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1. CURRICULARSTATEMENT: To understand about the problem of circle and its importance in
mathematics through observation, organization of charts and by analyzing prepared notes of the pupil.
CONTENT ANALYSIS
New term : congruent triangle.
Fact : If two angles of two triangle are equal, then the third angle also will be equal.
Concept : Concept of problem on circle.
Process : process of understanding the problem of circle.
Name of the teacher : Neethu Krishnan
Subject : Mathematics
Unit : circle
Subunit : circle -Theorem
Name of the school : Mount Tabor Training college
Pathanapuram
Standard : 9 .c
Strength : 44/47
Date :10/8/2015
Time :30 min

2. LEARNING OUTCOMES
The pupil will be able to:-
(1) recall the termbisector
(2) Recognize rightangle triangle.
(3) Conceptof bisectorof two equal chords.
(4) explain chordof equal length.
(5) Suggesta differentmethodtoprove chordsof equal lengths are atequal distance fromthe
Center.
(6) Identify the Conceptof bisectorof twoequal chords.
(7) Detecterror while provingchords of equal lengthare atequal distance fromthe center.
(8) Discussthe proof withotherstudents.
(9) Askquestiontoknowmore aboutthe proof.
(10) Observe all aspectsof provingchordsof equal lengthare atequal distance fromthe center
(11) recognize the qualitiesof exactnessandaccuracywhile drawingperpendicularfromcenterof
Middle pointof chord.
(12) Read chart quicklyandaccuratelyonprovingchordsof equal lengthare equal distance from
the center
PRE-REQUISITES : The students have knowledge on congruent triangle ,chord of circle,
perpendicular bisector
TEACHINGLEARNINGRESOURCES : Usual classroom aids.
LEARNINGSTRATEGES : Observation ,observing charts and explanation by the teacher.

3. Classroominteractionprocedure Expectedpupil responses
INTRODUCTION
ACTIVITY -1
Whengive a diameterfindthe radiusof the circle?
What isPythagorastheorem?
What do youknowabout bisectorof twoequal chords?
Throughthese questionsteacherleadsthe studentstothe
topic
PRESENTATION
ACTIVITY -2
Teacherdraw a figure onblack board
In the figure ABand CD are chordsof equal length,OPand
OQ are perpendicularsfromcentre Oof the circle to these
chords.
Teacherask to the studentsthathowcan we prove the
resultchordsof equal lengthare atequal distance fromthe
centre.
ACTIVITY -3
Here,these twoperpendicularsare fromcentre of circle.
So we get AP= ½ AB
CQ= ½ CD (BB)
We knowthatAB = CD.
So AP=CQ
What can you getfrom the rightangle trianglesOAPand
OCQ?
Here OA=OC,teacherask studentsthat,whyOA=OC?
Also,AP=CQ
I.e.,sowe get OP²=OQ²
Yes ,radiusis the half of the diameter
All studentsresponds
Some studentsrespondthat, bisectoristhe diameterof
circle
Pupil repliedthatthe perpendicularshave equal length
All studentsrespondthatperpendicularfromcentre
bisectthe chord
Studentrespondthatin ∆OAP,OP²=OA²-AP²
In ∆OCQ,OQ²=OC²-CQ²
Because OA and OC are radiusof the circle.

4. Classroom interaction procedure Expected pupil responses
I.e.,OP=OQ so chords of equal length are at equal
distance
From centre
ACTIVITY -4
Suppose AB and CD are not parallel, but they are
chords of equal length, so what we say?
Teacher draws a figure on black board.
(BB)
Since the perpendicular distance from O to lines AB and
CD are equal , the point O is on the bisector of the angle
formed by extending the lines AB and CD
ACTIVITY -5
Teacher explaining with the help of chart
“Chords of equal lengths are at equal
distance from the center.”
Teacher asks students to write proof on note book.
Students respond that perpendicular distance from O
to AB and CD are equal.
Students read carefully and write notes on book.

5. Classroom interaction procedure Expected pupil responses
CLOSURE
ACTIVITY-6
Teacher concludes the class by briefly explain the
results of chord of equal length are at equal distance
from the centre.
REVIEW
ACTIVITY -7
Teacher review the lesson by asking certain
questions from the proof.
FOLLOWUP ACTIVITY
In a circle of radius 5 cm two chords of lengths
6 cm, 8cm are drawn parallel to each other on either
side of the center .what is the distance between these
chords ?