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less than the triangle CAB; for, as the ratio , viz. that of the angle at the centre perimeter of every circumscribed poly. (13.) to four right angles, are to one anogon is greater than the circumference ther, alternando, as the whole circumA B (I. 10. Scholium), and therefore, as ferences (or circles), that is, by the probefore shown, the polygon itself greater position, as the radii (or the squares of than the triangle CAB, to which, how- the radii.) ever, it may be made to approach within Cor. 2. Similar segments of circles any given difference, because its peri- are as the squares of the radii (II. 22.). meter may be made (31. Cor. 1.) to ap- For they are the differences of similar proach to AB within any given difference; sectors, and similar triangles, (def. 13. so, because similar polygons may be in- and II. 32.), which sectors (Cor. 1.) as scribed and cicumscribed approaching to also the triangles (11. 42.) are as the one another more nearly than by any squares of the radii. given difference (31.) a polygon may be

PROP. 34. inscribed approaching to the triangle CAB within any given difference, that If K and L represent two regular is greater than the circle, if the circle be polygons of the same number of sides, supposed to be less than the triangle the one inscribed in, and the other cir CAB; which is absurd.

cumscribed about, the same circle, and if Therefore the circle is neither greater M and N represent the inscribed and nor less than the triangle' C A B, that is, circumscribed polygons of twice the it is equal to it.

number of sides ; M shall be a geomeTherefore, &c.

trical mean between K and L, and N Cor. Any circular sector is equal to shall be an harmonical mean between L half the rectangle under the radius of and M. the circle, and the arc upon which it Let C be the centre stands: for it is less than the circle in of the circle, AB a the ratio of that arc to the circum. side of the inscribed ference (13.).

polygon K, C D a

radius drawn per. Prop. 33. (Euc. xii. 2.)

pendicular to, and The circumferences of circles are as therefore (3.) bisectthe radi, and their areas are as the ing A B in the point squares of the radii.

1. Then, if EF be drawn, touching the Let R, 7, represent the radii of two circle in D, and terminated by C A and circles, C, c their circumferences, and CB produced ; EF will (27. Cor. 2.) be a A a, their areas: then C:c::R:r, side of the circumscriped polygon L of and A:a:: RP : re.

the same number of sides. Also, if AD For, in the first place, there may be be joined, and at the points A and B inscribed (31. Cor. 2.) two similar poly- tangents be drawn to meet E F in the gons, the perimeters of which approach points G and H; AD and G H will be more nearly to the perimeters C, c of the sides of the polygons M and N of twice two circles, than by any the same given the number of sides (27.). difference; and the perimeter of the one Now, because the triangles C AI, polygon (30.) is to the perimeter of the CED, CAD, CG H are severally conother, always in the same ratio, viz. as tained in the polygons K, L, M, N, the R tor: therefore, C:c::R:r (II. 28.). same number of times, viz. as often as

And, in the same manner, because the angle ACD, or GC H, is contained there may be inscribed in the circles two in four right angles, the polygons are similar polygons, the areas of which (31. one to another as those triangles(11.17.). Cor. 2.) approach more nearly to the But the triangle CAI is to the triangle areas A, a of the circles, than by any the CAD as CI to CD, (II. 39.) that is same given difference; and because the (because A I, E D are parallels), as CA area of the one polygon (30.) is always to CE, that is, again, as the triangle to the area of the other in the same ratio, CAD to the triangle CED (II. 39.). viz, as R2 to r2, A:a: R2 : 72 (II. 28.). And, because CAI is to CAD as CAD Therefore, &c.

to CED, K:M::M:L (II. 17. Cor. 2.) Cor. 1. Hence, similar arcs of circles and M is a mean proportional between are as their radii; and similar sectors are K and L. as the squares of their radii. For such Again, because the triangle C GH is arcs (or sectors) being to the whole cir- double of C G D, and, therefore, equal cumferences (or circles) in the same to the quadrilateral C AGD; the uri

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angle A EG is equal to the difference cumscribed figure of 8 sides to contain of the triangles CED, 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, CAD. But AEG 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 (11. 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 apmay be applied as follows.

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

the head of their respective columns. A B D E be drawn

rin. 2.

cir. 4. at right angles to

B (in. 2.8284271247 one another: the

Icir, 3.313708 1989 straight lines joining D

с

Sin. 3.06 14674589

i cir. ... 1825978780 their extremities will

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cir.

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22236300 the same a circumscribed square. Now it

in. . 12772509 is plain that the circumscribed square is

1 cir. equal to 4 times the square of the radius,

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{ cir. .6320807 and the inscribed to half the circum

5729404 scribed, that is, to twice the square of

i cir. .. 6025103

.5877253 the radius. Therefore, if the radius be

1 cir. ...5931177 assumed for the linear unit, the inscribed square will contain 2, and the circum

{cir. scribed 4 units of area. But the in

i cir. .28076 scribed figure of eight sides is a mean

fin.

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Icir. proportional between them : therefore,

.6343 the number of units of area which it con

cir. tains, will be a mean proportional be

487

1 cir. tween 2 and 4,= N2X4,=2.8284271247 &c. to the tenth decimal place inclusively.

i cir. And the number which is an harmo

i cir.

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in.

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

in. of superficial units in the circum

cir.

