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 Slide 1 : Similar Triangles
 Slide 2 : First the question arises what is similarity
 Slide 3 : Let’s see these figures and tell whether they are similar or not
 Slide 4 : Yes, these figures are similar to each other as both of them are squares but now check are these two figures similar to each other.
 Slide 5 :
 Slide 6 : No, these two figures are not similar as one is square where as another is circle.
 Slide 7 : So ,from the above observation we came to know that similar figures are those figures whose shape is exactly same.
 Slide 8 : Now ,next question is that is the similar figures are congruent. View these four figures and tell which is congruent and which is similar.
 Slide 9 :
 Slide 10 : From the above observation it can be concluded that all congruent figures are similar but all similar figures need not be congruent.
 Slide 11 : In the picture Fig. A is congruent to Fig. B but Fig. C is not congruent to Fig. D whereas the Fig. A is similar to Fig. B as well as Fig. C. is similar to Fig. D.
 Slide 12 : There are certain conditions for two polygon of same number of sides to be similar. If their corresponding angles are equal and Their corresponding sides are in the same ratio or proportion. (this can easily be proved by a simple activity)
 Slide 14 : Similarity of Triangles Two triangles are similar, if There corresponding angles are equal and Their corresponding sides are in the same ratio(or proportion)
 Slide 15 : BPT(Basic Proportionality Theorem) or Thales Theorem If the corresponding angles of two triangles are equal, then they are known as equiangular triangles. A famous Greek Mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: The ratio of any corresponding sides in two equiangular triangles is always the same. It is believed that he had used a result called Basic Proportionality Theorem (now known as the Thales Theorem) for the same.
 Slide 16 : To understand the Basic Proportionality Theorem, let us perform the following activity.
 Slide 17 : Activity Draw any angle XAY and on its one arm AX, mark points(say five points) P,Q,D,R and B such that AP=PQ=QD=DR=RB. Now, through B, draw any line intersecting arm AY at C. Also, through the point D, draw line parallel to BC to intersect AC at E. Measure AD/DB and AE/EC.
 Slide 18 : Observation We observe that both AD/DB and AE/EC is equal to 3/2. We can see that in ?ABC, DE|IBC and AD/DB =AE/EC. It is all due to Basic Proportionality Theorem.
 Slide 19 : There are several condition for two triangles to be similar.
 Slide 20 : Three conditions for Triangles to be similar. 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.(If two angles of one triangle are respectively equal to two angles of another are similar). If in two triangles, sides of one triangle are proportional(i.e. in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. If one angle of a triangle is equal to one angle of the other triangle and the sides including these sides are proportional, then the two triangles are similar.
 Slide 21 : Let prove these theorems by simple activity
 Slide 22 : Theorem-1 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.(If two angles of one triangle are respectively equal to two angles of another are similar).
 Slide 23 : Activity Draw two line segments BC and EF of two different lengths, say 3cm and 5cm respectively. Then, at the points B and C respectively, construct angles PBC and QCB of some measures, say 60o and 40o. Also, a the points E and F, construct angles REF and SFE of 60o and 40o respectively. Lets rays BP and CQ intersect each other at A and rays ER and FS intersect each other at D. In the two triangles ABC and DEF, you can see that ?B= ?E, ?C= ?F and ?A= ?D. That is , corresponding angles of these two triangles are equal and corresponding sides are BC/EF=3/5=.6. On measuring AB, DE, CA and FD, you will find that AB/DE and CA/FD are also equal to 0.6 (or nearly equal to 0.6). Thus, AB/DE=BC/EF=/CA/FD. You can repeat this activity by constructing several pairs of triangles having their corresponding sides in the same ratio(or proportion). This activity leads us to the following criterion for similarity of two triangles.
 Slide 24 : Theorem-2 If in two triangles, sides of one triangle are proportional(i.e. in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.
 Slide 25 : Activity Draw two triangles ABC and DEF such that AB=3cm, BC=6cm, CA=8cm, DE=4.5cm, EF=9cm and FD=12cm. So, we have AB/DE=BC/EF=CA/FD (each equal to 2/3) Now measure ?A, ?B, ?C, ?D, ?E and ?F. You will observe that ?B= ?E, ?C= ?F and ?A= ?D, i.e., the corresponding angles of the two triangles are equal. You repeat this activity by drawing several such triangles (having their side in the same ratio). Every time you shall see that their corresponding angles are equal. It is due to the following criterion of similarity of two triangles.
 Slide 26 : Theorem-3 If one angle of a triangle is equal to one angle of the other triangle and the sides including these sides are proportional, then the two triangles are similar.
 Slide 27 : Activity Draw two triangles ABC and DEF such that AB = 2 cm, ? A = 50°, AC = 4 cm, DE = 3 cm, ? D = 50° and DF = 6 cm. Here, you may observe that AB/DE =AC/DF (each equal to 2/3 ) and ? A (included between the sides AB and AC) = ? D (included between the sides DE and DF). That is, one angle of a triangle is equal to one angle of another triangle and sides including these angles are in the same ratio (i.e., proportion). Now let us measure ? B, ? C, ? E and ? F. You will find that ? B = ? E and ? C = ? F. That is, ? A = ? D, ? B = ? E and ? C = ? F. So, by AAA similarity criterion, ? ABC ~ ? DEF. You may repeat this activity by drawing several pairs of such triangles with one angle of a triangle equal to one angle of another triangle and the sides including these angles are proportional. Every time, you will find that the triangles are similar. It is due to the following criterion of similarity of triangles
 Slide 28 : This is all about the similar triangles.
 Slide 29 : Thanks for viewing it.
 Slide 30 : Made under the guidance of Ms. Poonam Aggarwal (Maths teacher)
 Slide 31 : Made by:- Shubham Bansal X-C Roll No.2