very very very very very very excellent job sir ...
im also a student of class IX and i can evidently and surely say that you should be paid to increase the TRP of this website
-regards prince joshi
email - chasekamal@gmail.com [also on orkut :P]

Geometry - IX Class
Chapter No. 3
CONGRUENCE OF TRIANGLES
Presented by
Suresh Borgaonkar
Nagpur

Slide 2:

Introduction :
If you go through different syllabi of middle school and high school of Maharashtra state board, you will find this chapter “Congruence of triangles” included in the course of middle school as well as high school, right from 1975, even before that.
Obviously, according to NCF 2005 this chapter finds a core-importance in IX class up-graded mathematics. It is written as per the aims and objectives of NCF 2005.
The “Constructivist” approach adopted in this chapter will help the students to cultivate the habit of finding and constructing knowledge for self, through activities, observations, reasoning and analysis.
The streamline flow of explaining the concepts, the eye-catching headings “Can you recall”, think it over, remember this, help though understanding of the concept.
The “think it over” article may cultivate higher order thinking and develop the capacity to design new problems and riders, i.e. constructing knowledge for self.

Slide 3:

Units and sub-units
3.1 Congruent figures
3.2 One-one correspondence and congruence of triangles
3.3 Properties of congruency in triangles
3.4 Congruence in different types of triangles
3.5 Sufficient conditions for congruency of triangles
3.6 S-A-S Test - Verification through activity
3.7 Isosceles triangle theorem
3.8 A-S-A Test - Proof
3.9 S-A-A Test - Proof
3.10 S-S-S Test - Proof
3.11 Hypotenuse - side theorem

Slide 4:

3.12 300 - 600 - 900 theorem
3.13 Converse of 300 - 600 - 900 theorem
3.14 Converse of isosceles triangles theorem
3.15 Theorem :
Relation between the length of hypotenuse and median drawn to it.
3.16 Perpendicular bisector of a segment
3.17 Perpendicular bisector theorem
3.18 Angle bisector and theorem on it
3.19 Locus
3.20 Properties of triangles based on inequalities
3.21 Shortest segment theorem
3.22 Relation between the three sides of the triangle. (Sum and difference)

Slide 5:

Congruent Figures
Look at the figures, and tell which of them have same shape and size

Slide 6:

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.

Slide 7:

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.

Slide 8:

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 ?

Slide 9:

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.

Slide 10:

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.

Slide 11:

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?

Slide 12:

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.

Slide 13:

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

Slide 14:

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.

Slide 15:

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)

Slide 16:

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?

Slide 17:

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

Slide 18:

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

Slide 19:

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.

Slide 20:

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

Slide 21:

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.

Slide 22:

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.

Slide 23:

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.

Slide 24:

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

Slide 25:

Now think
If AB = AC then
Can AF = BF
When this is true?

Slide 26:

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.

Slide 27:

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?

Slide 28:

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.

Slide 29:

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?

Slide 30:

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 ?

Slide 31:

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.

Slide 32:

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.

Slide 33:

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.

Slide 34:

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

Slide 35:

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.

Slide 36:

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.

Slide 37:

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.

Slide 38:

Historical Problems
Show
This method of trisection of angle was suggested by Archimedes

Slide 39:

Application based Problems
(1)
Show : Ray PQ ┴ line l
(2)
Show : Ray BM is angle bisector of
ABC

Slide 40:

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

Slide 41:

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

Slide 42:

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.

Slide 43:

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.

Slide 44:

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.

Slide 45:

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.

Slide 46:

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.

Slide 47:

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.

Slide 48:

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

Slide 49:

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

Slide 50:

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.

Slide 51:

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.

Slide 52:

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.

THANKS FOR ATTENTION :

THANKS FOR ATTENTION

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By: pralaicadu (40 month(s) ago)

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