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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: 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, in X (X-n) = n x (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 multi

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 polye.

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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

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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 1 a geometrifraction when expressed in decimals, ha

cal mcan betwcen l and m. 1663

To demonstrate this: significant figure in the first ten decimal places, the Let AB be a side, and C the centre of any regular harmonical mean 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 cirSince b=3.14, &c., this rule may be used when cle, and CA the radius, l, 4 6

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 b 2
EA, EB: from C draw CF

F G

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

draw FG parallel to AB, and

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

cnlar to ED, which it cuts in 86

the point H. 4o. When is not found in the last decimal Then, because the angle AEC is equal to half the

angle ACD (I. 19.), the angle AEB, or FEG, is place, (or 8 r2 in the last but two) neither the har. equal to half the angle ACD also, because EF is monical nor the geometrical mean 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 directly found to the re

circle. quireil number of places by the following rule.

But, because EF is equal to half EA, EH is * Let an inseribed polygon be the last computed; (11. 30.) equal to half E1), or to half the sum 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 2 m, which is next less than it ; multiply the quotient

the hypotenuse EC, EF is a mean proportional be

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

mean proportional betweep I and m. by

and add twice the product to the area of 3

Therefore, &c. the inscribed polygon, placing the units of the

Hence, heginning with the square or hexagon, we

product under the last decimal place of the area ; the

may proceed, by alternate arithmetical and geometri. sum shall be the circular area required."

cal means, to determine these radii for a regular polyThus, in the preceding table of areas, the difference

gon, the number of whose sides shall exceed 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 intermediate value of the radius

of a circle which has the sane given perimeter. 32768 is 1.128; the number by which this is to be 32768

There is yet a third theorem, nearly related to the multiplied -- 1) or 5461; the product to preceding, which may be applied to the purpose of

this approximation. the nearest unit 6160; and 14215, together with the If k and 1 represent the radii of the circles which double of this product, is 26535, which has the re are circumscribed about any regular polygon, and maining digits in question.

inscribed in it, and m an arithmetical mean between The second, third, and fourth of these rules may them; and if k' and ' represent these radii for a be established by the assistance of the binomial 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, k' shall be a mean propor. a series of quantities, each of which is an arithmetical tional between k and l,' 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 radius is unit; this being evidently the same polygon; let CD be drawn 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 represents (1), the superficial area of the cumscribed circle, and CD circle where the unit of superticies 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, although 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 theorem.

Then, because CG bisects the angle FCH, and FG If kand I represent the radii of the circles which is perpendicular to CG, the triangle FCH is isosare inscribed in any regular polygon, and circum celes (I. 5.); and, because CHCF is equal to scribed about it; and if m and ii represent these radii CAXCD, the triangle CFH is eqnal to the triangle, for a regular polygon which has twice as many sides CAD (I1. 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 inand 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 proporsemicircumference 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 if may also be con2 R in the ratio of r:l) is = 2 * R: and veniently used in cases not requiring a its area (greater than R2 in the ratio of great degree of approximation. The *:1) is = „R?.

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 versâ ; 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 the 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, DBC 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, , of the circumscribed circle, and CG the

BC (I. 27.), and let radius, l', of the inscribed circle

the triangle A B C be But, by the construction, CF is a mean proportional isosceles: the triangle between CA and CD; that is, k' is a mean proportional between k and l. And, again, because the tri. ABC shall have a less angle CGF is sitnilar to CDE, the triangle CGF is to

perimeter than the trithe triangle CDE as CG to CDP (II. 42. Cor.); but, because 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 (11.39.), or, because CE bisects the angle ACD as half AD, and produce it to F, so that EF CA+CD to CD (II. 50.); therefore (II. 12.) CG?; CD2 ::

CA+CD.
.: CD, and (11. 37. Cor. 2.) CG is a may be equal to EB: and join AF,

DF. Then, because the triangles BEA,

CA+CD mean proportional between CD and

that F E A have two sides of the one equal to

two sides of the other, each to each, is, it is a mean proportional between I and m. Therefore, &c.

and the included angles B EA, 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

to A B (I. 4.) and the angle F A E greater than CE, and not equal to it, as is wrongly represented in the figure : for, if 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.) ACB. But the anCG is greater than CEP by a square which is to P2 as QA O CD.

gles ACB, E A C are together equal to

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

PROP. 36. EAC are likewise equal to two right angles, and (I. 2.) FA, 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 D B; place, if it have not and it was shown that A F 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, A C; 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, and In the next place, let G B C be ano upon the same base BE. Then the ther triangle upon the same base B C, whole figure a B C D E is equal to the and between the same parallels, _but whole ABCDE; and because (35.) having the angle B G C less than B DC: 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 BC. A, B, which are unequal. And, because À 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 Ad equal to A D. Then, because AK (11.51.) 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 by lie in the cortherefore the point d is in the circum- responding side BC, that ference of the circle.

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is, between B and C, join Now, because the angle BGC is less ab: the figure abCDE 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 (I. 10. Cor. 1.) than Ab (II. 29.). Therefore (1. 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 FDCd 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, D C 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, a b together are less than the the triangle G B C is greater than the sides B A, B b together; and the figure perimeter of the triangle D BC. a b C DE has a less perimeter than the Therefore, &c.

figure ABCDE.

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