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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.
AB the radius.

CD the diameter.
Note, The diameter is equal to twice the radius.

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
A pentagon 5
A hexagon 6
A heptagon 7
An octagon 8
An eneagon 9
A decagon 10, &c.

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.

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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

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