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angle A EG is equal to the difference cumscribed figure of 8 sides to contain of the triangles C ED, CGH, and the 3.3137084989&c. superficial units. In the triangle AGD to the difference of the same manner, from these inscribed and triangles CGH, C A D. But A E G is circumscribed figures of 8 sides, are to be to AGD as E G to G D, that is, (II. 50.) obtained the inscribed and circumscribed because the line GC bisects the angle figures of 16 sides; and so on. This ACD, as EC to CD or CA, that is again, process leads us, after 18 times doubling as the triangle CED to the triangle the number of sides,* to the following CAD (II. 39.). And, because CED is to values of the inscribed and circumCAD as the difference of CED, CGH scribed polygons of 1,048,576 sides. to the difference of CGH, CAD, L:

3.1415926535 &c. M ::L-N:N-M (II. 17. Cor. 2.); 3.1415926535 &c. values which that is, N is an harmonical mean be- differ from one another by a quantity tween L and M (def. 17.).

which does not appear in the tenth deciTherefore, &c.

mal place. But the circle is greater than Scholium.

one, and less than the other of these The proposition which has been just demonstrated, affords one of the most * In the following table of polygonal areas, sucsimple methods of approximating to the cessively computed as in the text, the letters, A, B, area of the circle : to which purpose it 32, &c. sides; and, to show the progress of the ap

C, D, &c. indicate the regular polygons of 4, 8, 16, may be applied as follows.

proximation, dots are substituted for the figures at Let the diameters

the head of their respective columns. AB, D E be drawn

in. 2. А

i cir. 4. at right angles to

B \in. 2.8284271247 one another: the

i cir, 3.3137081989

in. 3.06 14674589 straight lines joining D


i cir. .. 1825978780 their extremities will

1214451522 include an inscribed

i cir.

..517249074 E

in. ...365484905 square, and the tan

cir. ...441183852 gents drawnthrough


in. ... 403311569

i cir. 22236300 the same a circumscribed square. Now it

in. 12772509 is plain that the circumscribed square is

i cir. .17503692

in. 5133011 equal to 4 times the square of the radius,

i cir.

.6320807 and the inscribed to half the circum

.5729404 scribed, that is, to twice the square of

, cir. 6025103 K

.5877253 the radius. Therefore, if the radius be

1 cir. 5931177 assumed for the linear unit, the inscribed

{ cir.

32696 square will contain 2, and the circum


.23456 scribed 4 units of area. But the in

{ cir. .28076 'scribed figure of eight sides is a mean

5766 N

i cir. .6921 proportional between them : therefore,


.6343 the number of units of area which it con

o {

.632 tains, will be a mean proportional be

.487 cir.

.560 tween 2 and 4,= N2X4,=2.8284271247



Q &c. to the tenth decimal place inclusively.

i cir. And the number which is an harmo

i cir.

.37 nical mean between 4 and 2.828427 1247


{ cir. will, in like manner, be the number

T of superficial units in the circum

i cir. scribed figure of 8 sides. Now, to of a calculation which has attracted so much atten, find such a mean x between two num

tion, it is not impossible that the student may be

curious enough to revise the steps, or even push it to bers m, n, we have this proportion, a still greater degree of approximation. m: n :: m X: - n (II. def. 17);

In doing this by the method here given, his labour

will be considerably abridged by attending to the whence, multiplying extremes and

following rules. means, m X (X-n) = n < (m - x); 1o. Annex one more to the decimal places which transposing, mx + nx = 2 mn; and

are required to be exactly ascertained, and with this

additional place, use the abbreviated modes of inulti

2 mn dividing by m+n, x=

that is, an plying, dividing, and extracting the square root, viz.

by inverting the multiplier, cutting off successively mtn

the figures of the divisor, and dividing out when the harmonical mean between two numbers root is obtained to half the required number of places is obtained by dividing twice their pro

(See Arith. art, 167. 185.).

