A B, and measure an isosceles triangle, RG K: then, at the point, P, lay down the triangle, O P Q, A K G B R с P Q equal to RG K; and PQ will be parallel to G K. 23. To divide a given line, A B, into a proposed number of equal parts; suppose six. From the extremities, A and B, draw two lines, parallel to each other, forming any con venient angle with AB; on these lines set off five equal a 3 B 3 5 24. Diagonal scales, which are constantly used, are thus constructed. Suppose a scale to 12ths of a line, A B, is required. C Having divided AB into three equal parts, draw two parallel lines, A H, BK, making any A convenient angle with A B; on these lines take four equal distances, suppose from A to D and from B to R; and through the points of division » draw four lines, parallel to AB; next H n PO I R K divide D R into three equal parts; then, if the points of division, in A B and D R, are joined diagonally, the scale is constructed. For by similar triangles, RB: BS :: RO: OP; therefore, RO being of R B, O P will be of BS, or of (or) of BA; and the next division, n n, is, &c. If Q R = CB = BA is the scale of a foot, O P is an inch, n n = 2 inches, I P = 13 inches, &c. But if we divide A B into 4 equal parts, only 3 must be taken on A A and B K to make 12ths of A B (because 4 × 3 = 12.) Generally, resolve the number to which the divisions are to be extended into two factors; then divide the given line (A B) into as many equal parts as there are units in one factor, and take as many equal parts on the other lines (A H, B K) as there are units in the other. Thus, if A B is divided into 3 equal parts, and 5 are taken on A H, B K; or if A B is divided into 5, and 3 are taken on A H, B H, in either case the scale gives 15ths of A B. On the common plain scales with mathematical instruments the equal parts on each line are 10, which give the divisions in 100ths. 25. To find a fourth proportional to three given lines, A B, BC, CD. 26. This Problem is of very extensive use in the reduction of scales, plans, and maps. We shall subjoin examples. = R S K parts, set off OD 1558, and OR = 1000; join D S, and parallel to it draw R Q, then OQ is a scale of 1000 yards. This divided and subdivided is the scale, nn, in which each of the least divisions is 100 yards. Or, without the construction, thus: the distance, OP, measured on a scale is 1.53 inches. Then, as 1558:1.53::1000: 0.98 of an inch, the length of n n, the scale of 1000 yards. 27. Let the plan in the last example be reduced to a scale of 1 inch to a mile. A 1 inch (the two scales) make the isosceles triangle, ACC; then because any two corresponding dis- A tances on the plan must be in the a same proportion as the two scales, if A R be made equal to the length of the plan, A BCD, and AS - AB its breadth, RN and SP (which are parallel to CC) will be the length and breadth of the reduced plan, a b c d. 28. Suppose a map is laid down to the scale, A B, of 4000 toises; and let it be required to adapt a scale (PK) of English miles (4 for example) of the same map. join S G, and draw R Q parallel thereto; then QO (PQ) is a scale of 4 miles. Or thus:-The length of the scale, A B, is 1.73 inches: therefore, as 4.84 m. 1.73 in.:: 4m.: 1.43 in., the length of the 4 mile square, P Q. And the map, or any part of it, may be enlarged or diminished to a proposed scale, after the manner seen in 26 for we can suppose A B CD to be a given part of a large plan. 29. To reduce a given trapezium, ABCD, to a triangle of equal area. Draw the diagonal, DB, and parallel to it draw CE, meeting A B produced in E. Join the points, DE; so shall D A B the triangle, AD E, be equal to the trapezium, ABCD; for the triangle, DBE, is equal to the triangle, DCB, having the same base and altitude. 30. To reduce an irregular polygon, A BCDE, of five sides, to a triangle of equal area. Extend the side, A E, both ways at pleasure; and draw the diagonals, CE, CA. Parallel to these diagonals draw B A the lines, D F and BG; join the points, C F, CG; and GCF will be the triangle required. Note. This and the former problem may be applied in finding the areas of trapeziums and irregular polygons, by first reducing them to triangles. |