www.aydinyayinlaricom.tr ¦&.#&37&%\"÷3& 5. MODÜL 11. SINIF )(1/m6(/(5m1(<q1(/m. ÖRNEK 16 ÷LJ¦FNCFSJO0SUBL5FôFUMFSJ E ôFLJMEFLJ¿FNCFSMFS\" x EBEõUBOUFóFUWF#$ D 7$1,0%m/*m 50° PSUBLEõUFóFUUJS A A m ( % ) = 50° BDA B d1 C D d2 BC :VLBSEBLJWFSJMFSFHÌSF m ( A%EC ) =x LBÀEFSFDFEJS \" # $ %UFóFUEFóNFOPLUBMBSPMNBLÐ[FSF d1WFE2EPóSVMBSOBJLJ¿FNCFSJOPSUBLEöUF- && ôFUEPôSVMBSEFOJS m (AB) = 100°, m (AC) = 2x° +Y=j x = r |\"#| = | $%| ÖRNEK 17 B d1 AD A Fx K 50° C K ED CB d2 \"#WF&%EõUBOUFóFU¿FNCFSMFSJOPSUBLEõ teóFU- MFSJEJS % ) = 50° PMEVôVOB HÌSF m ( A%FE ) =x m ( BCD \" # $ %UFóFUEFóNFOPLUBMBSPMNBLÐ[FSF LBÀEFSFDFEJS d1WFE2EPóSVMBSOB JLJ¿FNCFSJOPSUBLJÀUF- ) ) ôFUEPôSVMBSEFOJS m (AKE) + m (BCD) = 360°CVMVOVS r |\",| = | ,$| |KD | = |KB| |\"#| = |$%| 2x +=j x = A EB d1 ÖRNEK 18 d2 A d dPóSVTV [$%] K T B d WF [DE] ¿BQM C FD 7 3 ZBSN¿FNCFSMF- d3 SF\"WF#OPLUB- öLJ¿FNCFSCJSCJSJOFEõUBOUFóFU UFóFUOPLUBT C H D KE ,PMTVO#VEVSVNEBE1WFE2EPóSVMBSOBPS- MBSOEBUFóFUUJS UBLEõUFóFUEPóSVMBS E3EPóSVTVOBPSUBLJÀ UFôFUEPôSVTVEFOJS | | | |[\")] m [$E] [BK] m [$&] \") =CS BK = 3 br | |PMEVóVOBHËSF ,) LBÀCJSJNEJS | |0SUBLJÀUFôFUÀJ[JMJSTF \"),#ZBNVôVOEB 5% PSUBUB- | |CBO 5% = 5 r |\"#| = | $%| A #VEVSVNEB r |\"&| = |EB| = |EK| = |FK| = |$'| = |FD| 4 10 | |\"# =CS M x B A&BM EFQJTBHPS 3 3 r %% m (AK) + m (BK) = 180° H x K x = 2 21 r %% m (CK) + m (DK) = 180° 49 16. 40 17. 130 18. 2 21
TEST - 20 ¦FNCFSEF5FôFUWF6[VOMVLm** 1. 4. 01WF02NFSLF[MJ ¿FNCFSMFSJO ZBS- A O1 O2 ¿BQMBS TSBTZMB O1 O2 WFJMFPSBOUMES KL | |0102 = 8 br PMEVôVOBHÌSF ÀFNCFSMFSJOZBS- B ÀBQMBSUPQMBNOOBMBCJMFDFôJLBÀGBSLMUBNTB- ôFLJMEFLJ01WF02NFSLF[MJ¿FNCFSMFS\"WF#OPL- ZEFôFSJWBSES UBMBSOEBLFTJõNFLUFEJS \" # $ % & | |[0102][,-]WF 0102 =DNPMEVôVOB | |HÌSF ,- LBÀDNEJS 5. 