Mathematical Questions and Solutions in Continuation of the Mathematical Columns of the Educational Times Volume 32
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Mathematical Questions and Solutions in Continuation of the Mathematical Columns of the Educational Times Volume 32

Mathematical Questions and Solutions in Continuation of the Mathematical Columns of the Educational Times Volume 32


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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1880 Excerpt: ...yg parallel to b$, meeting BC in m and -; yh parallel to dD, and Sk parallel to eC, meeting CD in h and k; and SI parallel to eE, and cm parallel to dD, meeting DE in I and m; then A Abe = &A0y--b$yc, and AABC = AA + ffy$f+ 2 A0B/ = A A07 + bBye + 2A0B/; therefore AAbe and ABC = 2AA07 and 0Bf, that is, A123 and ABC = 2AA/87 and 0B/. So A134 and ACD = 2AA?5 and yCh, and A145 and ADE = 2AA8e and SBl. Therefore, adding, pentagon 12345+pentagon ABCDE = twice pentagon A/S.ySe + twice the pentagon made up by A /SB/, yCA, SDl, which last pentagon is of course also equal to the one made up of A ygO WD, cmE. Thus the dissections are along the full lines in the figures; and the rearrangements are A$y$c for one polygon; 12345, with gy&f, kSyh, and me$l, for a second equal one; fiBf, yGh, and SDl for a third; and ygC, WD, and emE for a fourth equal to it. Note 1.--Though stated for pentagons only, the state of things is just the same for two hexagons, heptagons, or any similar polygons whatever. Note 2.--The above is the simplest dissection, such as wanted, but is, of course, only one of an infinite number of such possible dissections. For, instead of being placed with a vertex A and two sides common, the two polygons might have been put anyhow homothetically, and the division of them been similarly into triangles radiating from their then centre of similitude. III. Solution by the Rev. F. D. Thomson, M.A. Let any two similar polygons ABC..., abc... be placed one within the other, so that the homologous sides AB, ah, &c are parallel. Then Aa, B4, Gc, ... will meet in a point O. Bisect -A in P, iB in Q, &c, and take Op = -P, --7 Oq = Q, &c Then PQ..., pq... will be two R 5. JSJ. polygons similar and similarly situated to the jet-v ..".


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Product Details
  • ISBN-13: 9781130135404
  • Publisher: Rarebooksclub.com
  • Publisher Imprint: Rarebooksclub.com
  • Height: 246 mm
  • No of Pages: 30
  • Spine Width: 2 mm
  • Width: 189 mm
  • ISBN-10: 1130135403
  • Publisher Date: 01 Mar 2012
  • Binding: Paperback
  • Language: English
  • Returnable: N
  • Weight: 73 gr


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