the As BH M, BMH, C MR, and CRM are equal, $ 209 (being measured by halves of equal arcs) ; ..the A BHM and C M R are equal, $ 107 (having a side and two adjacent of the one equal respectively to a side and two adjacent of the other). 1. ZB=LC, (being homologous & of equal a ). .. the polygon A B C D, etc., is equiangular. (being homologous sides of equal isosceles 8). is 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, y the apothem, a one side, A one angle, and C the angle at the centre; show that 1. In a regular inscribed triangle a = R V3, r = 1 R, A = 60°, C = 120°. 2. In an inscribed square a = RV2, p= } RV2, A = 90°, C = 90°. 3. In a regular inscribed hexagon a = R, r = 1 R V3, A = 120°, C = 60°. _ R (V5 – 1) 4. In a regular inscribed decagon a = " 2 = { R V10 + 2 V5, A = 144°, C = 36°, 2_ 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. D 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 D H through the centre 0, and draw 0 A. HD is I to the chord A B at its middle point C, $ 60 (two points, 0 and D, equally distant from the extremities, A and B, determine the position of a I to the middle point of A B). The 2 HA D is a rt. L, § 204 (being inscribed in a semicircle), . AD= D H XDC, § 289 (the square on one side of a rt. A is equal to the product of the hypotenuse by the adjacent segment made by the I let fall from the vertex of the rt. Z). Now and DH= 2 R, · AD= 2 R (R – CO). then 1D*=2 R (R – V+ 2? - 1B), = 2 R – R (V4 R – AB"). :. AD=V2 F2 – (V4 Pe – AB). = 2 R2 – Q. E. F. 402. COROLLARY. If we take the radius equal to unity, omes the equation 4 D=V2P – R (V4 ? – AB”) becomes AD=V2 – V4-AB. 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 2 R' No. when R= 1, 1 = 2: We make the following computations by the use of the formula obtained in the last proposition, AD=V2 – V4 - A B?, Length of Side. Perimeter. 12 AD=V2 - V1 – 12 .51763809 6.21165708 24 A D=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 - V4 – 0.03272346)2.01636228 6.28311544 768 AD=V2 – V4 – 0.01636228)2.00818121 6.28316941 Hence we may consider 6.28317 as approximately the circumference of a O whose radius is unity. .., which equals Donal. C 6.28317 2' = 2 ..7 = 3.14159 nearly. 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 maximum. В Let the triangles A B C and EBC have the sides A B and BC equal respectively to E B 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 ED, $ 326 (S having the same base are to each 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= AB, Hyp .. E D is < A B. .:. A ABC > A EBC. Q. E. D. |