the BHM, BMH, CMR, and CRM are equal, § 209 (being measured by halves of equal arcs); .. the ABHM and CMR are equal, § 107 (having a side and two adjacent of the one equal respectively to a side and two adjacent of the other). ..LB=LC, (being homologous ▲ of equal ▲ ). = In like manner we may prove Z CZ D, etc. .. the polygon A B C D, etc., is equiangular. § 241 (being homologous sides of equal isosceles §). .. the sides A B, BC, C D, etc. are equal, Ax. 6 and the polygon A B C D, etc. is equilateral. Therefore the circumscribed polygon is regular and similar to the given inscribed polygon. $ 372 Q. E F. Ex. Let R denote the radius of a regular inscribed polygon, r the apothegm, a one side, A one angle, and C the angle at the centre; show that 1. In a regular inscribed triangle a = R √3, r = 60°, C: = 120°. 2. In an inscribed square a C = 90°. R, 3. In a regular inscribed hexagon a = A=120°, C = 60°. 4. In a regular inscribed decagon a 7= = 1 R √10 + 2 √5, A = 144°, C=36°. 1) PROPOSITION XXI. PROBLEM. 401. To find the value of the chord of one-half an arc, in terms of the chord of the whole arc and the radius of the circle. Let A B be the chord of arc A B and AD the chord of one-half the arc A B. It is required to find the value of A D in terms of A B and R (radius). From D draw DH through the centre 0, and draw O A. HD is to the chord A B at its middle point C, § 60 (two points, O and D, equally distant from the extremities, A and B, determine the position of a 1 to the middle point of A B). 403. To compute the ratio of the circumference of a circle to its diameter, approximately. Ө Let C be the circumference and R the radius of a It is required to find the numerical value of π. We make the following computations by the use of the formula obtained in the last proposition, (.51763809)2 .26105238 6.26525722 24 AD=√2 −√4 — (.51763809)2 48 AD 2-√4-(.26105238)2 .13080626 = 6.27870041 96 AD=√2-√4-(.13080626)2 .06543817 6.28206396 192 AD=√2-4-(.06543817)2 .03272346 6.28290510 384 AD= 2-4-(.03272346)2 .01636228 6.28311544 768 AD=√2-√4-(.01636228)2 .00818121 6.28316941 Hence we may consider 6.28317 as approximately the circumference of a O whose radius is unity. 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 Let the triangles A B C and EBC have the sides A B and BC equal respectively to EB and BC; and let the angle ABC be a right angle. The AABC 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 each other as their altitudes). Now ED is < EB, $ 52 But (a is the shortest distance from a point to a straight line). EB A B, .. ED is < A B. ..A ABC>A EBC. Hyp. Q. E. D. |