than, there are two different stellate heptagons (411). Since 4 and 2 are both prime to 9 and less than, there are two different stellate nonagons. Since 5, 4, 3, 2, are each prime to 11 and less than, there are four different stellate undecagons. Since 6, 5, 4, 3, 2, are each prime to 13 and less than 12, there are 5 stellate 13-gons. 588. Let the sides AB, DC, of the regular pentagon ABCDE be produced to meet in F. Since ∠ABC, an exterior angle of FBC, = 108°, ∠ FBC = 72° ; .·. ∠ F = 36°. 589. The interior angles of an octagon = (8−2) × 180° = 1080°; of a decagon = (10 − 2) × 180° = 1440°; of a pentadecagon = (15 – 2) x 180° = 2340°. Each interior angle of a regular polygon of n sides = n-2 Now if n becomes n straight angles indefinitely great, = (1-2) straight angles. 2 becomes indefinitely small, and 1 2 tends towards 1 as limit. Hence each interior angle tends towards a straight angle as limit. Since each exterior angle is the supplement of the adjacent interior angle, as the latter tends towards a straight angle as limit, the former tends towards 0° as limit. 590. C = 10 in. × π= 31.42 in. S=(52×π) = 78.54 sq. in. 591. C = 30 in. × 2 π = 188.50 in. S = (302 × π) = 2827.44 sq. in. 592. π. R2=100; ... R = 10 ÷ √ ; C = 2π• R = 2 π× 10 ÷ √ = 20 √π = 35.45 in. 593. π. R2 = 6 ; ... R = √√6 ÷ π = 1.38 ft. 594. π× 2 R = 12; .. R = 12 ÷ 2 π = 1.91. 595. π. R2 = nᅲR2 ; .·. R' = √nR. 596. π× 72 − π× 52 = π× 24 = 75.398 sq. ft. 597. π. R2 - π. γ2 = 3 π• R2 ; π. γ2 = 3 π• R2 ; .*. r = √ R. 598. 25: x = 42 : 72 ; . •. x = 76.5625. Ans. 76.5625 sq. in. 599. π. 32 – (√9 + 9)2 = 9(π – 2) = 10.27. Ans. 10.27 sq. in. 600. 42 – π× 22 = 16 – 12.57 = 3.43. Ans. 3.43 sq. in. 601. Since (415) AC = R√3 = 10√3, AC = 17.32 in. 602. Since 10 = R√3, R = 10 + √3 = 5.77 in. 603. Since (417) AB = RV10-2√5 = 4√5.528, AB = 9.40 in. 604. Since 9 = RV10 – 2√5, R= 18 + 2.35 = 7.66 in. 605. Since (419) AE = R√2 - √2 = 6 x .765, AE = 4.59 in. 606. Since 10 = R√2 - √2, R = 10 ÷ .765 = 13.07 in. 607. Since (420) BF = R(√5 – 1) = 5 × 1.236, BF = 6.18 in. 608. Since 3 = R(√5 − 1), R = 6÷ 1.236 = 4.85 in. THEOREMS, p. 222. 609. Let A and A' denote an angle of the polygons of nand n + 2 sides, respectively. Then ∠ A : ∠A' = n − 2 : n n n+2 = n24: n2. 610. For that point is equidistant from all the sides (101). 611. Let the regular hexagon ABCDEF and the equilateral A ACE be inscribed in O. Join OA, OC, OE. It is easily shown that BO is a parallelogram bisected by AC; etc. 612. Let AD be the altitude on BC of equilateral ABC. On BC as diameter describe circumference BECF, cutting AC in E. Through E draw EF || to AD, cutting DC in G, and join BE. Since BEC is a right angle, AC = 2 EC; .·. AD = 2EG B (148). But EF = 2 EG (172) ; .·. AD = EF, and EF is a side of an inscribed triangle, since ∠ CEG = ∠CAD; whence CF = circumference. A E DG C F 613. For (see preceding diagram) BG = BD + DG = 3 DG, and BD = DC = 2DG; etc. 614. It is easily shown (Ex. 151) that the circumscribed equilateral triangle E4 E', the inscribed equilateral triangle. But H, the inscribed hexagon, = 2 E' (Ex. 611); ... E = 2 H; .·. E : H = H : E'. 615. In the diagram for Ex. 612, EC is the side of an inscribed hexagon, and EF that of an inscribed equilateral triangle. EF2 = 4 EG2, and EC2 = EG2 + CG2 = EG2 + 1EC2; .:. EC2 = EG2; .