2o. When the calculation has proceeded so far duct by their sum. Thus we find the cir- that (x being the difference of the two preceding poly:





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42 .32






polygons: therefore, the area of the by the Greek letter «, being the first circle is correctly denoted by 3.1415926 letter of the Greek word which signifies 535 as far as the tenth decimal place circumference.* For the same number inclusively.

which represents the area of a circle This number is commonly represented when the radius is taken for unit, re




gons, and b the lesser of the two), the quotient of the as the former, and an equal perimeter; m shall be an

arithmetical mean between k and I, and n a geometrifraction when expressed in decimals, has no

cal mean betwcen l and m. 1663

To demonstrate this: significant tigure in the first ten decimal places, the Let AB be a side, and C the centre of any regular harmonical inean may be found by taking half the polygon ; let CD be drawn perpendicular to AB, and sum or arithmetical mean, and subtracting therefrom join CA, CB: then CD is the

radius, k, of the inscribed cir

E 46 Since b=3.14, &c., this rule may be used when cle, and CA the radius, 1,

of the circumscribed circle. 2 x4 does not appear in the last place but three.

From DC produced cut off CE

equal to CA or CB, and join 3o. And in like manner, when

is not found
16 62
EA, EB: from Cdraw CF



perpendicular to EA, and in the last decimal place (or which is the same thing Therefore (1. 6. Cor. 3.) binearly, 7.x3 in the last place but three), the geome- secting EA, and through F trical inean inay be obtained by taking the arithme

draw FG parallel to AB, and

therefore (1. 14.) perpenditical mean and subtracting therefrom

cular to ED, which it cuts in S b

the point H. .22 149. When

is not found in the last decimal Then, because the angle AEC is equal to half the 46

angle ACD (1. 19.), the angle AEB, or FEG, is place, (or 8.ra in the last but two) neither the har- equal to half the angle ACD: also, because EF is ironical nor the geometrical inean will differ appa- equal to the half of EA, FG is equal (11. 30. Cor. 2.) rently from the arithmetical, which may therefore

to the half of AB: therefore FG is the side of a rebe taken for them.

gular polygon, which has twice as many sides as the Or, when this comes to be the case, instead of former, E its centre, EH the radius, m, of the inscribed computing the intermediate polygonal areas, the circle, and EF the radius, n, of the circumscribed area of the circle may be directiy found to the re

circle. quireil number of places by the following rule.

But, becarise EF is equal to half EA, EH is “ Let an inscribed polygon be the last computed; (11. 31.) equal to half ED, or to half the sun of CD take the difference between its area and that of the

and CA; that is, m is an arithmetical mean between preceding circumscribed : divide this difference

k and l. And, again, because from the right angle (considered as a whole number) by that powerof 2, say

F of the triangle EFC, FH is drawn perpendicular to 2m, which is next less than it ; inultiply the quotient

the hypotenuse EC, EF is a mean proportional be

tween EC and EH (II. 34. Cor.); that is, n is a 2 1

mean proportional betweep I and m. by

and add twice the product to the area of Therefore, &c. 3 the inscribed polygon, placing the units of the pro

Hence, leginning with the square or hexagon, we duct under the last decimal place of the area ; the

may proceed, by alternute arithmetical and geometric suin shall be the circular area reqnired.”

caliceans, to deterinine these radii for a regular polyThus, in the preceding table of areas, the difference

gon, the number of whose sides shall excee.I any given between the inscribed polygon L and the circum

number; in which process it is evident that the values scribed polygon K is 36962; the power of 2, which is

of the radii will continually approach to one another, next less, is 32768; the quotient of 36962 divided by

and, therefore, to the interinediate value of the radius 32768 is 1.128; the number by which this is to be

of a circle which has the san:e given perinneter. 1 32768

There is yet a third theorem, nearly related to the multiplied 3

- 1) or 5461; the product to preceding, which may be applied to the purpose of 2

this approximation. the nearest unit 6160; and 14215, together with the

If k and I represent the radii of the circles which double of this product, is 26533, which has the re- are circumscribed about any regular polygon, and maining iligits in question.

inscribed in it, and m an arithmeticul mean between The second, third, and fourth of these rules may them; and if l' and l' represent these radii for a be established by the assistance of the binonial theo- regular polygon which has twice as many sides as the rem: the last is derived from the algebraical form of former, and an equal area, h' shall be a mean propor. a series of quantities, each of which is an arithmetical tional between K and 1, and a mean proportional mean between the two preceding.

