Triangles (Similarity)

Maths (CLASS 10) Chapter Triangles PPT
thales theorem
similar triangles
phyathagoras theorem ,etc
In this ppt all theorem are proved solution are gven
there are videos also
all topic cover

1. MATHS F.A.- 2 ACTIVITY (GROUP) SESSION : 2016-2017 TOPIC : TRIANGLE (SIMILARITY) MADE BY : GROUP 4 CLASS : 10 E VIDYA BAL BHAWAN SR. SEC. SCHOOL

2. ACKNOWLEDGEMENT We wish to express our deep gratitude and sincere thanks to the Principle Sir, Dr. S.V. Sharma for his encouragement and for all the facilities that he provided for this Project. We sincerely appreciate this magnanimity by taking us into his fold for which we shall remain indebted to him. We extend our hearty thanks to Mrs. Shweta Bhati , Maths teacher, who guided us to the successful completion of this Project. We take this opportunity to express our deep sense of gratitude for her invaluable guidance, constant encouragement, immense motivation, which has sustained our effort at all the stages of this project work… We can`t forget to offer our sincere thanks to parents and also to all group members who helped us to carry out this activity successful and for their valuable advice and support, which we received from them time to time ……..

3. Certificate This is to certify that this project on “TRIANGLES (SIMILARITY)” is made by Group – 4 and the members are Abhishek Mahto , Lakshya Kumar, Mohan Kumar, Ritik Kumar, Vivek Singh of class X E . We are thankful to Principal sir ‘Dr. S.V. Sharma’ for giving us this opportunity and concerned Subject teacher ‘ Shweta Bhati ’. Principal Sign. Teacher Sign. Marks Obtained VIDYA BAL BHAWAN SR. SEC. SCHOOL

4. Content CONCEPT OF SIMILARITY SIMILAR POLYGONS SIMILAR TRIANGLES AND THEIR PROPERTIES SOME BASIC RESULTS ON PROPORTIONALITY THALE ` S THEOREM CONVERSE OF BPT CRITERIA FOR SIMILARITY OF TRIANGLES AREAS OF SIMILAR TRIANGLES PYTHAGORAS THEOREM

5. INTRODUCTION In general, there are several objects which have something common between them. Observing them closely, let see that some of them have same shape but may have different or same size. For example, if the photographs a person developed from same negative, they all look same in all respect except for their size. Such objects are called similar objects. Two line segments of different sizes, two circles of different radii, two squares of different sizes, two rectangles of different dimensions - come under similar figures. One smaller circle can be got by shrinking a larger circle. One bigger square can be got by stretching a smaller square. Then, what about the similarity of triangles? Is it true to say any two given triangles are similar? The answer is NO. This is true only when the triangles are equilateral.

6. TRIANGLES A B C

7. CONCEPT OF SIMILARITY In earlier classes, we have learnt about congruent figures. Two geometric figures having the same shape and size are known as congruent figures. Note that congruent figures are alike in every respect. Geometric figures having the same shape but different sizes are known as similar figures. Two congruent figures are always similar but similar figure not be congruent.

8. ILLUSTRATION 1 : Any two line segments are always similar but they need not be congruent. They are congruent, if their lengths are equal. ILLUSTRATION 2 : Any two circles are similar but not necessarily congruent. They are congruent if their radii are equal.

9. ILLUSTRATION 3 : Any two squares are similar. Any two equilateral triangles are similar.

10. SIMILAR POLYGONS DEFINITION : Two polygons are said to be similar to each other, if ( i ) Their corresponding angles are equal (ii) The length of their corresponding

11. If two polygons ABCDE and PQRST are similar, then : Angle at A = Angle at P, Angle at B = Angle at Q, Angle at C = Angle at R, Angle at D = Angle at S, Angle at E = Angle at T And, AB BC CD DE EA PQ QR RS ST TP IF two polygons ABCDE and PQRST, are similar, we write ABCDE ̴ PQRST

12. Here, ABCDE ̴ PQRST D E A B C R T Q P S 130⁰ 100⁰ 120⁰ 90⁰ 130⁰ 100⁰ 120⁰

13. SIMILAR TRIANGLES AND THEIR PROPERTIES DEFINATION : Two triangles are said to be similar, if their Corresponding angles are equal Corresponding sides are proportional It follows from this defination that two triangles ABC and DEF are similar, if

14. ( i ) A= D, B= E, C= F and, (ii) AB BC AC DE EF DF C B A F E D

15. NOTE 1: In the later part of this PPT we shall show that two condition given in previous slide are not independent. In fact, if either of the two conditions hold, then the oyher holds automatically. So any one of the two conditions can be used to define similar triangles. NOTE 2: If corresponding angles of two triangles are equal, then they are known as equiangular triangles

16. SOME BASIC RESULTS ON PROPORTIONALITY In this section, we shall discuss some basic results on proportionality. Let us do the following activity. ACTIVITY : Draw any angle XAY and mark points P 1 ,P 2 ,D,P 3 and B on its arm Ax such that A P 1 = P 1 P 2= P 2 D = DP 3 B = 1 units. Through point B, draw any line intersecting arm AY at point C. Also, through point , draw a line parallel to BC to intersect AC at E.

