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32. Around every regular polygon a circle can be drawn in such a manner, that all the vertices of the polygon shall be at the circumference of the circle.

33. Two regular polygons of the same number of sides are similar figures.

34. The sums of all the sides of two regular polygons of the same number of sides, are to each other in the same ratio, as the radii of the inscribed or circumscribed circles.

35. The areas of two regular polygons of the same number of sides, are to each other as the areas of the squares, constructed upon the radii of the inscribed or circumscribed circles.

36. The area of a regular polygon is found by multiplying the sum of all its sides by the radius of the inscribed circle, and dividing the product by 2; or we may at once multiply half the sum of all the sides by the radius of the inscribed circle, or half that radius by the sum of all the sides.

37. If the arcs subtended by the sides of a regular polygon, inscribed in a circle, are bisected, and chords drawn from the extremities of these arcs to the points of division, the new figure thus inscribed in the circle, is a regular polygon of twice the number of sides as the one first inscribed.

38. The circumference of a circle differs so little from the sum of all the sides of a regular inscribed polygon of a great number of sides,

(several thousands for instance) that, without perceptible error, the one may be taken for the other.

39. The circumferences of two circles are in proportion to the radii of these circles; that is, a straight line, as long as the circumference of the first circle, is as many times greater than a straight line as long as the circumference of the second circle, as the radius of the one is greater than the radius of the other.

40. The area of two circles are in proportion to the squares constructed upon their radii; that is, the area of the greater circle is as many times greater than the area of the smaller circle, as the area of the square upon the radius of the one, is greater than the area of the square upon the radius of the other.

41. The area of a circle is found by multiplying the circumference, given in rods, feet, inches, &c. by half the radius given in units of the same kind.

42. The circumference of a circle, whose radius is 1, is equal to the number 6,2831852, and the circumference of any other circle is found by multiplying the number 6,2831852 by the length of the radius.

43. The length of 1 degree in a circle, whose radius is 1, is equal to the number 0,0174533 The length of 1 minute

0,0002909

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44. The length of an arc given in degrees, minutes, seconds, &c. is found by multiplying the degrees by 0,0174533, the minutes by 0,0002909, the seconds by 0,0000048, &c. then adding these products together, and multiplying their sum by the radius of the circle.

45. The area of a circle, whose radius is 1, is equal to 3,1415926 square units; and the area of any other circle is found by multiplying the number 3,1415926 by the square of the radius.

46. The area of a semicircle is found by dividing the area of the whole circle by 2.

47. The area of a quadrant is found by dividing the area of the whole circle by 4.

48. The area of a sector is found by multiplying the length of the arc by half the radius.

49. In order to find the area of a segment, we first draw two radii to the extremities of the arc of that segment; then calculate the area of the sector, formed by the two radii and that arc, and subtract from it the area of the triangle formed by the two radii and the chord of the segment; the remainder is the area of the segment.*

*

* The teacher ought now to ask his pupil to demonstrate these principles.

SECTION V.

APPLICATION OF THE FORECOING PRINCIPLES TO THE SOLUTION OF GEOMETRICAL PROBLEMS.

PART I.

Problems relative to the drawing and division of lines and angles.

PROBLEM I. To construct an equilateral triangle upon a given straight line AB.

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arc of a circle, and from the point B, with the same radius AB, another arc cutting the first.

2. From the point of intersection C draw the lines AC, BC; the triangle ABC will be equilateral.

DEMONSTRATION. The three sides AB, AC, BC, of the triangles ABC, are all equal to each other; because they are radii of equal circles.

Remark. In a similar manner can an isosceles triangle be constructed upon a given basis."

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PROBLEM II. From a given point in a straight line, to draw a perpendicular to that line.

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I. SOLUTION. Let MN be the given straight line, and D the point in which another straight line is to be drawn perpendicular to it.

1. Take any distance BD on one side of the point D, and make DA equal to it.

2. From the point B, with any radius greater than BD, describe an arc of a circle, and from the point A, with the same radius, another arc cutting the first.

3. Through the point of intersection C and the point D draw a straight line CD, which will be perpendicular to the line MN.

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