Plane and solid geometry . 6. 7. 8. 9. 10. AT All All A III A 111 a_ ^ .2 a 12 cV_e d aa 12 .J2 d d^d AIAl A II A III All A III A I + A II -f A III _ A I _a^ A I+ A II+ A IIIP AV f2 a- a 12 a Eeasons 1. § 54, 15. 2. § 439. % § 503. 4. § 503. 5. § 503. 6. § 419. 7. § 54, 13. 8. §54,1. 9. § § 309. 506. Cor. Tjvo similar polygons are to each otiier as thesquares of any two homologous diagonale. Ex. 879. If one square is double another, what is the ratio of theirsid^s ? Ex. 880. Divide a given hexagon into two equivalent parts so thatone part shall be a hexagon similar to the given h
Plane and solid geometry . 6. 7. 8. 9. 10. AT All All A III A 111 a_ ^ .2 a 12 cV_e d aa 12 .J2 d d^d AIAl A II A III All A III A I + A II -f A III _ A I _a^ A I+ A II+ A IIIP AV f2 a- a 12 a Eeasons 1. § 54, 15. 2. § 439. % § 503. 4. § 503. 5. § 503. 6. § 419. 7. § 54, 13. 8. §54,1. 9. § § 309. 506. Cor. Tjvo similar polygons are to each otiier as thesquares of any two homologous diagonale. Ex. 879. If one square is double another, what is the ratio of theirsid^s ? Ex. 880. Divide a given hexagon into two equivalent parts so thatone part shall be a hexagon similar to the given hexagon. Ex. 881. The areas of two similar rhombuses are to each other asthe squares of their homologous diagonals. Ex. 882. Oup side of a polygon is 8 and its area is 120. The homol-ogous side of a similar polygon i^ 12 ; find its area. 232 PLANE GEOMETRY Proposition XIL Theorem 507. The square described on the hypotenuse of aright triangle is equivalent to the sum of the squaresdescribed on the other two sideso. Given rt. A ABC, right-angled at C, and the squares describedon its three sides. To prove square AD =0= square BF + square Cfl*. Argument 1. rrom C draw CM l^AB, cutting AB at L and KD at if. 2. Draw CK and BH, 3. Aacg, BCA, and FCB are all rt A. 4. ,\ ACF and GCB are str. lines, 6. In A CAK and HAB, CA = HA, AK = AB. 6. Z CAB = Z CAB. 7. Z BAK = Z ^^C. 8. .. Z C^iT = Z /f^5. 9. .-. Ac^^= AiLiJS. 10. A CAKBJid rectangle AM have the samebase AK and the same altitude, theJ_ between the lis AK and CM, Reasons 1. § 155. 2. § 54, §76. 5 §233. 6. By idert 7. § 64. 8. § 64, 2 9. § §235. 1 BOOK IV 233 Argument 11. .*. A CAK =0= ^ rectangle AM. 12. Likewise A HAB and square CH have the same base HA and the same alti-tude, the J_ between ils HA and GB, 13. . HAB =0= i square CH 14. But A CAK = A HAB. 15. .*. i rectangle AM=o=^ square CH. 16. .*. rectangle AM=o^ square CH. 17. Likewise, by drawing CD and AE, it may be proved that rectangl
Size: 1691px × 1478px
Photo credit: © The Reading Room / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1910, booksubjectgeometr, bookyear1912
