angle A EG is equal to the difference of the triangles CED, CGH, and the triangle AGD to the difference of the triangles CG H, CAD. But A EG is to AGD as EG to GD, that is, (II. 50.) because the line GC bisects the angle ACD, as EC to CD or CA, that is again, as the triangle CED to the triangle CAD (II. 39.). And, because CED is to CAD as the difference of CED, CGH to the difference of C GH, CAD, L: ML-N: N-M (II. 17. Cor. 2.); that is, N is an harmonical mean between L and M (def. 17.). Therefore, &c. C D cir. E cir. ...441183852 F in. ...403311569 cir.....22236300 square, and the tangents drawnthrough the same a circumscribed square. Now it is plain that the circumscribed square is equal to 4 times the square of the radius, and the inscribed to half the circumscribed, that is, to twice the square of the radius. Therefore, if the radius be assumed for the linear unit, the inscribed square will contain 2, and the circumscribed 4 units of area. But the inscribed figure of eight sides is a mean proportional between them: therefore, the number of units of area which it contains, will be a mean proportional between 2 and 4,= √2×4, 2.8284271247 &c. to the tenth decimal place inclusively. And the number which is an harmonical mean between 4 and 2.8284271247 will, in like manner, be the number of superficial units in the circumscribed figure of 8 sides. Now, to find such a mean x between two numbers m, n, we have this proportion, m: n::m- x : x − n (II. def. 17); whence, multiplying extremes and means, m x (x−n) = n x (m-x); transposing, mx + nx = 2mn; and dividing by m+n,x= ; m+ n harmonical mean between two numbers is obtained by dividing twice their product by their sum. Thus we find the cir 2 m n that is, an 17503692 5138011 cir. .6320807 in. 5729404 I cir. 6025103 (in. K ..5877253 cir. .5951177 Leir. in. .32696 (in. 23456 M Of a calculation which has attracted so much attention, it is not impossible that the student may be curious enough to revise the steps, or even push it to a still greater degree of approximation. In doing this by the method here given, his labour will be considerably abridged by attending to the following rules. 1o. Annex one more to the decimal places which are required to be exactly ascertained, and with this additional place, use the abbreviated modes of multiplying, dividing, and extracting the square root, viz. by inverting the multiplier, cutting off successively the figures of the divisor, and dividing out when the root is obtained to half the required number of places (See Arith. art, 167. 185.) 2o. When the calculation has proceeded so far that (a being the difference of the two preceding poly. H a 3 16 62 in the last decimal place (or which is the same thing nearly, 7x3 in the last place but three), the geometrical mean may be obtained by taking the arithmex 2 tical mean and subtracting therefrom 14°. When 36 x 2 46 place, (or 8 r2 in the last but two) neither the harmonical nor the geometrical mean will differ apparently from the arithmetical, which may therefore be taken for them. is not found in the last decimal Or, when this comes to be the case, instead of computing the intermediate polygonal areas, the area of the circle may be directly found to the required number of places by the following rule. "Let an inscribed polygon be the last computed; take the difference between its area and that of the preceding circumscribed divide this difference (considered as a whole number) by that powerof 2, say 2m, which is next less than it; multiply the quotient m-1 2 by 3 1 and add twice the product to the area of the inscribed polygon, placing the units of the product under the last decimal place of the area; the sum shall be the circular area required." Thus, in the preceding table of areas, the difference between the inscribed polygon L and the circumscribed polygon K is 36962; the power of 2, which is next less, is 32768; the quotient of 36962 divided by 32768 is 1.128; the number by which this is to be 1 32768 multiplied -1) or 5461; the product to 3 2 the nearest unit 6160; and 14215, together with the double of this product, is 26535, which has the remaining digits in question. The second, third, and fourth of these rules may be established by the assistance of the binomial theorem: the last is derived from. the algebraical form of a series of quantities, each of which is an arithmetical mean between the two preceding. *The letter is, however, more generally understood to represent the semicircumference of a circle whose radius is unit; this being evidently the same number which represents the circumference when the diameter is assumed for unit. In fact represents (1), the superficial area of the circle where the unit of supernicies is the square of the radius; (2) the linear value of the circumference, where the diameter is the unit of length; and (3) the linear value of the semicircumference, where the radius is the unit of length. The last of these is the meaning most commonly attached to the symbol. In the method of approximation which is adopted in the text, although the principle is perhaps more obvious, the computation is not so concise as in another method, which may be derived from the following elegant theorem. If k and I represent the radii of the circles which are inscribed in any regular polygon, and circumscribed about it; and if in and represent these radii for a regular polygon which has twice as many sides by the Greek letter, being the first letter of the Greek word which signifies circumference.