CURRICULAR STATEMENTS 
hr¯¯nsâ Npäfhv Ip] nSn¡p¶p. 
CONTENT ANALYSIS 
Symbols : ss] π, hr¯¯nsâ Bcw ‘r’ 
Facts : hymkw IqSpt¼mÄ hr¯¯nsâ Npäfhv IqSp¶p. GsXmcp hr¯¯nsâbpw Npäfhns hymkw sImv lcn¨m In«p¶ kwJy Ft¸mgpw Øncambncn¡pw. 
Name of the Teacher : Ashabai. B.S Std : IX 
Name of the Subject : Mathematics Str : 40 
Name of the Unit : hr¯¯nsâ AfhpIÄ Time : 45 minutes 
Name of the Topic : hr¯¯nsâ Npäfhv Date :
Concept : hr¯¯nsâ Npäfhv F¶ Bibw 
Formula : hr¯¯nsâ Npäfhv = 2πr ; r → hr¯¯nsâ Bcw 
LEARNING OUTCOME 
The student will be able to; 
1) Recall the information about circles. 
2) Describe the peculiarities of a circle 
3) Interpret about the perimeter of a circle 
4) Explain the concept of perimeter of a circle 
5) Use the above concept in another familiar 
6) Situation 
7) Organise different elements fit within the solution 
8) Judge the appropriateness of the above concept in a given problem 
9) Accept the beauty of maths. 
PRE-REQUISITES 
Ip«nIÄ¡v hr¯w, hr¯ tI{μw, Bcw, hymkw F¶nhsb¡pdn¨pff Adnhv. 
TEACHING LEARNING RESOURCES 
Models and Charts: 
1. hr¯¯nsâ amXrI 
2. hr¯¯nsâ Npäfhv ImWmpff kq{XhmIyw FgpXnb NmÀ«v 
3. {]iv ImÀUpIÄ
Class room Interaction Procedure 
Expected pupils response 
Introduction 
Hcp IYbneqsS ¢mÊv Bcw`n¡p¶p. A¸phnsâ hoSnv ASp¯v Hcp ¢ºv Dv. AhnsS HmW¯nv [mcmfw ]cn]mSnIÄ Act§dmdpv. HmWw hcmdmbn A¸p ¢ºnse ]cn]mSnIÄ ImWm³ Im¯ncn¡pI bmWv. A§s HmWw F¯n ¢ºnse ]cn]mSnIÄ XpS§n. ¢ºn DdnbSn, kpμcn¡v s]m«psXmSoÂ,A¯¸q¡f aÕcw, ItkcIfn, apXemb Hcp]mSv aÕc§Ä Dmbncp ¶p. Ah³ FÃm aÕc§fpw nco£n¨p. ItkcIfn aÕcambncp¶p ASp¯ Cw . ItkcIÄ tNÀ¯n«v Hcp {]tXyI BIrXnbn aÕcmÀ°nIÄ S¡p¶Xv Ah³ {i²n¨p. AXv GXv BIrXnbn Bbncn¡psa¶v n§Ä¡v ]dbmtam? 
hfsc icnbmWv. A¸phnv hopw Hcp kwibw. aÕcmÀ°nIÄ B hr¯w Hcp XhW Id§n hcpt¼mÄ AhÀ F{X ZqcamWv k©cn¨Xv F¶v F§s Is ¯pw? 
Ah³ Hcp]mSv BtemNn¨p. ho«n F¯n A¸p Xsâ tN¨ntbmSv CsX¸än tNmZn¨p. At¸mÄ tN¨n ]dª adp]Sn F´mbncn¡psa¶v n§Ä¡v Duln¡mtam? 
hr¯w 
hr¯¯nsâ Npäfhv Is ¯n bm aXn F¶v Ip«nIÄ ]dªp.
Class room Interaction Procedure 
Expected pupils response 
Presentation 
Ip«nIsf {Kq¸mbn Xncn¨v Hmtcm {Kq¸npw hr¯¯nsâ amXrIbpw, qepw kvsIbnepw ÂIn. F¶n«v AXnsâ Npäfhv , hymkw F¶nh ImWm³ Bhiys¸Sp¶p. In«nb Npäfhns hymkw sImv lcn¨v Xmsg X¶ncn¡p¶ ]«nI ]qÀ¯nbm¡m³ Bhiys¸Sp¶p. 
Npäfhv 
hymkw 
Npäfhv/hymkw 
XpSÀ¶pÅ NÀ¨bneqsSbpw hniZoIcW§fneqsSbpw hr¯¯nsâ Npäfhns hymkw sImv lcn¨m In«p¶Xv Ft¸mgpw Hcp Ønc kwJy Bbncn¡psa¶pw AXns π F¶v kqNn¸n¡p¶psh¶pw A²ym]nIbpsS klmbt¯msS hnZymÀ°nIÄ Is ¯p¶p. 