17. We have, AD=AP 1 +P 1 P 2 +P 2 D=3 units And, DB=DP 3 +P 3 B=2 units AD 3 DB 2 Now, measure Ae and EC and find AE EC You will find that AE 3 EC 2 AD AE DB EC Thus, we observe that in ABC if De II BC, then AD AE DB EC We prove this result as a theorem known as basic proportionality theorem or Thale`s Theorem. C x B P 3 D P 2 P 1 A E Y

18. THALE`S THEOREM (Basic Proportionality Theorem or Thale`s Theorem) If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio

19. Given : A triangle ABC in which DE II BC, and intersects Ab at D AC in E. To prove : AD AE DB EC Construction : Join BE,CD and draw EM I BA and DN I CA Proof: Area of triangle Similarly, Hence,

20. Similarly , Triangles BDE and DEC are on the same base, i.e. DE and between same parallels, i.e. DE and BC . Hence, From above equations, it is clear that;

21. PRESS ENTER AFTER VIDEO PLAYBACK FOR NEXT SLIDE

22. CONVERSE OF BPT If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

23. Construction : ABC is a triangle in which line DE divides AB and AC in the same ratio. This means: To prove : DE || BC Let us assume that DE is not parallel to BC . Let us draw another line DE' which is parallel to BC . Proof : If DE' || BC, then we have; According to the theorem ; Then according to the first theorem; E and E' must be coincident. This proves: DE || BC

24. PRESS ENTER AFTER VIDEO PLAYBACK FOR NEXT SLIDE

25. CRITERIA FOR SIMILARITY OF TRIANGLES AAA or AA (ANGLE- ANGLE) SSS (SIDE- SIDE- SIDE) SAS (SIDE-ANGLE-SIDE)

26. AA CRITERIA If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This is also called AAA (Angle-Angle-Angle) criterion.

27. Construction: Two triangles ABC and DEF are drawn so that their corresponding angles are equal . This means : ∠ A =∠ D, ∠ B = ∠ E and ∠ C = ∠ F To prove : Draw a line PQ in the second triangle so that DP = AB and PQ = AC Proof : Because corresponding sides of these two triangles are equal This means; ∠ B = ∠ P = ∠ E and PQ || EF

28. This means; Hence; Hence ,

29. SSS CRITERIA If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. This is also called SSS (Side-Side-Side) criterion.

30. Construction: Two triangles ABC and DEF are drawn so that their corresponding sides are proportional. This means : To prove: ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F And hence ;  ABC ~  DEF In triangle DEF, draw a line PQ so that DP = AB and DQ = AC Proof : Because corresponding sides of these two triangles are equal This means;

31. This also means; ∠ P = ∠ E and ∠ Q = ∠ F We have taken; ∠ A = ∠ D, ∠ B = ∠ P and ∠ C = ∠ Q Hence; ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F From AAA criterion ;  ABC ~  DEF proved

32. SAS CRITERIA If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This is also called SAS (Side-Angle-Side) criterion.

33. Construction: Two triangles ABC and DEF are drawn so that one of the angles of one triangle is equal to one of the angles of another triangle. Moreover, two sides included in that angle of one triangle are proportional to two sides included in that angle of another triangle. This means ; ∠ A = ∠ D and To prove :  ABC ~  DEF Draw PQ in triangle DEF so that , AB = DP and AC = DF Proof : Because corresponding sides of these two triangles are equal (GIVEN) ∠ A = ∠ D

34. Hence ; (   from SSS criterion ) Hence ; Hence;  ABC ~  DEF proved

35. AREAS OF SIMILAR TRIANGLES THEOREM: The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

36. Construction: Two triangles ABC and PQR are drawn so that,  ABC ~  PQR. To prove : Draw AD ⊥ BC and PM ⊥ PR Proof : Hence ;

37. Now, in  ABD and  PQM; ∠ A = ∠ P, ∠ B = ∠ Q and ∠ D = ∠ M (because  ABC ~  PQR) Hence;  ABD ~  PQM Hence; Since,  ABC ~  PQR So , Hence ; Similarly , following can be proven:

38. THEOREM: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other .

39. Construction: Triangle ABC is drawn which is right-angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC. To prove :  ABC ~  ADB ~  BDC Proof : In  ABC and  ADB; ∠ ABC = ∠ ADB ∠ BAC = ∠ DAB ∠ ACB = ∠ DBA From AAA criterion;  ABC ~  ADB In  ABC and  BDC; ∠ ABC = ∠ BDC ∠ BAC = ∠ DBC ∠ ACB = ∠ DBC From AAA criterion;  ABC ~  BDC Hence;  ABC ~  ADB ~  BDC proved.

40. Pythagoras Theorem In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

41. Construction: Triangle ABC is drawn which is right angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC. To prove : Proof : In  ABC and  ADB; Because these are similar triangles (as per previous theorem ) In  ABC and  BDC; Adding equations (1) and (2), we get ; Proved .

42. THANK YOU PRESENTATION BY NAME ROLL NO. ABHISHEK MAHTO 1 LAKSHYA KUMAR 13 MOHAN KUMAR 15 RITIK KUMAR 23 VIVEK SINGH 38