* For the same number which represents the area of a circle when the radius is taken for unit, re as the former, and an equal perimeter; m shall be an arithmetical mean between k and 1, and n a geometrical mean between 1 and m. To demonstrate this: Let AB be a side, and C the centre of any regular polygon; let CD be drawn perpendicular to AB, and join CA, CB: then CD is the radius, k, of the inscribed circle, and CA the radius, 1, of the circumscribed circle. From DC produced cut off CE equal to CA or CB, and join EA, EB: from C draw CF perpendicular to EA, and therefore (I. 6. Cor. 3.) bisecting EA, and through F draw FG parallel to AB, and therefore (I. 14.) perpendicular to ED, which it cuts in the point H. H Then, because the angle AEC is equal to half the angle ACD (I. 19.), the angle AEB, or FEG, is equal to half the angle ACD also, because EF is equal to the half of EA, FG is equal (II. 30. Cor. 2.) to the half of AB: therefore FG is the side of a regular polygon, which has twice as many sides as the former, E its centre, EH the radius, m, of the inscribed circle, and EF the radius, n, of the circumscribed circle. But, because EF is equal to half EA, EH is (II. 30.) equal to half ED, or to half the sum of CD and CA; that is, m is an arithmetical mean between k and l. And, again, because from the right angle F of the triangle EFC, FH is drawn perpendicular to the hypotenuse EC, EF is a mean proportional between EC and EH (II. 34. Cor.); that is, n is a mean proportional between / and m. Therefore, &c. Hence, beginning with the square or hexagon, we may proceed, by alternate arithmetical and geometrical means, to determine these radii for a regular polygon, the number of whose sides shall exceed any given number; in which process it is evident that the values of the radii will continually approach to one another, and, therefore, to the intermediate value of the radius of a circle which has the same given perimeter. There is yet a third theorem, nearly related to the preceding, which may be applied to the purpose of this approximation. Ifk and 1 represent the radii of the circles which are circumscribed about any regular polygon, and inscribed in it, and m an arithmetical mean between them; and if k' and 'represent these radii for a regular polygon which has twice as many sides as the former, and an equal area, shall be a mean propor tional between kand 1, and a mean proportional between land m. To demonstrate this: Let AB be a side, and C the centre of any regular polygon; let CD be drawn perpendicular to AB, and join CA, CB: then CA is the radius, k, of the circumscribed circle, and CD the radius, 7, of the inscribed circle. Draw the straight line CE bisecting the angle ACD; in CD produced take CF a mean proportional between CA and CD; from F draw FG perpendicular to CE, and produce it to meet CA in H. H E Then, because CG bisects the angle FCH, and FG is perpendicular to CG, the triangle FCH is isosceles (I. 5.); and, because CHXCF is equal to CAX CD, the triangle CFH is equal to the triangle CAD (II. 40. Cor.): therefore FH is the side of a re presents also the circumference when the diameter is taken for unit, because the area of a circle, being equal to the rectangle under the semicircumference and radius (32.), bears to the square of the radius the same ratio which the semicircumference bears to the radius, or the circumference to the diameter. And hence if R be the radius of any circle, its circumference (greater than 2 R in the ratio of : 1) is 2 R: and its area (greater than R2 in the ratio of T: 1) is = TR2. It remains to observe that the circumference of a circle is incommensurable with its diameter, for which reason their ratio can never be exactly represented by numbers. This was for the first time demonstrated in the year 1761 A. D. by Lambert. During the long period for which it was only matter of conjecture, the quest of the exact numerical ratio (and that by methods not more expeditious than the above) occupied many laborious calculators. Could they have assigned any such, it is evident that they might likewise have assigned the exact value of the area of a circle, whose radius is given, and vice versâ ; because that area is (32.) equal to half the product of the radius and circumference. But the hope of arriving at a term of the approximation is now demonstrated to have been vain, and accordingly an exact solution of the celebrated problem of squaring the circle, that is, of finding a straight line, the square of which shall be exactly equal to a given circle, impracticable. At the same time, the gular polygon which has twice as many sides as the former, and an equal area, C its centre, CF the radius, k', of the circumscribed circle, and CG the radius, l', of the inscribed circle But, by the construction, CF