Successfully reported this slideshow.

Triangles ix


Published on

Published in: Education, Technology
  • Be the first to comment

Triangles ix

  1. 1. Geometry - IX Class Chapter No. 3 CONGRUENCE OF TRIANGLES
  2. 2. Congruent Figures Look at the figures, and tell which of them have same shape and size A B C D C D P Q R A B C P R R
  3. 3. It is a bit difficult to determine it by mere observation. One may think of using i) divider ii) a trace paper or iii) drawing the figures on the plane paper and cut along the boundary and superpose the pieces. Such method of Superposition is not a proper way to determine it. We say figures of the same shape and size are congruent figures. A B C D Q RS P A B C P Q R
  4. 4. Association with real number and congruency We associate a unique real number with the figures of the same type, which helps to determine the congruency. (i) The real number associated with the segment is its length and that associated with angle is its measure. We know l(AB) = l (CD) seg AB ≈ seg CD and m ABC = m PQR (ii) To determine the congruency of triangles, we suggest on activity leading to the concept of one to one correspondence.
  5. 5. One to one correspondence and congruency of triangles. Activity : To determine whether the given ∆ABC and ∆ PQR are congruent ...... (i) Draw the given triangles on plane papers. (ii) Cut the triangular pieces along the boundary. (iii) Put one triangular piece over the other and try to match. (iv) See whether the triangles cover each other exactly. (v) Try to find out all possible ways of keeping one triangular piece over the other. (vi) What do you find ? B A C P RQ
  6. 6. There are six different ways of putting one triangle cover the other i.e. six different one to one correspondence between the vertices of the triangles. ABC ↔ PQR , ABC ↔ QPR , ABC ↔ QRP ABC ↔ RQP , ABC ↔ RPQ , ABC ↔ PRQ Out of six different one to one correspondences between the vertices of triangles if there exist at least one, in which triangles cover each other exactly than we say the triangles are congruent.
  7. 7. Congruent Triangles Out of the six one to one correspondence between the vertices of two triangles, if there exist at-least one, one to one correspondence such that, the sides and angles of one triangle are congruent to corresponding sides and angles of other triangle then the triangles are said to be congruent with respect to that correspondence and the property known as congruence.
  8. 8. We have ABC ↔ QPR such that i ) A ≈ Q ii ) B ≈ P iii ) C ≈ R iv) side AB ≈ side QP v) side BC ≈ side PR vi) side AC ≈ side QR then ∆ABC ≈ ∆ QPR i.e. To determine the congruency of triangle we require the six conditions. But do we really need all the six condition?
  9. 9. Sufficient conditions for congruency of the triangles Three out of six conditions if property chosen are sufficient to determine the congruency of the two triangles. When these three conditions are satisfied then the other three automatically get satisfied, hence the triangles become congruent. This fact was first proved by Euclid the father of Geometry. These sufficient conditions are referred as SAS, SSS, ASA, SAA tests. SAS test is taken for granted i.e. accepted as postulate and other are proved.