01WF02NFSLF[MJ¿FNCFSMFS,WF-OPLUBMBSOEB \" # $ % & LFTJõNFLUFEJS K 2. d1 C A O1 A B O2 O1 T O2 B L D d2 | | | |[,-] a [0102] = {B} \"# =CS 02B = 2 br PM- 01WF02NFSLF[MJ¿FNCFSZBZMBSCJSCJSJOF5OPL- UBTOEB E1WFE2EPóSVMBS02NFSLF[MJZBZB$WF EVôVOBHÌSF O1O2,ÑÀHFOJOJOBMBOLBÀCJSJN- %OPLUBMBSOEBUFóFUUJS LBSFEJS \" 50 6 B) 80 2 $ | |O2NFSLF[MJÀFNCFSJOZBSÀBQCS \"$ =CS D) 75 5 E) 75 6 PMEVôVOBHÌSF O1NFSLF[MJÀFNCFSJOZBSÀBQ 6. ôFLJMEFLJ\"#$%EJLEËSUHFOJOJOJ¿JOF¿J[JMFO\"NFS- LBÀCJSJNEJS LF[MJ¿FZSFL¿FNCFSWF[#$]¿BQMZBSN¿FNCFS5 \" # $ % & OPLUBTOEBUFóFUUJS 3. T DC N A O1 B O2 C 4 T ôFLJMEF01WF02NFSLF[MJZBSN¿FNCFSMFS#OPL- UBTOEBCJSCJSJOFUFóFU \" # $OPLUBMBSEPóSVTBM- A EB ES | | | |\"% = 4 br PMEVôVOBHÌSF EB LBÀCJSJNEJS [025 01 NFSLF[MJ ¿FNCFSF 5 OPLUBTOEB UFóFU \" 3_ 2 - 1 i B) 4_ 2 - 1 i | | | |\"# = DN WF #$ = 4 cm PMEVôVOB HÌSF $ 3_ 5 - 3 i D) 4_ 3 - 1 i | |5/LBÀDNEJS \" # $ % & E) 4_ 5 - 1 i 1. \" 2. C 3. B 50 4. \" 5. \" 6. B
¦FNCFSEF5FôFUWF6[VOMVLm** TEST - 21 1. ôFLJMEF 02 4. C T E ôFLJMEFLJ ¿FNCFSMFS $ NFSLF[MJ ¿FN- A D OPLUBTOEB J¿UFO UFóFU- O2 D CFS 01 mer- B LF[MJ ZBSN UJS[\"&]J¿UFLJ¿FNCFSF x ¿FNCFSF $ WF P %OPLUBTOEBUFóFUUJS ) % OPLUBMBSOEB A O1 C 1 B UFóFUUJS m (APC) = 120° ) m (DTC) = 140° | |01NFSLF[MJ¿FNCFSJOZBS¿BQCJSJNWF #$ = 1 br PMEVôVOBHÌSF 02NFSLF[MJÀFNCFSJOZBSÀBQ :VLBSEBLJWFSJMFSFHÌSF m ( % ) = x LBÀEF- LBÀCJSJNEJS CAE \" 4 B) 5 $ 6 D) 7 8 SFDFEJS 5 6 7 8 E) \" # $ % & 9 2. A ôFLJMEFLJ¿FNCFSMF a % & ' OPLUBMBSO- 5. D E C C T Di EBUFóFUUJS E F m ( D%AE ) = a b % = b m ( EBF ) B maD%CFk = i :VLBSEBLJWFSJMFSFHÌSF a + b + i UPQMBNLBÀ AB EFSFDFEJS ¥FWSFTJ CJSJN PMBO \"#$% LBSFTJOJO J¿JOEFLJ # \" # $ % & NFSLF[MJ¿FZSFL¿FNCFSJMF[DE]¿BQMZBSN¿FN- CFS5OPLUBTOEBUFóFUUJS 3. \"#$%EJLEËSUHFOJõFLMJOEFLJCBI¿FOJO[\"%]WF[#$] | | #VOBHÌSF DE =YLBÀCJSJNEJS ¿BQMZBSNEBJSFCJ¿JNJOEFLJCËMHFMFSJOFNTSEJLJM- \" # $ % & NJõWF,OPLUBTOBCJSJQMF CJSJOFLCBóMBONõUS DC 6. D [\"#] WF [$#] E ¿BQM ZBSN ¿FNCFSMFSEF AK B <%0><\"#> | | | | | | | |\", = ,# \"# = DN BC = DN PM- A CO |\"0| = |0#| EVôVOBHÌSF JOFôJONTSFLJMJCÌMHFMFSFHJSNF- B EFOPUMBZBCJMFDFôJBMBOOFOGB[MBPMNBTJÀJOJQ V[VOMVôVLBÀNFUSFPMNBMES | | | | | |DE =DN &0 = 3 cmPMEVôVOBHÌSF \"$ \" # $ % & LBÀDNEJS \" # $ % & 1. D 2. C 3. B 51 4. \" 5. E 6. B
TEST - 22 ¦FNCFSEF5FôFUWF6[VOMVLm** 1. D C ôFLJMEF CÐZÐL ¿FN- )(1/m6(/(5m1(<q1(/m. CFS EJLEËSUHFOJOJO п LFOBSOB Fõ PMBO JLJ 4. A B d1 LпÐL ¿FNCFS JLJõFS aF d2 LFOBSOBUFóFUUJS E 74° O1 C O2 AB D | |#$ = DN PMEVóVOB HËSF LÌöFMFSJ CV ÀFN- ôFLJMEFLJE1WFE2EPóSVMBS 01WF02NFSLF[- CFSMFSJONFSLF[MFSJPMBOÑÀHFOJOBMBOLBÀDN2 EJS \" 8 3 B) 8 2 $ MJ¿FNCFSMFSFUFóFUUJSMFS\" # $ %UFóFUEFóNF D) 4 3 E) 4 2 OPLUBMBS WF % = 74° PMEVôVOB HÌSF m ( AEC ) % = a LBÀEFSFDFEJS m ( BFD ) \" # $ % & 2. ôFLJMEF J¿UFO UFóFU JLJ 5. \"WF#PSUBLEõUFóFUMFSJOEFóNFOPLUBMBSE S ¿FNC FSJONFSLF[MFSJ0 O 2T M 6 WF.EJS[,5] 0NFS- LF[MJ¿FNCFSF5OPLUB- TOEBUFóFUUJS K | | | |05 = DN WF 5, = DN PMEVôVOB HÌSF AB | |OM LBÀDNEJS ¦FNCFSMFSJO NFSLF[MFSJ BSBTOEBLJ V[BLML \" 5 # $ 7 9 DN ZBSÀBQMBS DN WF DN PMEVôVOB HÌSF 2 2 D) 4 E) | |\"#LBÀDNEJS 2 \" # $ D) 8 2 E) 6 3 3. A 0 NFSLF[MJ ¿FZSFL 6. \"# WF $% JLJ ¿FNCFSJO PSUBL Eõ UFóFUMFSJ [,-] WF[KM]¿FNCFSMFSFTSBTZMB/WF5OPLUBMBSO- 2 ¿FNCFSWF[0#]¿BQ- EBUFóFUUJS MZBSN¿FNCFSWFSJM- CD NJõUJS A K B 30° B N T O CL MD % | | | |m(BOD) = 30° \"0 = 2 br PMEVôVOBHÌSF CD LBÀCJSJNEJS ,-.ÑÀHFOJOÀFWSFTJCJSJNPMEVôVOBHÌ- \" 2 - 3 B) 2 $ 3 - 1 | | | |SF \"# + CD UPQMBNLBÀCJSJNEJS D) 3 E) 2 - 2 \" # $ % & 1. B 2. E 3. \" 52 4. B 5. B 6. C