is a mean proportional between CA and CD; that is, k' is a mean proportional between k and l. And, again, because the triangle CGF is similar to CDE, the triangle CGF is to the triangle CDE as CG2 to CD? (II. 42. Cor.); but, because the triangle CGF is equal to half CHF, that is to half CDA, CGF is to CDE as half DA to DE (II. 39.), or,because CE bisects the angle ACD as half CA+CD to CD (II. 50.); therefore (II. 12.) CG2: CA+CD CD2 :: .: CD, and (II. 37. Cor. 2.) CG is a CA+CD 2 2 mean proportional between CD and is, is a mean proportional between / and m. Therefore, &c. that This theorem is applied in the same manner as the preceding. It is necessary to observe that CG is greater than CE, and not equal to it, as is wrongly represented in the figure: for, if P be taken a third proportional to EC and ED, it may be shown that CG2 is greater than CE2 by a square which is to P2 as CA to CD. approximate solution exhibited in the number 3.1415926535 &c. is sufficient for every useful purpose. If the ratio be considered as expressed by the integer and first ten decimal places, the error committed will bear a less proportion to the whole circumference than an inch to the circuit of the earth. Instead of the number 3.1415 &c. the fractions and may also be conveniently used in cases not requiring a great degree of approximation. The first (discovered by Archimedes) will be found to fail in the third decimal place: the other (due to Metius, and remarkably made up of the odd numbers 1, 3, 5) fails in the seventh decimal place only. SECTION 5.- The circle a maximum of area, and a minimum of perimeter. In the present section it is proposed to show that of all plane figures having equal perimeters, the circle contains the greatest area; and consequently, of all plane figures containing equal areas, has the least perimeter; in other words, as it is announced in the title of the Section, that the circle is a maximum of area and a minimum of perimeter. PROP. 35. Of equal triangles upon the same base, the isosceles has the least perimeter; and, of the rest, that which has the greater vertical angle has the less perimeter. TA dD Let the triangles ABC, DBC be upon the same base B C, and between the same parallels AD, BC (I. 27.), and let the triangle ABC be isosceles: the triangle ABC shall have a less perimeter than the triangle D B C. E B K From B draw BE perpendicular to AD, and produce it to F, so that EF may be equal to EB: and join AF DF. Then, because the triangles BE A, FEA have two sides of the one equal to two sides of the other, each to each, and the included angles BE A, FEA equal to one another, A F is equal to A B (I. 4.) and the angle FÀ E to the angle B A E, that is, to ABC (I. 15.) or (I. 6.) A C B. But the angles AC B, EAC are together equal to two right angles (I. 15) ; therefore FAE, EAC are likewise equal to two right angles, and (I. 2.) FÁ, A C are in the same straight line. And because the triangles BED, FED have the two sides BE, ED of the one equal to the two FE, ED of the other, each to each, and the included angles equal to one another (I. 4.) DF is equal to DB; and it was shown that A F was equal to AB. But DF, DC are greater than FC (I. 10.); therefore D B, DC are greater than AB, AC; and, BC being added to each, the perimeter of the triangle D B C is greater than the perimeter of the triangle A B C. In the next place, let G B C be another triangle upon the same base B C, and between the same parallels, but having the angle B G C less than B D C: the perimeter of the triangle G B C shall be greater than the perimeter of the triangle D B C. Bisect B C in K, and join A K. Then, because ABC is an isosceles triangle, AK is (I. 6. Cor. 3.) at right angles to B C. And, because AK bisects BC at right angles, it passes (3. Cor. 2.) through the centre of the circle which is circumscribed about the triangle DBC (5. Cor. 2.) Take Ad equal to AD. Then, because AK passes through the centre of this circle, and bisects the chord BC, it bisects also the chord which passes through the point A parallel to BC (3. Cor. 1.); and therefore the point d is in the circumference of the circle. Now, because the angle BGC is less than BDC, the point G must lie without the circle, (15. Cor. 3.) that is, G must be some point in the line D d produced, and does not lie between the points D, d. But if it lie upon the same side of FC with the point D, FG, GC together must be greater (I. 10. Cor. 1.) than FD, DC together; and therefore, because FG is equal to BG, and FD to BD, (I. 4.) the perimeter of the triangle GBC must be greater than the perimeter of the triangle D B C. And if it lie upon the other side of FC, FG, GC together will be greater than Fd, dC together. But because the diagonals FC, Dd bisect one another (I.22.) the figure FDCd is a parallelogram, and (I. 22.) the sides Fd, d C together are equal to the sides FD, DC together. Therefore FG, GC together are greater than FD, DC together, and, as before, the perimeter of the triangle G B C is greater than the perimeter of the triangle D B C. Therefore, &c. PROP. 36. If a rectilineal figure A B C D E have not all its sides equal and all its angles equal, a