AXntmsSm¸w Xs¶ Npäfhv hymkw 
F¶pw hr¯¯nsâ Npäfhv Ip] nSn¡m³ π F¶ Ønc kwJysb hymkw sImv KpWn¡Ww F¶pw A²ym]nIbpsS klmbt¯msS hnZymÀ°nIÄ Is ¯p¶p. 
AXmbXv hr¯¯nsâ Npäfhv = 2πr F¶v A²ym]nI t{ImUoIcn ¡p¶p. 
Ip«nIÄ ]«nI ]qÀ¯nbm¡p¶p. 
= π,
Class room Interaction Procedure 
Expected pupils response 
hr¯¯nsâ Npäfhv ImWp¶Xnsâ kq{XhmIyw FgpXnb NmÀ«v {]ZÀin¸n¡p¶p. 
hr¯¯nsâ Npäfhv 
‘r’F¶ Bcapff hr¯¯nsâ Npäfhv = 2πr 
Application/review 
Hmtcm {Kq¸npw {]iv ImÀUpIÄ ÂIp¶p. 
Ip«nIÄ NmÀ«v {i²tbmsS hmbn¡p¶p. 
X¶n«pff hneIÄ r = 10 cm 
hr¯¯nsâ Npäfhv 
= 2πr 
= 2 x 3.14 x 10 
= 62.8 cm2 
X¶n«pff hneIÄ d = 8 cm 
. . . r = 4 cm 
hr¯¯nsâ Npäfhv 
= 2πr 
= 2 x 3.14 x 4 
= 25.12 cm 
hr¯¯nsâ Bcw 10 cm BsW¦n AXnsâ Npäfhv Ip] nSn¡pI? 
(π = 3.14) 
8 cm hymkapff Hcp hr¯¯nsâ Npäfhv Ip] nSn¡pI 
(π = 3.14)
FOLLOW UP ACTIVITIES 
1. 10 cm hymkapff Hcp hr¯w hcbv¡pI. AXnp tijw AXnsâ Npäfhv Ip] nSn¡pI. 
2. Xmsg X¶ncn¡p¶ hr¯¯nsâ Npäfhve Is ¯pI 
4 cm

Record1

  • 1.
    CURRICULAR STATEMENTS hr¯¯nsâNpäfhv Ip] nSn¡p¶p. CONTENT ANALYSIS Symbols : ss] π, hr¯¯nsâ Bcw ‘r’ Facts : hymkw IqSpt¼mÄ hr¯¯nsâ Npäfhv IqSp¶p. GsXmcp hr¯¯nsâbpw Npäfhns hymkw sImv lcn¨m In«p¶ kwJy Ft¸mgpw Øncambncn¡pw. Name of the Teacher : Ashabai. B.S Std : IX Name of the Subject : Mathematics Str : 40 Name of the Unit : hr¯¯nsâ AfhpIÄ Time : 45 minutes Name of the Topic : hr¯¯nsâ Npäfhv Date :
  • 2.
    Concept : hr¯¯nsâNpäfhv F¶ Bibw Formula : hr¯¯nsâ Npäfhv = 2πr ; r → hr¯¯nsâ Bcw LEARNING OUTCOME The student will be able to; 1) Recall the information about circles. 2) Describe the peculiarities of a circle 3) Interpret about the perimeter of a circle 4) Explain the concept of perimeter of a circle 5) Use the above concept in another familiar 6) Situation 7) Organise different elements fit within the solution 8) Judge the appropriateness of the above concept in a given problem 9) Accept the beauty of maths. PRE-REQUISITES Ip«nIÄ¡v hr¯w, hr¯ tI{μw, Bcw, hymkw F¶nhsb¡pdn¨pff Adnhv. TEACHING LEARNING RESOURCES Models and Charts: 1. hr¯¯nsâ amXrI 2. hr¯¯nsâ Npäfhv ImWmpff kq{XhmIyw FgpXnb NmÀ«v 3. {]iv ImÀUpIÄ
  • 3.
