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Draw the diameter CD perpendicular to the chord AB; and draw AC and AD; CAD is a right-angled triangle (23); therefore (II. 26)

CE:AEAE: ED

A

E

D

B

that is, ED is the same part of A E that AE is of CE. But AE is half the infinitely small chord AB (16), and AB is infinitely small in comparison with CE; therefore ED is infinitely small in comparison with A E, that is, the point E is on D, and the chord AB coincides with the arc A D B.

THEOREM VIII.

26. A circle is a regular polygon of an infinite number of sides.

If the circumference of a circle is divided into equal arcs, each infinitely small, the infinitely small chords of these ares would form a regular polygon (24) of an infinite number of sides; and as each chord would coincide with its arc (25), the polygon would be the circle itself.

27. Scholium. It might be supposed that although the difference between each chord and its arc is infinitesimal, yet as there is an infinite number of these differences their sum would not be infinitesimal and ought not to be neglected; that is, that the perimeter of the polygon and the circumference of the circle differed by a finite quantity. But each chord is infinitely small compared with the diameter of the circle, or is equal to ; and the difference between each chord and its Inf.

D

arc is infinitely smaller than the chord itself, or is equal to D ; and an infinite number of these differences is equal

Inf. X Inf.

to

D

Inf. X Inf.

D

× Inf. = ;

Inf. that is, the difference between the perimeter of the polygon and the circumference of the circle is infinitesimal.

THEOREM IX.

28. Circumferences of circles are to each other as their radii, or as their diameters.

For circles are regular polygons of an infinite number of sides (26); and if the circumferences of circles are divided into the same infinite number of arcs, the polygons formed by their chords, that is, the circles themselves, are regular polygons of the same number of sides and are therefore similar (II. 33); and the apothems of the polygons are the radii of the circles; therefore the circumferences of the circles are as their radii (II. 36), or as twice their radii, that is, as their diameters.

29. Cor. 1. If C and c denote the circumferences, R and r the corresponding radii, and D and d the corresponding diameters, we have

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That is, the ratio of the circumference of every circle to its radius or to its diameter is the same, that is, is constant. The constant ratio of the circumference to its diameter is denoted by (the Greek letter p).

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31. The area of a circle is equal to half the product of its circumference and its radius.

The area of a regular polygon is half the product of its perimeter and its apothem (II. 37); a circle is a regular polygon

of an infinite number of sides (26); the circumference of the circle is the perimeter of the polygon, and its radius is the apothem; therefore the area of a circle is half the product of its circumference and its radius.

32. Corollary. If Cthe circumference, D the diameter, R = the radius, and A = the area of a circle, we have A = CX R

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33. The side of a regular hexagon inscribed in a circle is equal to the radius of the circle.

In the circle whose centre is C draw the chord A B equal to the radius; A B is the side of a regular hexagon inscribed in a circle.

Draw the radii CA and CB; CAB is an equilateral, and therefore an equiangular triangle; hence the angle C is equal to

A

B

one third of two right angles, or one sixth of four right angles; that is, the arc A B is one sixth of the whole circumference, or the chord AB the side of a regular hexagon inscribed in the circle (12 and 24).

34. Corollary. The chord of half the arc A B would be the side of a regular dodecagon; the chord of one quarter of the arc A B, the side of a regular polygon of twenty-four sides; and

so on.

PROPOSITION XII.

PROBLEM.

35. The chord of an arc given to find the chord of half the arc.

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36. Cor. 1. If C denote the given chord, c the chord of half the arc, the equation becomes

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37. Cor. 2. If the diameter D, that is, 2 R, is unity, the equation in (36) becomes

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PROPOSITION XIII.

PROBLEM.

38. To find the arithmetical value of the constant .

From (30) C=D; if D = 1, this equation becomes С =π. If then we can find the circumference of a circle whose diameter is unity, we shall have the value of π.

If the diameter is unity, radius is one half, and the side of a regular hexagon inscribed in the circle is one half (33), and the perimeter of the hexagon is 6 X = 3.

As the diameter is unity, and the side of the inscribed hexagon one half, we can find the side of the regular inscribed dodecagon from the equation in (37):

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The perimeter of the inscribed dodecagon is therefore .25883.105+.

12

=

By using the side of the dodecagon .2588+, as C, or

=

.067 = C2, from the same equation we can find the side of a regular inscribed polygon of twenty-four sides :

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The perimeter of the inscribed polygon of twenty-four sides is

therefore 24 X .13038 3.129.

By continuing this process we approximate to the circumference, that is, to the value of π.

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