  10. 10. Activity : Verification of SAS Test : i) Construct ∆ABC and ∆ PQR of the given measures as shown in the fig. ii) Cut the triangular pieces along the boundary. iii) Place triangular piece ABC over triangular piece DEF such that A fall on D and AB falls along DE. iv) Since AB = DF , so B falls on E v) Since A = D , so AC will fall along DF vi) But AC = DF C will fall on F Thus AC will coincide with DF and BC will coincide with EF ∆ ABC coincides with ∆ DEF Hence ∆ ABC ≈ ∆ DEF A B C P Q R
  11. 11. Included Angle is a must for SAS Test : Activity : ’ i) Construct another ∆DEF of the same measures i.e. EF = 5 cm, ED = 3.8 cm, C = 450 ii) See, we get two different triangles of the same measures. ∆ABC ≈ ∆DEF ∆ABC ≈ ∆D‘EF ∆DEF ≈ ∆D’EF i.e. if the angle is not included then the triangles may or may not be congruent. A B C 5cm 3.8cm D E F 5cm 3.8cm D’
  12. 12. Think it over i) ∆ABC is isosceles triangle with seg AB ≈ seg BC We know that ∆ABC ≈ ∆ABC ---- (reflexivity) State another congruent correspondences. ii) ∆ ABC is equilateral triangle We have ∆ABC ≈ ∆ABC ---- (reflexivity) State another congruent correspondences. iii) ∆ ABC is Scalene triangle ∆ABC ≈ ∆ABC ---- (reflexivity) State another congruent correspondence if exit. Also find another congruent correspondences between the vertices of triangles in each case i) ∆ABC ≈ ∆PQR and AB = AC (i.e. isosceles) ii) ∆ABC ≈ ∆PQR and AB = BC = AC (i.e. equilateral) iii) ∆ABC ≈ ∆PQR (scalene)
  13. 13. Isosceles Triangle Theorem and its Converse In ∆ABC , seg AB ≈ seg BC C = A Here we suggest activity for the verification i) Construct ∆ABC where AB = BC and measure C and A ii) Construct ∆ABC where C = A and measure AB and BC iii) Construct ∆ABC where AB ≠ BC and measure C and A iv) Construct ∆ABC where C ≠ A and measure AB and BC What relations you get?
  14. 14. Proof of the theorem The proof can be given by two ways i) ∆ABC ≈ ∆CBD (SAS Test) A ≈ C ii) ∆ABC is isosceles triangle and AB = BC ∆ABC ≈ ∆CBA A ≈ C B A CD
  15. 15. Think it over We find subtitle “Think it over” on many pages in this chapter. This will motivate the students to think differently and develop the higher order thinking skill. (HOTS) Some instances are given from the text. E.g. (i) Show : ABP ≈ ACP This is the solved problem in the text book. Then it is asked to think it over if the point is in the exterior of ∆ ABC Show : ABP ≈ ACP A B C P B C A P
  16. 16. Think it over (2) Solved problem To prove side AC || side DB Think it over if Here Pair of congruent sides are changed then think, can side AC and DB be parallel. A B D C O O A B D C
  17. 17. Think it over Solved Problem Show : Ray AD || Side BC Think it over if AB ≠ AC then state whether the ray AD remains parallel to side BC R D B C A E
  18. 18. Think it over The proof of 300 - 600 - 900 theorem is given i.e. in 300 - 600 - 900 triangle, length of side opposite to angle 300 is half the hypotenuse. i) In turn it is asked to find the length of side opposite to angle 600 ii) Again it is asked to modify the theorem. iii) Think it over the converse of 300 - 600 -900 theorem.
  19. 19. Think it over Do we really need two different tests ASA and SAA, for congruency? When two angles of triangle are congruent with corresponding two angles of another triangle, then the third pair also becomes congruent. Hence actually we do not need two different tests ASA and SAA.
  20. 20. Think it over (1) Medians of isosceles triangle Solved problem : Given : AB = AC BD and CE are medians Show : BD = CE Now it is asked to think it over can medians AE and BD be equal. A B C DE A B C D E
  21. 21. Think it over Altitudes of isosceles triangle Solved example : Given : Altitude BE = Altitude DC Show : AB = BC (1) Asked to think over the converse of above rider If AB = AC Show : Altitude DC = Altitude BE A B C ED ┐ ┌ A B C ED F ┐ ┌
  22. 22. Now think If AB = AC then Can AF = BF When this is true? A B C ED F ┐ ┐ ┌
  23. 23. Think it over If two triangles are congruent then corresponding altitudes are equal. Then think it over the areas of congruent triangles. Whether areas of congruent triangles are equal.
