Chapter 14 Answers - BISD Moodle

(((((((((((Chapter , Review(Propositions through ), the regular hexagon(Proposition )—concluding with the regular-gon (Proposition ). In his commentary onthe Elements (Cambridge University Press, ),Sir Thomas Heath wrote: “Proclus refers to thisproposition in illustration of his statement thatEuclid gave proofs of a number of propositionswith an eye to their use in astronomy. ‘With regardto the last proposition in the fourth Book in whichhe inscribes the side of the fifteen-angled figure ina circle, for what object does anyone assert thathe propounds it except for the reference of thisproblem to astronomy? For, when we haveinscribed the fifteen-angled figure in the circlethrough the poles, we have the distance from thepoles both of the equator and the zodiac, sincethey are distant from one another by the side ofthe fifteen-angled figure.’ This agrees with whatwe know from other sources, namely that up tothe time of Eratosthenes (c.– B.C.) ° wasgenerally accepted as the correct measurement ofthe obliquity of the ecliptic.”Although the penny farthing bicycle reachedits peak of popularity in about , it is stillpopular in some parts of Europe. At the time ofthis writing, Hammacher Schlemmer was offeringa version of the bicycle handmade in the CzechRepublic for $,.!Concerning the Gothic arch, J. L. Heilbronwrote in Geometry Civilized—History, Culture,and Technique (Clarendon Press, ): “Theconstruction of the equilateral triangle, whichfigures so prominently in Euclid’s proofs, leads toa figure of great beauty and importance. . . . Thecurvilinear shape ACB is that of the basic Gothicarch. You can see it everywhere in churches builtduring the later middle ages. . . . At the thresholdof the study of geometry, in connection with thevery first proposition of Euclid, you haveencountered one of the most important andversatile elements of architecture.”Molecule.1. Regular hexagons and a regular pentagon.•2. It must be convex, equilateral, and equiangular.3. It becomes more circular.A Regular 15-gon.5. mAD = 72°. ( .)•6. mDB = 48°.(mDB = mAB – mAD = 120° – 72° = 48°.)7. mBE = 24°.(mBE = mDE – mDB = 72° – 48° = 24°.)8. B and E, C and F.(mBE = mCF = 24° =360°. BecausemEF = 72° = 3(24°), BE and CF are sides ofthe same regular 15-gon.)•9. 3.12. (N = 15 sin = 15 sin 12° ≈ 3.12.)•10. 3.05. (M = 15 sin 12° cos 12° ≈ 3.05.)11. π.( (•12. 6.24r. [p = 2Nr ≈ 2(3.12)r = 6.24r.]13. 3.05r 2 . (A = Mr 2 ≈ 3.05r 2 .)14. 0.42r. (s ≈ ≈ 0.42r.)Penny Farthing Bicycle.•15. About 420 turns. [ ≈ 420.]π16. About 3.77 ft. [c = 2π(0.6) = 1.2π ≈ 3.77.]17. About 1,401 turns. ( ≈ 1,401.)SAT Questions.•18. 2π 2 . [c = 2π(π) = 2π 2 .]19. . (1 = πd, d = .)ππ•20. 4π. [4π = 2πr, r = 2, A = π(2) 2 = 4π.]Semicircles on the Sides.π•21. a 2 π, b 2 π, and c 2 . [A = π( ) 2 π= a 2 , etc.]22. Yes. Because ∆ABC is a right triangle,a 2 + b 2 = c 2 . Multiplying each side of thisπ πequation by gives a 2 π+ b 2 π= c 2 .•4. mAB = 120°. ( = 120°.)