between land m. * The letter T is, however, more generally under. To demonstrate this: stood to represent the semicircumference of a circle Let AB be a side, and C the centre of any regular whose radins is unit; this being evidently the same polygon; let CD be drawu perpendicular to AB, and number which represents the circumference when the join CA, CB: then CA is diameter is assuined for unit.

the radius, k, of the cirIn fact a represents (1), the superficial area of the cumscribed circle, and CD circle where tlie unit of superficies is the square of the the radius, l, of the inradius ; (2) the linear value of the circumference, scribed circle. Draw the where the diameter is the unit of length; and (3) the straight line CE bisecting linear value of the semicircumference, where the the angle ACD; in CD radius is the unit of length. The last of these is the produced take CF a mean meaning most commonly attached to the symbol. proportional between CA

In the method of approximation which is adopted in and CD; from F draw FG the text, althongh the principle is perhaps more ob- perpendicular to CE, and vious, the computation is not so concise as in another produce it to meet CA method, which may be derived from the following

in H. elegant theorein.

Then, because CG bisects the angle FCH, and FG if ki and I represent the radii of the circles which is perpendicular to CG, the triangle FCH is isosare inscribed in any regular polygun, and circum- celes (I. 5.); and, becanse CHxCF is equal to scribed about it; and if in and u represent these radii CAXCD, the triangle CFH is equal to the triangle for a regular polygon which has twice as many sides CAD (II. 40. Cor.): therefore FH is the side of a re

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presents also the circumference when approximate solution exhibited in the the diameter is taken for unit, because number 3.1415926535 &c. is sufficient the area of a circle, being equal to the for every useful purpose. If the ratio rectangle under the semicircumference be considered as expressed by the in. and radius (32.), bears to the square teger and first ten decimal places, the of the radius the same ratio which the error committed will bear a less propor. semicircumference bears to the radius, tion to the whole circumference than an or the circumference to the diameter. inch to the circuit of the earth.

And hence if R be the radius of any Instead of the number 3.1415 &c. the circle, its circumference (greater than fractions and {{{ may also be con2 R in the ratio of r:1) is = 2 * R: and veniently used in cases not requiring a its area (greater than R’ in the ratio of great degree of approximation. The 7:1) is FR2.

first (discovered by Archimedes) will It remains to observe that the circum- be found to fail in the third decimal ference of a circle is incommensurable place: the other (due to Metius, and with its diameter, for which reason their remarkably made up of the odd num. ratio can never be exactly represented bers 1, 3, 5) fails in the seventh decimal by numbers. This was for the first time place only. demonstrated in the year 1761 A. D. by Lambert. During the long period for SECTION 5.- The circle a maximum of which it was only matter of conjecture, area, and a minimum of perimeter. the quest of the exact numerical ratio (and that by methods not more expe- In the present section it is proposed ditious than the above) occupied many to show that of all plane figures having laborious calculators. Could they have equal perimeters, the circle contains the assigned any such, it is evident that greatest area ; and consequently, of all they might likewise have assigned the plane figures containing equal areas, has exact value of the area of a circle, whose the least perimeter; in other words, as radius is given, and vice versa; because it is announced in the title of the Secthat area is (32.) equal to half the pro- tion, that the circle is a maximum of duct of the radius and circumference, area and a minimum of perimeter. But hope of arriving at a term of the approximation is now demonstrated to

PROP. 35. have been vain, and accordingly an exact solution of the celebrated problem of

Of equal triangles upon the same base, squaring the circle, that is, of finding the isosceles has the least perimeter; a straight line, the square of which and, of the rest, that which has the shall be exactly equal to a given circle, greater vertical angle has the less peimpracticable. At the same time, the rimeter.

Let the triangles ABC, D B C be upon the same base

B C, and between the gular polygon which has twice as many sides as the

same parallels AD, former, and an equal area, C its centre, CF the radius, k', of the circumscribed circle, and CG the

BC (I. 27.), and let

d ADG radins, l', of the inscribed circle

the triangle A B C be E But, by the construction, CF is a mean proportional isosceles: the triangle between CA and CĐ; that is, k' is a mean proportional between k and l. And, again, becanse the tri

ABC shall have a less angle CGF is similar to CDE, the triangle CGF is to the triangle CDE as CG2 to CD! (11. 42. Cor.); buto perimeter than the tribecause the triangle CGF is equal to half