·. EF2 : EC2 = 4 : 4 = 3:1. Now 616. Let ABC be the inscribed equilateral triangle. Join DE, the mid points of arcs AB, AC, and let DE cut AB, AC in F, G respectively. Since DB = arc EC (Const.), DE is parallel to BC; .·. AFG is an equilateral triangle; .·. AF = AG = FG. Now & D, FAD, E, and EAG are equal, being measured by equal arcs; .·. FD = FA, GE = AG; ... DF=FG= GE. 617. Let ABCDE be a regular pentagon, and AC, AD diagonals drawn from any vertex A to the opposite vertices C, D. Circumscribe a circumference about ABCDE. Since BC = CD = DE, arc BC = arc CD = arc DE ; .·. ∠BAC = ∠ CAD = ∠ DAE. 618. We prove, as in the preceding exercise, that the angles formed by the diagonals are equal, as being measured by equal arcs; and there are n 2 of them, since they are subtended by all the n equal arcs, except the two whose chords form the divided angle A. 619. Let ABCDE be the regular pentagon, and abcde the figure formed by the intercepts of the diagonals. The figure is equiangular, since its angles are measured by equal arcs; Za for example, by one half the sum of the arcs of the circumscribing circle which are subtended by the chords AB, CD, DE. The figure is also equilateral, its sides being the bases of equal isosceles triangles. 620. Let the figure abcde in the diagram for the preceding exercise be the given pentagon. The isosceles triangles formed on each side by producing the alternate sides are easily proved equal, hence the lines AB, BC, etc., joining their vertices are easily proved equal, and the ▲ ABC, BCD, etc. are also equal. 621. In the diagram for Ex. 619, AB, bBC are easily proved equal and isos., ... aA=aB=bB=bC. Again, ∠BAa = ∠bBa, and LAbB is common to AbB and Bab; ... Ab :bB = bB: ab; or Ab : Aa = Aa : ab; ... Ab + Aa : Aa=Aa+ab: ab; ... Ab+bC: Aa = Ab: ab; or AC: Ab=Aa: ab=Ab: Aa. 622. Circumscribe a circle about the pentagon ABCDE; then ∠D = ∠CFE, each being measured by equal arcs, i.e., by the sum of three of the arcs subtended by the sides of the pentagon. Similarly ∠DEF = ∠DCF, each being measured by the sum of two of those arcs. Hence FCDE is a parallelogram (141). 623. Let ABCDE be a regular pentagon. Join AC, BE, intersecting in F. FD is a parallelogram (Ex. 622), and EF=DC=EA; ... ZEAF = ∠EFA = supplement of ∠ For ∠D or ∠EAB. Hence if trapezoid ACDE be folded back, with AE as axis, AC will take the position AC' in the same straight line with AB, and ED the position ED' parallel to C'B. In the same way trapezoid BEDC may be folded back so as to have BE in the same line with BA, and CD in the same line with CD'. It is evident that the strip D'E' may be folded back so as to coincide with the pentagon by drawing from E, any convenient point in one of its edges, EA, making ∠EAC = 72°, making AB = AE, ZABC = ∠ BAE; etc. 624. P=R√10-2√5 (417); H = R (418); D = 1R(√5 – 1) (420);... P= (10-2√5) = R2+ (6-2√5) = H2 + D2. R2 4 R2 4 625. Let P be the point from which perpendiculars Pa, Pb, Pc..... Pk, are drawn to the n equal sides of the polygon Q. Join P with the vertices, dividing into n triangles with bases each equal to a side. Then Q(Pa AB + Pb • AB + ... Pk • AB) OR being the apothem; ... Pa + Pb + ... Pk = n· OR. 627. Let ABC, DEF have the common center O, and AB be a chord of ABC tangent to DEF at D. Join OB, OD. OD is I to BD (191). Since © ABC = π· OB2, and © DEF = ᅲ. OD2, 2 ABC - DEF≈ᅲ(OB2 – OD2)BD2; and BD is the radius of a whose diameter is AB. 628. Let ABC, DAB, and EAC be semicircles described upon the hypotenuse BC and the arms AB, AC of rt. ABC. Now ©ABC=ᅲ·BC2, ¦© DAB=ᅲ• ΑΒ2, AB, and EAC = π· } AC2 (398) ; D B B 0 D C A C F E 0 F C DAB+ EAC ABC. From each of these equals take the sum of the figures FAB, GAC; then ADBF+ AECG≈△ABC. REMARK. - If the triangle is isosceles, 2 ADBF isosceles right triangle ABCAB2. These figures, known as the lunes of Hipparchus, are interesting as being probably the first curvilinear figures whose quadrature was effected. PROBLEMS, p. 224. The first eight of these problems are easily solved by means of the constructions given in Arts. 406-409, and in Art. 385. Yet as the inscribed or circumscribed trigon, pentagon, and hexagon are so frequently required in demonstration, it may be useful to suggest the easiest way of producing them. To construct an equilateral triangle inscribed in a circle, with any convenient radius OA describe a circumference. If a point A on this circumference is to be a vertex, Fmark the extremities A, B of the diameter from A; then with Bas center, and radius BO, describe an arc cutting the circumference in C and D. A, FD C, D, are the required vertices. If an inscribed hexagon is required, from both B and A as centers, with radius BO, describe arcs cutting the circum A E 0 C B ference in C, D, E, and F. Then A, E, C, B, D, F, are the required vertices. 0 C To construct a pentagon inscribed in a circle, from any center O draw a line perpendicular to the direction in which the base is to lie. Lay off any convenient equal lengths OA, AB, and describe a circumference with radius OB. Draw OC perpendicular to OA and equal to OA; join CB, and lay off CD equal to OC. From Bas center, with radius BD, describe an arc cutting the circumference in E and F. E, F are vertices of the required pentagon. A D E F B To obtain these polygons circumscribed, we first find the vertical points as above, then 629. See Art. 407. 630. See Art. 409. proceed by Art. 385. 633. See Arts. 407 and 385. 634. See Arts. 409 and 385. 635. See Arts. 409 and 385. 636. See Arts. 409 and 385. 637. Let ABCD be the given rectangle. At E, the mid point of AB, erect EF perpendicular to AB and equal to AE. Produce FE to meet CD in G, and produce farther to H so that GH = GD. Join FA, FB, and draw HDK, HCL to meet FA, FB produced in Kand L respectively. FLHK is easily shown to be a square. D F A B C E 638. Let ABCD be the given square, and AEF the required equilateral triangle with a vertex in A. Since AB = AD, and AE = AF, rt. A ABE = rt. △ ADF; .·. ∠ BAE = ∠ DAF. But ∠BAE BAE + + ∠ DAF or 2∠ BAE = 90° - 60° = 30° ; .. ∠BAE=15°. Hence trisect ∠BAD (Ex. 246); then bisect the two outer parts by AE, AF, and join EF; etc. 639. Let AB be a side of the given equilateral triangle. Find a line CD such that CD : AB = 1 : √2 (363); then the equilateral triangle constructed upon CD as base is the triangle required (342). 640. Let AB be a side of the given square. Find a line CD such that |