a 9. Can you construct a four-sided figure with opposite angles equal, such pairs of angles having any magnitude ? If there be any restriction, what is it? 10. Show how to bisect a straight line by means of a set-square (or other triangular shape) and ruler. 11. Using protractor and ruler, on a given diagonal AB construct a four-sided figure, such that AB bisects the angles at A and B, these angles being equal. What is the figure? Apply tests. Give proof. 12. Give proof that the diagonals of a parallelogram or rhombus are in general unequal. When are they equal ? 13. At what angle do the diagonals of a rhombus intersect? Apply test. Give proof. 14. Draw AB, CD intersecting in 0, and make OA, OB, OC, OD all equal to one another. What is the figure OCBD? Apply tests and give proof. 15. If two railway tracks of the same gauge cross one another at any angle, what special kind of parallelogram is formed by the rails ? Apply tests. Give proof. 16. In the preceding question, if the tracks be of different gauges, can this special kind of parallelogram be formed ? 17. Describe a circle, and drawing any two diameters, join their extremities. What is the figure so formed? Apply tests and give proof. 18. Construct a parallelogram with angles 120° and 60°, and sides 110 and 50 millimetres. Bisect the angles of the parallelogram. What is the figure formed by the bisecting lines ? Apply test. Give proof. 19. Show that every straight line through the intersection of the diagonals of a parallelogram divides the parallelogram into two equal areas. 20. D is any point lying in the angle BAC. Construct a parallelogram ABEC, such that D may be the intersection of the diagonals. 21. D is any point lying in the angle BAC. Through D draw a line bisected at D and terminated by AB, AC. 22. On any line, with the dividers mark off equal lengths AB, BC, CD, DE....; and through A, B, C, D, .... draw, with the parallel rulers, in any direction, parallel lines cutting any line in K, L, M, N, What do you note as to the lengths of KL, LM, MN. CHAPTER VIII. A E А E D F Certain Relations in Area between Parallelograms and Triangles. 1. If two parallelograms have equal bases and equal heights, or altitudes, they are equal in area. For if such be placed, or constructed, on the same base, we shall get one of the three following cases : (1) The parallelograms may lie A as ABCD and DBCE, and the triangle EDC can be cut out, pushed to the left, and made to cover exactly the triangle DAB. Thus the area DBCE is made to coincide with the area ABCD, and they are equal. (2) The parallelograms may lie as ABCD and EBCF, and the triangle FDC can be cut out, pushed to the left, and made to cover exactly the triangle EAB. Thus the area EBCF is made to coincide with the area ABCD, and they are equal. (3) The parallelograms may lie A as ABCD and EBCF. Draw GHK parallel to BC or AF. The triangle KHC can be cut out, pushed to the left, and made to cover B exactly the triangle HGB. Next draw GL parallel to BE or CF, and KM parallel to AB or CD. The figure EHKM can now be cut out, pushed to the left, and made to cover exactly A F D 2 E 8 6 one 5 4 나 3 the figure LGHD; and the triangle FMK can be cut out, moved to the left, and made to cover exactly the triangle LAG. Thus the area EBCF is made to coincide with the area ABCD, and they are equal. If the parallelo А grams be much inclined from 6 5 another, more than one line correspond 3 ing to GHK must 2 2 be drawn. The accompanying figure illustrates how EBCF must be cut up so that its sections may exactly make up ABCD. Corresponding numbers are placed on the figures, which are to be placed on one another. It will be noticed, however, that all the triangles, on both sides, numbered from 1 to 6, can be made to coincide with one another. Several pairs of parallelograms should be constructed, each pair with the same base and between the same parallels, and the cutting just described should be done so as to show that each pair may be made to coincide. Care should be taken to illustrate the different cases that may occur. The figures, of course, must be accurately and completely drawn before the cutting is proceeded with. 2. If any triangle be cut along a straight line through the centres of two of its sides, the two parts of the triangle can be formed into a parallelogram, of course equal in area to the triangle. For, let D and E be the middle points of two sides, so that А D B the cutting is made along DE. Then let the triangle ADE be turned about E, through 180°, in the direction indicated by the arrow head. The point A arrives at C, and D at F. Since the alternate angles ADE, EFC are the same, DB is parallel to CF, and they are equal. Hence (Ch. VII., 2) DBCF is a parallelogram. Since D is the middle point of AB, it is (Ch. VII., 4) mid-way on the perpendicular through D to each of the parallels GAH and BC. Hence the triangle ABC has twice the altitude of the parallelogram DBCF into which it has been converted, G А 3. If two triangles have the same or equal bases, and equal altitudes, they are equal in area. For, let the triangles ABC, GBC, upon the same base BC, have the same altitude. The triangle ABC can, by a sec K DA F tion along DE, be converted into the parallelogram DBCF, whose altitude is half that of the triangle. Also, the triangle GBC can, by a section along HK, be converted into the parallelogram LBCK, whose altitude is half that of the triangle. Hence the parallelograms DBCF, LBCK, being on the same base and with equal altitudes, are (Ch. VIII., 1) equal in Hence the triangles are equal in area. The parallelograms DBCF, LBCK may be cut up (Ch. VIII., 1) so that the one exactly coincides with the other. Hence, in so cutting and placing the parallelograms, the original triangles are made to exactly coincide with each other, area. н we E 4. The triangle ABC is half of F GA the parallelogram ABCD, and, therefore, half of the rectangle HBCG. Hence if find E, the bisection of BC, and draw EF perpendicular to BC, the tri B angle ABC is equal to either of the rectangles HBEF or FECG. Hence to construct a rectangle equal to a triangle, bisect the base of the triangle, and on the half-base construct a rectangle of the same altitude as the triangle. Construct rectangles equal in area to the following triangles : Sides 80, 90, 140 millimetres. 8 5 5. To find the area of a rectangle we multiply the length by the breadth. Thus the adjoining rectangle being 8 units in length, and 5 units in breadth, the area is evidently 8 x 5 = 40 square units. Since the area of a triangle is half that of the rectangle on the same base and with same altitude, we may find the triangle's area by multiplying the base by the perpendicular height and dividing by 2. Calculate approximately, in square millimetres, the areas of the triangles in 4, by finding the lengths of the sides of the rectangles which have been constructed equal to the triangles. |