CHF, that is angle D B C. to half CDA, CGF is to CDE as half DA to DE From B draw BE perpendicular to (II. 39.), or, because CE bisects the angle A CD aschal? AD, and produce it to F, so that E F CA+CD to CD (II. 50.); therefore (II. 12.) CG2:

may be equal to EB: and join A F. .: CD, and (II. 37. Cor. 2.) CG is a

DĚ. Then, because the triangles BEA,

CA+CD mean proportional between CD and

FE A have two sides of the one equal to

two sides of the other, each to each, is, l' is a mean proportional between 1 and m. Therefore, &c.

and the included angles B E A, FEA This theorem is applied in the same manner as the equal to one another, A F is equal preceding. It is necessary to observe that CG is greater than CE, and not equal to it, as is wrongly to A B (I. 4.) and the angle FÅ E represented in the figure : for, if P be taken a third to the angle B A E, that is, to ABC proportional to EC and ED, it may be shown that (I. 15.) or (I. 6.) AC B. But the anCG2 is greater than CE2 by a square which is to P2 as

gles ACB, E A C are together equal to





CD2 : :





CA to CD

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two right angles (I. 15); therefore FAE,

Prop. 36. EAC are likewise equal to two right angles, and (I. 2.) FĂ, A C are in the If a rectilineal figure A B C D E have same straight line. And because the not all its sides equal and all its angles triangles BED, FED have the two equal, a figure of equal area may be sides B E, E D of the one equal to the found, which shall have the same numtwo FE, E D of the other, each to each, ber of sides and a less perimeter. and the included angles equal to one For, in the first another (I. 4.) DF is equal to B; place, if it have not and it was shown that AF was equal to all its sides equal, AB. But DF, DC are greater than there must be at least FC (I. 10.); therefore DB, DC are two adjacent sides greater than A B, AC; and, B C being which are unequal. added to each, the perimeter of the Let these be AB, AE, triangle D B C is greater than the pe- and join BE: and let a B E be an rimeter of the triangle A B C.

isosceles triangle of equal area, andIn the next place, let G B C be ano- upon the same base B E. Then the ther triangle upon the same base B C, whole figure a BCDE is equal to the and between the same parallels, _but whole ABCDE; and because (35.) having the angle B G C less than BDC: a B, a E together are less than AB, AE the perimeter of the triangle G B C shall together, the figure a B C D E has been be greater than the perimeter of the found of equal area with the figure triangle D BC.

ABCDE, and having a less perimeter. Bisect B C in K, and join A K. Then, Next, if it have not all its angles because ABC is an isosceles triangle, AK equal, there must be two adjacent angles is (I. 6. Cor. 3.) at right angles to B C. A, B, which are unequal. And, because A K bisects B C at right And, first, let the sides angles, it passes (3. Cor. 2.) through the AE, BC, meet one anocentre of the circle which is circumscribed ther in a point P. Take about the triangle DBC (5. Cor. 2.) Take Pa a mean proportional A d equal to A D. Then, because AK (II.51.) between PA, PB, passes through the centre of this circle, and make Pb equal to and bisects the chord BC, it bisects Pa. Then, in the first also the chord which passes through the place, if one of these point A parallel to BC (3. Cor. 1.); and points, as b, lie in the corTherefore the point d is in the circum- responding side BC, that ference of the circle.

is, between B and C, join Now, because the angle BGC is less ab: the figure ab CDE than BDC, the point G must lie without shall be of equal area the circle, (15. Cor. 3.) that is, G must with the figure ABCDE, be some point in the line D d produced, and shall have a less peand does not lie between the points rimeter. For, because D, d. But if it lie upon the same side of PA is to Pa as Pa or FC with the point D, FG, GC together Pb to PB, a B joined is parallel to must be greater (1. 10. Cor. 1.) than Ab (II. 29.). Therefore (I. 27.) the FD, DC together; and therefore, be- triangle a A b is equal to the triangle cause FG is equal to BG, and FD to BD, BAb, and the figure abCDE is equal to (I. 4.) the perimeter of the triangle GBC the figure ABCDE. And because the must be greater than the perimeter of triangle Pab is isosceles, the angle Eab the triangle D BC. And if it lie upon is equal to the angle Cba (I. 6. or I. 6. the other side of FC, FG, GC together Cor. 2.); but the two E ab, Cba are togewill be greater than Fd, dC together. ther equal to the two EAB, CBA (I. 19.), But because the diagonals FC, D d bi- of which one, viz. EAB, is the greater; sect one another (I.22.) the figure FD Cd therefore the angle E ab is greater than, is a parallelogram, and (I. 22.) the sides the other CBA. And these latter angles Fd, dC together are equal to the sides are the vertical angles of the equal trianFD, DC together. Therefore FG, GC gles a A b, B A b, which stand upon the together are greater than FD, DC toge- same base Ab: therefore (35.) the ther, and, as before, the perimeter of sides a A, ab together are less than the the triangle G B C is greater than the sides BA, B b together; and the figure perimeter of the triangle D BC. a b C D E has a less perimeter than the Therefore, &c.

