PROPOSITION XXII. PROBLEM. 403. To compute the ratio of the circumference of a circle to its diameter, approximately. Let C be the circumference and R the radius of a circle. § 376 No. when R = 1, 1 = 5. We make the following computations by the use of the formula obtained in the last proposition, AD=V2 – V4 – ABS, Length of Side. Perimeter. 12 AD=V2 - VĂ – 12 .51763809 6.21165708 24 AD=V2 - V4 — (.51763809)2 .26105238 6.26525722 48 AD=V2 - V4– (.26105238)2 .13080626 6.27870041 96 AD=V2 -V4-(.13080626)2.06543817 6.28206396 192 AD=V2-V4 – (.06543817)2 .03272346 6.28290510 384 AD=V2-14-.03272346)2.01636228 6.28311544 768 AD=V2-14 – 0.01636228)2 .00818121 6.28316941 Hence we may consider 6.28317 as approximately the circumference of a O whose radius is unity. ..t, which equals C. _ 6.28317 Q. E. F. ON ISOPERIMETRICAL POLYGONS. — SUPPLEMENTARY. 404. Def. Isoperimetrical figures are figures which have equal perimeters. 405. DEF. Among magnitudes of the same kind, that which is greatest is a Maximum, and that which is smallest is a Minimum. Thus the diameter of a circle is the maximum among all inscribed straight lines; and a perpendicular is the minimum among all straight lines drawn from a point to a given straight line. PROPOSITION XXIII. THEOREM. 406. Of all triangles having two sides respectively equal, that in which these sides include a right angle is the maxi mum. Α ---------- B D and BC equal respectively to EB and BC; and The A ABC and E B C, having the same base B C, are to each other as their altitudes A B and E D, § 326 (A having the same base are to cach other as their altitudes). Now E D is < EB, § 52 (a I is the shortest distance from a point to a straight line). But EB= A B, Hyp. ... E D is < A B. :: A ABC>A EBC. Q. E. D. PROPOSITION XXIV. THEOREM. 407. Of all polygons formed of sides all given but one, the polygon inscribed in a semicircle, having the undetermined side for its diameter, is the maximum. Let A B, BC, C D, and D E be the sides of a polygon inscribed in a semicircle having A E for its diameter. We are to prove the polygon A B C D E the maximum of polygons having the sides A B, BC, C D, and D E. From any vertex, as C, draw C A and C E. § 204 Then the Z A C E is a rt. 2, Now the polygon is divided into three parts, A BC, C D E, and A C E The parts A B C and C D E will remain the same, if the ZAC E be increased or diminished; but the part AC E will be diminished, § 406 (of all A having two sides respectively equal, that in which these sides in clude a rt. Z is the maximum). .:. A B C D E is the maximum polygon. Q. E. D. PROPOSITION XXV. THEOREM. 408. The maximum of all polygons formed of given sides can be inscribed in a circle. Н Let A B C D E be a polygon inscribed in a circle, and A' B'C'D' E' be a polygon, equilateral with re- Draw the diameter A H. Join C H and D H. Upon C'D' (=C D) construct the AC'H' D'= ACHD, . and draw A' H. Now the polygon A B C H > the polygon A' B'C' H', $ 407 (of all polygons formed of sides all given but one, the polygon inscribed in a semicircle having the undetermined side for its diameter, is the maximum). And the polygon A EDH > the polygon A' E' D'H'. § 407 Add these two inequalities, then the polygon A BCHDE > the polygon A'B'C' I' D' E'. Take away from the two figures the equal A CH D and C'H'D'. Then the polygon A B C D E > the polygon A' B'C' D' E': Q. E. D. PROPOSITION XXVI. THEOREM. 409. Of all triangles having the same base and equal perimeters, the isosceles triangle is the maximum. Let the A AC B and A D B have equal perimeters, and let the A ACB be isosceles. We are to prove AACB>A AD B. Draw the Is C E and D F. § 326 A ACB C E A ABD – DF Produce A C to H, making C H= A C. Draw H B. The Z A B H is a rt. 2, for it will be inscribed in the semicircle drawn from C as a centre, with the radius C B. |