44. Find the ratio between the areas of regular octagons inscribed in and circumscribed about the same circle.

45. (a) The apothem of a regular octagon is 10 feet. Find the width of a uniform strip laid off around it which occupies its area.

(b) Show how to construct this strip geometrically without first computing its width. Prove that the inside figure is a regular octagon.

46. A regular octagon ABCDEFGH is to be divided into four parts of equal area by means of octagons as shown in the figure. (a) If AB = 12 inches find a side of each octagon.

(b) Show how to construct the figure without first computing the

sides.

(c) Show how to construct such a figure if the four parts I, II, III, IV, are to be in any required ratios.

(d) Measure the sides of the two inner parts of the ceiling pattern and hence find the ratio of their areas.

47. The middle points, A, B, C, D, of alternate sides of a regular octagon are joined as shown in the figure. AH is perpendicular to AE and equal to it. BG, CK, and DL are constructed in the same manner.

(a) Prove that ABCD and HGKL are squares.

(b) Find the areas of ABCD and HGKL, if EF = 10 inches.

(c) What fraction of the whole octagon is occupied by the square

HGKL.

See the accompanying tile border.

MEASUREMENT OF THE CIRCLE.

349. If in a circle a regular polygon is inscribed, its perimeter may be measured.

For example, the perimeter of a regular inscribed hexagon is 6r if r is the radius of the circle. See § 334, Ex. 10.

If the number of sides

of the inscribed polygon be doubled, the resulting perimeter may be measured or computed in terms of r. See § 357.

If again the number of sides be doubled, the resulting perimeter may be computed in terms of r, and so on. In a similar manner a regular polygon, say a hexagon, may be circumscribed about a circle and its perimeter expressed in terms of r.

If the number of sides of the circumscribed polygon be doubled, its perimeter may again be computed, and so on as often as may be desired.

350. By continuing either of these processes it is evident that the inscribed or the circumscribed polygon may be made to lie as close to the circle as we please.

351. The word length has thus far been used in connection with straight line-segments only. Thus, the perimeter of any polygon is the sum of the lengths of its sides.

352. We now assume that a circle has a definite length and that this can be approximated as nearly as we please by taking the perimeters of the successive inscribed or circumscribed polygons.

The length of a circle is often called its perimeter or circumference.

It is evident that approximate measurement is the only kind possible in the case of the circle, since no straight line unit of measure, however small, can be made to coincide with an arc of a circle.

353. Comparison of the Lengths of Two Circles. In each of two circles, o and o', let regular polygons of 6, 12, 24, 48, 96, 192, etc., sides be inscribed. Call the perimeters of the polygons in OO, P6, P12, P24, etc., and those in Oo', P's P P' etc. Let the radii of the circles be R and

69

129

249

R' respectively.

Then by the theorem of § 347, we have

however great the number of sides of the inscribed polygons. Show that the same relations hold if polygons are circumscribed about the circle.

From these considerations we are led to the following:

354. PRELIMINARY THEOREM. The lengths of two circles are in the same ratio as their radii.

355. Hence, if c and care the circumferences of two circles, R and R' their radii, and D and D' their diameters,

Hence, the ratio of the circumference to the diameter in one circle is the same as this ratio in

any other circle.

356. This constant ratio is denoted by the Greek letter π, pronounced pi.

The argument used above shows that a theorem like that of § 354 holds for every pair of polygons used in the approximation process, and hence it is established for all purposes of practical measurement or computation.

357. PROBLEM. To compute the approximate value

of π.

B

E

SOLUTION. Suppose a regular polygon of n sides is inscribed in a circle whose radius is r and let one of the sides AB be called Sn

We first obtain in terms of s, the length AC, or San, of a side of a regular polygon of 2n sides inscribed in the same circle.

Let AB be a side of the first polygon. Bisect AB at C and draw AC and BC. Then AC is a side of a regular inscribed polygon of 2n sides (Why?).

or

Then in the figure

2

(S2n)2 = AC2 = ( } AB)2 + h2 = (} §„)2 + h2. (Why?) (1) But h × DE = h (2 r − h) = AD × DB = AD2, (Why?)

Taking only the negative sign, since h<r, and squaring,

Hence, from (1), Sqn2 = 2 r2 — r √4 r2 — 8„2,

-

-

and

Sqn = √2 r2 − r √± r2 — S„2.

Since, by § 355, the value of π is the same for all circles, we take a circle whose radius is 1.

If the first polygon is a regular hexagon, then s。= 1.

2- √4 −1 = 0.51763809.

Hence, S12= √2

=

Denoting the perimeter of a regular inscribed polygon of n sides by Pn, we have,

P1212(0.51763809)= 6.21165708.

In the formula (3) let n = 12.

Hence, P24 24(0.26108238)=6.26525722.

Computing 848, P48, etc., in a similar manner, we have,

358. The Length of the Circle. By continuing this process it is found that the first five figures in the decimal remain unchanged. Hence 6.28317 is an approximation to the circumference of a circle whose radius is 1.