figure ABCDE.





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But, in the second place, if neither than a certain line, viz. the least possible of the points a, b lie in a side of the by which, under the aforesaid condition, figure, but both of them

the given area can be inclosed. But it in the sides produced,

is shown in the proposition that, except take any point m in BC,

a figure have all its sides equal and all and join m A. Through

its angles equal, another may be found B draw B n parallel to

inclosing the same area under the same m A, and join mn:

15 number of sides and with a less perithe figure nm CDE

meter. Therefore, of all the above

A, shall be of equal area

figures there is one only which is conwith the figure ABC

tained by the least possible perimeter, DE, and shall have a less perimeter. and that one is the regular polygon (def. The figures are of equal area, because, 11.) B n being parallel to m A, the triangles Cor. 2. And hence a regular polygon n Am, B Am are (I. 27.) equal to one contains a greater area than any other another. And because the angle m n A rectilineal figure having the same numis (I. 8. Cor. 1.) greater than the angle ber of sides and the same perimeter : ma A,* much more is it greater than the for a similar polygon which should have angle b a A; but the latter angle being, the same area with the figure, would as in the preceding case, equal to ab C', have a less perimeter (Cor. 1.), and thereis (I. 8. Cor. 1.) greater than ABC; fore (30.) a less area than the regular therefore much more is the angle mnA polygon which has the same perimeter. greater than the angle ABC.


PROP. 37. hence it follows, as before, that n A, nm together (35.) are less than BA, B m toge- Of regular polygons having equal ther, and that the perimeter of the figure perimeters, that is greatest which has nm C D E is less than that of the figure the greatest number of sides. ABCDE.

Let A B, ab
The two cases in which the sides AE, be the sides of
B C are parallel, are easily demonstrated two regular po-

lygons having
equal perime-

ters, and let

the polygon

which has the
side ab have a greater number of sides
than the other. The polygon which has
the side ab shall be greater than the

Let the sides A B, ab, be placed in in a manner taken with little variation from the preceding, and readily to be the same straight line, and so that their apprehended by aid of the adjoined middle points may coincide, as at D. gures, in which ab is drawn through number of times than AB in the common

Then, because ab is contained a greater the middle point of AB perpendicular to AE and BC.

perimeter, ab is less than A B, and the Therefore, in every case, if a figure point a lies between A and D. From D have not, &c.

draw DC at right angles to A B, and Cor. 1. Of plane rectilineal figures therefore (3. Cor. 2. and def. 12.) passing having the same number of sides, and through the centres of both the polycontaining the same area, the regular gons: and let C be the centre of the polygon has the least perimeter.

polygon which has the side AB, and c For it is obvious, that a certain area

the centre of the polygon which has the

side ab; it being supposed as yet unbeing to be comprehended under a certain number of sides, the perimeter of known whether DC or Dc is the greater the containing figure cannot be less than of the two. Join CA, ca; through C of some certain length depending on the draw Co parallel to ca to meet AD in 0; extent of the area; that is, in other words, and lastly, with the centre C and radius

Co describe the arc mn cutting C A in of figures inclosing the same area, and

m, and CD produced in n. Then, behaving the same number of sides, the

cause AB is to the common perimeter of perimeters cannot, any of them, be less

the two polygons (II. 17.) as the angle * The line ma is not drawn in the figure.

ACB to four right angles, and the com




u n А.


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