    Class room InteractionProcedure Expected pupils response Introduction Hcp IYbneqsS ¢mÊv Bcw`n¡p¶p. A¸phnsâ hoSnv ASp¯v Hcp ¢ºv Dv. AhnsS HmW¯nv [mcmfw ]cn]mSnIÄ Act§dmdpv. HmWw hcmdmbn A¸p ¢ºnse ]cn]mSnIÄ ImWm³ Im¯ncn¡pI bmWv. A§s HmWw F¯n ¢ºnse ]cn]mSnIÄ XpS§n. ¢ºn DdnbSn, kpμcn¡v s]m«psXmSoÂ,A¯¸q¡f aÕcw, ItkcIfn, apXemb Hcp]mSv aÕc§Ä Dmbncp ¶p. Ah³ FÃm aÕc§fpw nco£n¨p. ItkcIfn aÕcambncp¶p ASp¯ Cw . ItkcIÄ tNÀ¯n«v Hcp {]tXyI BIrXnbn aÕcmÀ°nIÄ S¡p¶Xv Ah³ {i²n¨p. AXv GXv BIrXnbn Bbncn¡psa¶v n§Ä¡v ]dbmtam? hfsc icnbmWv. A¸phnv hopw Hcp kwibw. aÕcmÀ°nIÄ B hr¯w Hcp XhW Id§n hcpt¼mÄ AhÀ F{X ZqcamWv k©cn¨Xv F¶v F§s Is ¯pw? Ah³ Hcp]mSv BtemNn¨p. ho«n F¯n A¸p Xsâ tN¨ntbmSv CsX¸än tNmZn¨p. At¸mÄ tN¨n ]dª adp]Sn F´mbncn¡psa¶v n§Ä¡v Duln¡mtam? hr¯w hr¯¯nsâ Npäfhv Is ¯n bm aXn F¶v Ip«nIÄ ]dªp.
  • 4.
    Class room InteractionProcedure Expected pupils response Presentation Ip«nIsf {Kq¸mbn Xncn¨v Hmtcm {Kq¸npw hr¯¯nsâ amXrIbpw, qepw kvsIbnepw ÂIn. F¶n«v AXnsâ Npäfhv , hymkw F¶nh ImWm³ Bhiys¸Sp¶p. In«nb Npäfhns hymkw sImv lcn¨v Xmsg X¶ncn¡p¶ ]«nI ]qÀ¯nbm¡m³ Bhiys¸Sp¶p. Npäfhv hymkw Npäfhv/hymkw XpSÀ¶pÅ NÀ¨bneqsSbpw hniZoIcW§fneqsSbpw hr¯¯nsâ Npäfhns hymkw sImv lcn¨m In«p¶Xv Ft¸mgpw Hcp Ønc kwJy Bbncn¡psa¶pw AXns π F¶v kqNn¸n¡p¶psh¶pw A²ym]nIbpsS klmbt¯msS hnZymÀ°nIÄ Is ¯p¶p. AXntmsSm¸w Xs¶ Npäfhv hymkw F¶pw hr¯¯nsâ Npäfhv Ip] nSn¡m³ π F¶ Ønc kwJysb hymkw sImv KpWn¡Ww F¶pw A²ym]nIbpsS klmbt¯msS hnZymÀ°nIÄ Is ¯p¶p. AXmbXv hr¯¯nsâ Npäfhv = 2πr F¶v A²ym]nI t{ImUoIcn ¡p¶p. Ip«nIÄ ]«nI ]qÀ¯nbm¡p¶p. = π,
  • 5.
    Class room InteractionProcedure Expected pupils response hr¯¯nsâ Npäfhv ImWp¶Xnsâ kq{XhmIyw FgpXnb NmÀ«v {]ZÀin¸n¡p¶p. hr¯¯nsâ Npäfhv ‘r’F¶ Bcapff hr¯¯nsâ Npäfhv = 2πr Application/review Hmtcm {Kq¸npw {]iv ImÀUpIÄ ÂIp¶p. Ip«nIÄ NmÀ«v {i²tbmsS hmbn¡p¶p. X¶n«pff hneIÄ r = 10 cm hr¯¯nsâ Npäfhv = 2πr = 2 x 3.14 x 10 = 62.8 cm2 X¶n«pff hneIÄ d = 8 cm . . . r = 4 cm hr¯¯nsâ Npäfhv = 2πr = 2 x 3.14 x 4 = 25.12 cm hr¯¯nsâ Bcw 10 cm BsW¦n AXnsâ Npäfhv Ip] nSn¡pI? (π = 3.14) 8 cm hymkapff Hcp hr¯¯nsâ Npäfhv Ip] nSn¡pI (π = 3.14)
  • 6.
    FOLLOW UP ACTIVITIES 1. 10 cm hymkapff Hcp hr¯w hcbv¡pI. AXnp tijw AXnsâ Npäfhv Ip] nSn¡pI. 2. Xmsg X¶ncn¡p¶ hr¯¯nsâ Npäfhve Is ¯pI 4 cm