32. A circle is a figure bounded by one curve line, which is called the circumference. Plate 1. fig. 17. 33. The centre of a circle is a point A, within the figure, equidistant from every point in the circumference. 34. The radius of a circle is the distance between the centre and circumference. 35. The diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. A is the centre. CD the diameter. 36. An arch is any part of the circumference. 37. The chord of an arch is a straight line, drawn between the extremities of an arch. 38. The segment of a circle is that space contained between the chord and arch of the fame circle. 39. A regular polygon is that whose fides are all equal. 40. An irregular polygon is a figure whose fides are not all equal. 41. Polygons receive names' according to the number of their sides and angles. Thus, A trigon has 3 Gides. A tetragon 4 42. A mixed angle is that which is formed by one curved line meeting another straight line. 43. A curve-lined angle is that which is formed by the meeting of two curved lines. GEO A 2 GEOMETRICAL PROBLEMS. 1. To make an Equilateral Triangle upon a given line AB. FROM the centre A, at the distance AB, describe an arch; and from the centre B, with the same radius, describe another arch, cutting the former in C; join CA and CB. Platei. fig.18. PROBLEM II. To bisect any given line AB into two equal parts. Upon B for a centre, with a radius more than the half of AB, describe an arch; and on A for a centre, with the same radius, describe another arch, cutting the former in the points CD: Join CD, and CD will bisect AB in the point E. Plate 2. fig. 19. PROBLEM III. To erect a perpendicular from a given point A, in a given line AB. UPON any point, C for a centre, with the radius CA, de. scribe a circle, cutting the given line also in D; draw the diameter DCE, and join EA; then shall EA be the perpendicular. Plate 2. fig. 20. PRO. PROBLEM IV. To erect a perpendicular from a given point A, in a given line AB, a nother way. FROM the given point A, with any radius AC, describe an arch, cutting the given line in C; from C, with the same radius, cut the former arch in D and E; and upon these points as centres, describe arches cutting in R; join RA, and it will be the perpendicular required. Plate 2. fig. 21, PROBLEM V. From a given point C, to drop a perpendicular upon a given line AB. On C, the given point, as centre, with any convenient distance, sweep an arch, cutting the given line in the points DE; and from these points, with any radius more than half their distance, describe arches cutting each other either above or below the line ; join the point of intersection and C, and it will be the perpendicular. Plate 2. fig. 22. PROBLEM VI. To bifect a given angle ABC. From B the angular point as centre, describe an arch cutting the containing sides in D,F; on D,F for centres, describe arches of equal radii, cutting each other in E; join BE, which will bisect the angle ABC. Plate 2. fig. 23. PROPROBLEM VII. To trifect a right angle ABC. FROM the angular point B, with any radius describe the arch AC; from C as centre, with the same radius, cut the arch AC in D; and from the centre A, with the same radius cut the arch AC in E; then join DB, EB, and they will trifect the angle. Plate 2. fig. 24. FROM any two points, D and E, defcribe arches of equal radii; draw CF to touch these arches, and CF will be parallel to AB. PROBLEM IX. To divide a line AB into any number of equal parts. LET it be required to divide AB into seven equal parts, from A draw AD at any angle; and from B draw BC parallel to AD. On each of these parallel lines lay off as many equal parts as AB is to be divided into : Join the opposite points of division by ftraight lines, passing through AB, and they will divide AB as required. Plate 2. fig. 26. PROBLEM X. To find a fourth proportional to three given lines. MAKE any angle ABC: Set off the first term from B to D, the second from D to A, the third from B to E; join DE, and through through A draw AC parallel to DE; then EC will be the fourth proportional required. Plate 2. fig. 27. PROBLEM XI. To find a mean proportional between two given lines, AB, BC. MAKE AC equal AB BC; bisect the line AC in the point D, with the centre D, and radius DA, or DC, describe the fee micircle AEC; erect the perpendicular BE, and it will be the mean proportional required. Plate 2. fig. 28. PROBLEM XII. To make a triangle with three given lines, AB, BC, CA. TAKE any line AB for the base line; on the centre A, with the radius AC, describe an arch; on the centre B, with the radius BC, describe another arch, cutting the former in C; join CA and CB, and ABC is the triangle required. Plate 2. fig. 29. PROBLEM XIII. To measure any given angle from a line of chords." FROM the angular point A, with the chord of 60° for a radius, describe an arch cutting the containing fides, produced, if necessary, in the points D,E ; take the distance DE in your compasses, and apply it to the line of chords. Thus the quantity of any angle is obtained. I testi Noie, When the angle to be measured is obtuse, it must be taken off at twice. Thus, let the angle be 120°; first take 90° and 30°, or 60° and 60°, either of which will do. PRO |