(16) Ө (17). A regular polygon has all its sides equal, and all its angles of an equal opening. When such a polygon is inscribed in a circle, the sides are cords of equal arcs, and the points cut the circle into equal parts. 26. Inscribe a regular octagon in a circle. (fig.16.) Draw two diameters perpendicular to each other, then divide each quarter of the circle into halves by other diameters; then draw arcs from diameter to di ameter. 27. Inscribe a regular pentagon in a circle. (fig. 17.) It is difficult by the eye alone to divide the circumference into five equal parts, and the object of this problem is to exercise the pupils. (18) (19) 28. Make a triangle, and circumscribe a circle, (fig. 18.) First make a triangle, and then the object is to describe a circle which shall cut each of its three points. To do this, raise a perpendicular on the middle of one of the sides, and then raise another on another side. These perpendiculars will cross each other, and the point of section (that is, the point where they cut each other) will be the centre of the circle required. In the figure, the dotted lines show the perpendiculars and centre. 29. Make a circle, and draw a tangent triangle. (fig. 19.) Three tangents to a circle are easily made, but the monitor may increase the difficulty by giving directions to the tangent sides. Thus, let two sides be at right angles, obtuse or acute; let the triangle be equilateral, &c. 30. Draw a regular pentagon, and circumscribe it with a circle. (fig. 17.) 31. Draw a regular hexagon, and circumscribe it with a circle. (fig. 13.) 32. Draw a regular octagon, and circumscribe it with a circle. (fig. 16.) In the former problems, the circle was made first, now the polygon. 33. Inscribe a circle in a triangle. (fig. 19.) To find the centre of the circle, draw a line from the middle of either side of the triangle to the point opposite, then do the same by another side and its opposite point; the place where these two lines cross each other, will be the centre of the circle to be inscribed. See the dotted lines in fig. 19. 34. Make an arc which shall pass through two given points. (fig. 20.) (20) (21) After having marked two points, trace an arc of a circle which shall pass through both of them. The centre must be somewhere on a perpendicular to the middle of a cord which would join these two points. 35. Make several arcs pass through two given points. (fig. 21.) Draw one arc as in problem 34, then a cord from point to point, then a perpendicular to the cord, and then you may make any number of arcs pass through the two points, all of whose centres must be on the perpendicular. This problem will assist the pupil in drawing the meridians on a map of the globe. 36. Describe a circle, and circumscribe it with a hexagon. (fig. 22.) (22) (23) Cut the circumference into six equal arcs, as if you wished to inscribe the polygon. Then draw a radius to each point, and six tangents perpendicular to the radii will form the regular polygon required. 37. Inscribe and circumscribe a circle with regular and parallel hexagons. (fig. 23.) 38. Inscribe a circle in a regular hexagon. (fig. 22.) This problem is the inverse of the 36th. First draw the hexagon, then describe the circle, touching it on all sides. The centre of the circle may be found by raising perpendiculars on the middle of any two sides until they cross each other. The point where they cross, is the centre. 39. Make a triangle whose three sides are given. (fig. 24.) (24) Trace three right lines for sides. Take one of them, the longest if you please, for the base, and then make a point where you think the other two sides will reach. The difficulty is to ascertain exactly where this point should be. With dividers it may be easily found in the following manner. After you have drawn the base, open the dividers the length of the next side to be drawn, and placing one foot of the dividers on one end of the base, draw an arc with the other foot. Then taking the length of the third side, place one foot on the other end of the base, and draw an arc which shall cross the former arc the point where the arcs cross each other is the summit or apex required. If the two arcs cannot cross, the problem is said to be absurd: for no triangle can be made of the given sides. Each of the three sides must be shorter than the two others would be if united. (1) FOURTH CLASS. 1. Draw a right line tangent to two circles. (fig. 1.) Take either of the right lines in fig. 1. The circles may be placed more or less distant from each other, and may even intersect or cut each other. A radius drawn from the centre to the point of contact will be perpendicular to the tangent. 2. Draw four tangents to two circles. (fig. 1.) There may be two interiour and two exteriour tangents. The right line which joins the centres is also the point where the tangents must intersect each other. 3. Add two squares. (fig. 2.) This figure and figure 3, present two rectangular triangles, on whose sides three squares are constructed. The two small squares have this peculiarity, that one of their sides is exactly one side of the triangle, and another is merely a prolongation of the other side of the right angle. If a semicircle be drawn on the greatest side of the triangle it must touch the apex of the triangle. It is a fact in geometry, that the greatest of these three squares, contains a surface equal to the other two added together. |