Notes of lessons on the Herbartian method (based on Herbart's plan) . matter. IV. Recapitulate construction and proof in correct order,asking references and making class account for them. V. Refer to area of squares, and ask how to find lengthof side. Show in numbers that principle is true. 4, 3, 5, by questioning the class how this may become usefulin arithmetic, finding areas and distances, etc., etc. LESSON ON EUCLID, BOOK I., PRO-POSITION 32. Class—Age, 13 to 15 years. Time—Half an hour. Aim—Exerciseof reasoning to lead class to discover proof of Proposition 32. MATTER. Herbartian
Notes of lessons on the Herbartian method (based on Herbart's plan) . matter. IV. Recapitulate construction and proof in correct order,asking references and making class account for them. V. Refer to area of squares, and ask how to find lengthof side. Show in numbers that principle is true. 4, 3, 5, by questioning the class how this may become usefulin arithmetic, finding areas and distances, etc., etc. LESSON ON EUCLID, BOOK I., PRO-POSITION 32. Class—Age, 13 to 15 years. Time—Half an hour. Aim—Exerciseof reasoning to lead class to discover proof of Proposition 32. MATTER. Herbartian Steps. 1. Preparation. 1. Enunciation: If a side of a triangle be producedthen the exterior angle, etc. (see Euclid I. 32)./JTriangle. 2. Definitions to | Angle, exterior angle. be stated. j Interior opposite Right angle. Euclid, Book I., Proposition 32 201 II. Presentation. Analysis ofEnunciation. fGiven : Triangle and produced side. /(a) Exterior angle = sumof interior and oppositeangles.(b) Three interior angles= two right angles. Required toprove. 2. Construction. Through C draw CE parallel to BA (I. 31). ((a) Because BA and CE are parallel and AC meetsthem, therefore angle ACE = alternate angle CAB(I. 29). Again, because BAand CE are parallel and BDmeets them, therefore exteriorangle ECD = opposite in-terior angle ABC (I. 29);therefore whole exterior angleACD = sum of interior oppos-iteangleCABandangleABC.(b) Since angle ACD = angle CAB + angle ABC proved,to each add angle BCA, then angle BCA + angleACD = angle BCA + angle CAB + angle adjacent angles BCA and ACD are togetherequal to two right angles (I. 13). Therefore alsoangle BCA + angle CAB + angle ABC = two right I angles.—() III. Association. Definitions, axioms, references. IV. Recapitulation. Repetition of proof by class step accounted for. V. Application. 1. Deduce from proposition :— In any right-angled triangle the two acute angles arecomplementary.
Size: 1674px × 1492px
Photo credit: © The Reading Room / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1900, bookpublisherlondo, bookyear1902