  24. 24. Think it over l1 , l2 , l3 , l4 are different lines intersecting the seg AB in different points. Which line is perpendicular bisector of AB? What are the conditions to be imposed over a line to be perpendicular bisector of given segment? P A B l1 A B┐Q l2 A B R l3 A B┐S l4
  25. 25. Activity Leading to perpendicular bisector theorem. Measure the distances P1A , P1B P2A , P2B P3A , P3B P4A , P4B What condition is obeyed by the points P1 , P2 , P3 ------ the points on the perpendicular bisector of seg AB. P1 A B P2 P3 P4 P5
  26. 26. Activity Leading to angle bisector theorem Measure the distances P1Q1 , P1R1 P2Q2 , P2R2 P3Q1 , P3R1 What condition is obeyed by the points on the angle bisector of angle? B A C R1 P2 P3 P4 Q1 Q2 Q3 R4 R3 R2 Q4 P1
  27. 27. Activity Leading to shortest segment theorem Out of the segments PM, PC, PD, PA, PB which is shortest and what is the angle made by shortest segment with line l ? ┐ B A M C D l P
  28. 28. Locus i) Points on perpendicular bisector of segment. ii) Points on angle bisector. iii) Points on circle. The set of points which obey certain conditions is known as locus.
  29. 29. Locus Activity : AP . is constant ellipse AC + BC = constant When points move according to given conditions, certain path is traced, that path is known as locus. P A A B C
  30. 30. Activity Leading to Difference of remaining two sides < Length of a side of a triangle < Sum of remaining two sides Take sticks of different sizes and try to construct triangles. Think about the restrictions on the lengths of sides of triangles. Sets of sticks i)8 cm, 3 cm, 5 cm ii) 8 cm, 5 cm, 6 cm, iii) 8 cm, 3 cm, 2 cm.
  31. 31. Sum of two sides = Third side Sum of two sides > Third side Diff. of two sides = Third side Diff. of two sides < Third side Triangle is possible Sum of two sides < Third side Diff. of two sides > Third side 8 5 3 6 5 8 8 3 2
  32. 32. Up- gradation of Chapter In an attempt to upgrade the chapter i) Proofs of SSS, SAA, ASA tests and converse of 300 - 600 - 900 theorem are included. ii) Many activities are suggested to verify the situation. iii) Day to day life problems, hot problems, historical problems, application and activity based problems are included. iv) At some places the problems based on basic geometric construction are given, to enrich the concept of congruence and its application.
  33. 33. Day today life Problems (1) A’ B’ How will you find the distance between two places A and B when there is a big obstacle between them. (2) How will you find the breadth of the river without crossing it.
  34. 34. Activity based Problems (1) Two squares of the same size are kept on one another, as shown in the figure where O is the centre of one square. Find the area overlapped. (2) How will you use unmarked ruler with parallel edges to construct an angle bisector of given angle. A B C D P O R S
  35. 35. Historical Problems Show This method of trisection of angle was suggested by Archimedes CAB 3 1 P CAB 3 1 P P A B C
  36. 36. Application based Problems (1) Show : Ray PQ ┴ line l (2) Show : Ray BM is angle bisector of ABC A P B C D Q l R P B Q S M A C
  37. 37. Higher Order Thinking Skill (HOTS) Problems (1) Given : AB < AC Ray AP is angle bisector of BAC Show : BP < PC (2) In ∆ABD seg AD ┴ seg BC , BD < DC Show : AB < AC A B P C A B D C
  38. 38. Routine Problems For 1 mark each 1) ∆ABC ≈ ∆PQR , A = 400 , Find P 2) State the triangle Congruent to ∆ABP 3) In ∆ABC, seg AB ≈ seg BC , B = 400 find A 4) Name the greatest side of ∆ABC 5) ∆ABC ≈ ∆PQR and seg AB ≈ seg BC . State another correspondence between the vertices of ∆ ABC and ∆ PQR which is congruent. 6) PQ is shortest segment then find PQR. P B Q C A P T Q R S A B C 400 300