www.aydinyayincilik.com.tr ¦&.#&37&%\"÷3& 5. MODÜL 11. SINIF %\"÷3&/÷/¦&73&4÷7&\"-\"/* ÷MJöLJMJ,B[BONMBS 11.5.4.1 : %BJSFOJO¿FWSFWFBMBOCBóOUMBSOPMVõUVSVS %BJSFOJO¦FWSFTJWF:BZ6[VOMVôV ÖRNEK 2 ôFLJMEFLJEBJSFEF 7$1,0%m/*m Çemberin kendisi ile çemberin iç bölgesinin A m( % = 30° CJSMFõJNJOFdaire denir. 30° BAC ) SZBS¿BQMCJSEBJSFOJO¿FWSFV[VOMVóVOVO¿BQB B8 | BC | = 8 br PSBOÕZFFõJUUJS K ) C Buna göre, BKC kaç birimdir? 0.FSLF[ r S:BS¿BQ m^ ) h = 60° j |BC| = r = 8 br j >;; ? = 8π O Çevre BKC BKC 3 =π 2r #VEVSVNEBSZBS¿BQMCJSEBJSFOJO¿FWSFTJ ÖRNEK 3 0NFSLF[MJ¿FNCFS r Ç =ÕS ZBZMBSWFSJMNJõUJS O 0NFSLF[MJ SZBS¿BQMEBJSFOJO\"0#NFSLF[B¿- D | OD | = 6 cm % C B | BD | = 4 cm A TOOHËSEÐóÐZBZV[VOMVóV AB JMFHËTUFSJMJS & CD =ÕDN A 0.FSLF[ & :VLBSEBLJWFSJMFSFHÌSF AB kaç cm dir? r S:BS¿BQ a° % a° O AB = 2πr· B 360° 0PSUBLNFSLF[PMEVôVOEBO NFSLF[BÀÌMÀÑMFSJBZOES r 6 3π & = & AB = 5π 10 & AB ÖRNEK 1 ÖRNEK 4 0NFSLF[MJ¿FNCFSJOZBS¿BQ B |\"#| = | BC | = 8 br CSWFm % = 30° 120° ) ( AOB ) m (AKC) = 240° C O ) A r3 Buna göre, O merkez- 6 30° Buna göre, ACB kaç br dir? r r li çemberin çevresi kaç O birimdir? AB C ) a° = 2π .6· 30° K ACB = 2πr· =π 360° 360° %% m ( ABC ) = 120° ve m ( AOC ) = 120° |AB| = |AC| = r = 8 , Çevre =Ö=Ö 1. Ö 53 8π 3. Ö 4. Ö 2. 3
11. SINIF 5. MODÜL ¦&.#&37&%\"÷3& www.aydinyayincilik.com.tr ÖRNEK 5 ÖRNEK 7 0NFSLF[MJLпÐL¿FNCFS CÐZÐL¿FNCFSFUFóFUPMBDBL õFLJMEFTBBUZËOÐOEFEËONFLUFEJS B K ( m (AKB) = 120° O A | 0\"| = 3 cm Dik silindir biçiminde ZBS¿BQMBSFõJUWFCSPMBOпLÐ- ,ÑÀÑLÀFNCFS\"OPLUBTOEBO#OPLUBTOBJMLVMBöU- UÐLJQMFHFSHJOCJSõFLJMEFZVLBSEBLJHJCJCBóMBONõUS ôBOEBJLJUBNUVSBUUôOBHÌSF CÑZÑLÀFNCFSJOZB- ,VMMBOMBOJQJOV[VOMVôVen az kaç br dir? SÀBQLBÀDNEJS AK ,ÑÀÑLÀFNCFSCJSUBNUVSEBÀFWSFTJ Ö LBEBSZPMBMS ) 6 12 6 AKB = 2.6π = 12π L M #VEVSVNEB CÑZÑLÀFNCFSJOÀFWSFTJ >;; ? ,.