figure of equal area may be found, which shall have the same number of sides and a less perimeter. B For, in the first place, if it have not all its sides equal, there must be at least two adjacent sides which are unequal. Let these be AB, AE, and join BE: and let a B E be an isosceles triangle of equal area, and upon the same base B E. Then the whole figure a BCDE is equal to the whole ABCDE; and because (35.) a B, a E together are less than AB, AE together, the figure a B C D E has been found of equal area with the figure ABCDE, and having a less perimeter. Next, if it have not all its angles equal, there must be two adjacent angles A, B, which are unequal. a Ꭺ E α E P B And, first, let the sides AE, BC, meet one another in a point P. Take Pa a mean proportional (II. 51.) between PA, PB, and make Pb equal to Pa. Then, in the first place, if one of these points, as b, lie in the corresponding side BC, that is, between B and C, join ab: the figure ab CDE shall be of equal area with the figure ABCDE, and shall have a less perimeter. For, because PA is to Pa as Pa or Pb to PB, a B joined is parallel to Ab (II. 29.). Therefore (I. 27.) the triangle a A b is equal to the triangle BAb, and the figure abCDE is equal to the figure ABCDE. And because the triangle Pab is isosceles, the angle Eab is equal to the angle Cba (I. 6. or I. 6. Cor. 2.); but the two Eab, Cba are together equal to the two EAB, CBA (I. 19.), of which one, viz. EAB, is the greater; therefore the angle Ea b is greater than. the other CBA. And these latter angles are the vertical angles of the equal triangles a A b, B A b, which stand upon the same base Ab: therefore (35.) the sides a A, ab together are less than the sides BA, B b together; and the figure abCDE has a less perimeter than the figure ABCDE. E วน B m But, in the second place, if neither of the points a, b lie in a side of the figure, but both of them in the sides produced, take any point m in BC, and join m A. Through B draw Bn parallel to m A, and join mn: the figure nm CDE shall be of equal area with the figure A B C DE, and shall have a less perimeter. The figures are of equal area, because, Bn being parallel to m A, the triangles n Am, BAm are (I. 27.) equal to one another. And because the angle m n A is (I. 8. Cor. 1.) greater than the angle ma A,* much more is it greater than the angle ba A; but the latter angle being, as in the preceding case, equal to a b C', is (I. 8. Cor. 1.) greater than ABC; therefore much more is the angle mn A greater than the angle ABC. And hence it follows, as before, that n A, n m together (35.) are less than BA, Bm together, and that the perimeter of the figure nm C D E is less than that of the figure ABCDE. The two cases in which the sides AE, B C are parallel, are easily demonstrated have not, &c. Cor. 1. Of plane rectilineal figures having the same number of sides, and containing the same area, the regular polygon has the least perimeter. For it is obvious, that a certain area being to be comprehended under a certain number of sides, the perimeter of the containing figure cannot be less than of some certain length depending on the extent of the area; that is, in other words, of figures inclosing the same area, and having the same number of sides, the perimeters cannot, any of them, be less The line ma is not drawn in the figure. than a certain line, viz. the least possible by which, under the aforesaid condition, the given area can be inclosed. But it is shown in the proposition that, except a figure have all its sides equal and all its angles equal, another may be found inclosing the same area under the same number of sides and with a less perimeter. Therefore, of all the above figures there is one only which is contained by the least possible perimeter, and that one is the regular polygon (def. 11.) Cor. 2. And hence a regular polygon contains a greater area than any other rectilineal figure having the same number of sides and the same perimeter: for a similar polygon which should have the same area with the figure, would have a less perimeter (Cor. 1.), and therefore (30.) a less area than the regular polygon which has the same perimeter. side ab have a greater number of sides than the other. The polygon which has the side ab shall be greater than the other. Let the sides AB, ab, be placed in the same straight line, and so that their middle points may coincide, as at D. Then, because ab is contained a greater number of times than AB in the common perimeter, a b is less than A B, and the From D point a lies between A and D. draw DC at right angles to A B, and therefore (3. Cor. 2. and def. 12.) passing through the centres of both the polygons: and let C be the centre of the polygon which has the side AB, and c the centre of the polygon which has the side ab; it being supposed as yet unknown whether DC or Dc is the greater of the two. Join CA, ca; through C draw Co parallel to ca to meet AD in o; and lastly, with the centre C and radius Co describe the arc m n cutting CA in m, and CD produced in n. Then, because AB is to the common perimeter of the two polygons (II. 17.) as the angle ACB to four right angles, and the com |