  39. 39. For 2 marks 1) Arrange the sides of triangle in ascending order of their lengths. 2) ∆ABD ≈ ∆CDB To prove this some information is missing. State that minimum information required to prove it. A B C 400 300 A B C D
  40. 40. 3) PA = PB QA = QB RA = RB SA = SB TA = TB Name the path of points i.e. the locus of points P,Q,R,S............ 4) In 300 - 600 - 900 theorem, side opposite to angle 300 is 5 cm, find other two sides. 5) In ∆ ABC , AB = 5 cm , BC = 7 cm. then find the maximum possible length of third side of ∆ ABC. 6) With side 4, 3, 1 cm, state whether a triangle can be drawn, explain. 7) Perpendicular bisector of side AB and BC of ∆ ABC intersect each other at point P. If PA = 5 cm, find PB and PC. . P . Q . R . S . T A B
  41. 41. For 3 marks 1) Given : AB = AC line PQ || side BC Show : seg AP ≈ seg AQ. 2) Prove that corresponding altitudes of congruent triangles are congruent. 3) seg CP ┴ seg AB , segBQ ┴ seg AC seg AB ≈ seg AC Show : seg PC ≈ seg BQ. A P B C Q A B C QP ┐ ┌
  42. 42. 4) Given : AB = AC and B - C - D Show : AB < AD 5) Given : BP and CP are angle bisector of B and C respectively. Show : Ray AP is angle bisector of A. A B C D A B
  43. 43. For 4 marks 1) Given : Ray AD is angle bisector of A and is perpendicular on opposite side BC Show : AB = AC 2) B ≈ C and arcs are drawn with centers B and C keeping the radius constant. Show : seg PS || seg BC. B A CD ┐┌ A P B Q R S C
  44. 44. 3) ABC = 900 D mid point of AC Show : AD =DC = BD 4) Show that perpendicular bisector of angles of triangle are concurrent. 5) Arcs drawn with centre B, intersect the sides of ABC in points P, Q and R,S. Segments PQ and RS intersect at M. Show that Ray BM is angle bisector of ABC. A B C D R P B Q S M A C
  45. 45. For 5 marks 1) The length of any side of a triangle is greater that the difference between the lengths of remaining sides. 2) Perimeter of a triangle is greater than the sum of the three medians of triangle. 3) Show : BAC < BPC < BQC A B C P Q R
  46. 46. 4) Fig. ABCDE is a regular pentagon. If BP = CQ = DR = ES = AT then show that fig. PQRST is a regular pentagon. 5) ABCD is a square P, Q, R and S are the points on its sides Such that AP = BQ = CR = DS and ASP = 300 Show that (i) PQRS is quare (ii) If perimeter of PQRS is 16 cm then find perimeter ABCD A A B C D E S T P Q R D S A P B Q CR
  47. 47. Non-Routine Problems (1) P is a point inside the equilateral ∆ABC such that PA = 3 cm, PB = 4 cm, PC = 5 cm. Find the side of ∆ ABC. (2) Find the maximum value of C, so that the figure can be drawn as per the description. (3) Point P is in the interior of rectangle ABCD. Find another point Q in its interior such that line PQ will divide the rectangle in to two regions of equal areas. A B C P Q D A B C . P
  48. 48. Fallacy A right angle triangle is equilateral triangle Proof : Construction : (i) Draw angle bisector of C (ii) Draw perpendicular bisector of side AB (iii) Name their point of intersection as M (iv) Draw seg ML ┴ side AC seg MN ┴ side CB (v) Draw MA, MB. BA C L N R M
  49. 49. Proof : Point M is on angle bisector of C -------- (Construction) LM = NM ------- (i) Point M is on perpendicular bisector of seg AB -------- (Construction) AM = MB ------- (ii) ∆LMA ≈ ∆NMB ------- (side hyp. theorem) LA = NB ----------- (c.s.c.t.) ------- (iii) Similarly ∆LMC ≈ ∆NMC LC = NC ----------- (c.s.c.t.) ------- (iv) Adding (iii) and (iv) LA + LC = NB + NC AC = CB Hence proved Find the fallacy in the above proof. BA C L N R M