-EJLEÌSUHFO]\",]= |LM| = 24 %& = 3. AKB = 36π AL + KM = 2π .6 = 12π ÷QV[VOMVôVFOB[= 48 +Ö Ö3=Öj R = 18 ÖRNEK 6 ÖRNEK 8 \"SDIJNFEFT \"SõJNFU Õ TBZTOO EFóFSJOJ FMEF FUNFL N V[VOMVóVOEB CJS JQF CBóMBONõ ZBS¿BQ N PMBO J¿JO CJS ZËOUFN HFMJõUJSNJõUJS #V ZËOUFN #JS ¿FNCFSJO ¿FNCFS\"LPOVNVOEBZLFOTFSCFTUCSBLMQ#LPOVNV- ¿FWSF V[VOMVóV LËõFMFSJ CV ¿FNCFSJO Ð[FSJOEF PMBO O OBLBEBSVMBõZPS LFOBSM EÐ[HÐO ¿PLHFOJO ¿FWSF V[VOMVóV JMF LFOBSMBS CV¿FNCFSFUFóFUPMBOOLFOBSMEÐ[HÐO¿PLHFOJO¿FWSF O V[VOMVóVBSBTOEBES 88 #VOB HÌSF ZBSÀBQ CS PMBO CJS ÀFNCFSEF \"SDIJ- NFEFT ZÌOUFNJZMF Ö TBZTO IFTBQMBNBL JÀJO EÑ[- 22 HÑOBMUHFOLVMMBOMSTBÖTBZTJÀJOCVMVOBDBLBSB- M M' MôIFTBQMBZO[ AB FA ÷ÀFSEFLJ BMUHFOJO ÀFWSFTJ ¦FNCFSJONFSLF[JNZFSEFôJöUJSEJôJOFHÌSF NFS- = 36 br LF[JOBMEôZPMLBÀNFUSFEJS 6 Çemberin çevresi =ÖCS | |:FSEFôJöUJSNF ..h =CVEVSVNEBTFSCFTUCSBLM- E 6 B %öBSEBLJBMUHFOJOÀFWSFTJ NBZMBPMVöBOZBZONFSLF[J0 ZBSÀBQCS NFSLF[ O = 24 3 br BÀT \"SDIJNFEFThFHÌSF 60° 10π DC 36 <Ö< 24 3 NFSLF[JOBMEôZPM=Ö = m 360° 3 3 <Ö< 2 3 bulunur. 5. Ö 6. Ö 2 3 54 7. 18 10π 8. 3
www.aydinyayincilik.com.tr ¦&.#&37&%\"÷3& 5. MODÜL 11. SINIF %BJSFOJO\"MBO ÖRNEK 9 %m/*m \"MBO Ö CS2 PMBO EBJSFOJO MJL NFSLF[ BÀTOO r HÌSEÑôÑZBZOV[VOMVôVLBÀCJSJNEJS O %BJSFOJOBMBO=Ö=ÖS2 j r = 2 :BZV[VOMVôV=Ö 90° =π 360° 0NFSLF[MJ SZBS¿BQMEBJSFOJOBMBO ÖRNEK 10 0NFSLF[MJEBJSFEF r \"=ÕS2 C | |OB =DNWF %BJSF%JMJNJOJO\"MBO 7$1,0%m/*m 40° maA%CBk = 40° #JSEBJSFEFJLJGBSLMZBS¿BQOEBJSFEFOBZSEó O QBS¿BMBSOIFSCJSJOFdaire dilimi denir. 4 AB r A :VLBSEBLJ WFSJMFSF HÌSF UBSBM \"0# EBJSF EJMJN JOJO 0.FSLF[ BMBOLBÀDN2 dir? a° Or S:BS¿BQ %BJSFEJMJNJOJONFSLF[BÀÌMÀÑTÑEJS B 5BSBMBMBO=Ö2· 80° 32π = 360° 9 ôFLJMEFLJUBSBMBMBOCJSEBJSFEJMJNJEJS % = a° PMNBLÐ[FSF m ( AOB ) r %BJSFEJMJNJOJOBMBO=ÕS2 · a ÖRNEK 11 360° D %BJSF)BMLBTOO\"MBO 6 7$1,0%m/*m 0SUBLNFSLF[MJJLJEBJSFOJOBSBTOEBLBMBOCËM- 60° geye EBJSFIBMLBTdenir. A OB C 0.FSLF[ 0NFSLF[MJZBSN¿FNCFSJOZBS¿BQ6 3 DN [ CD çem- ôFLJMEFLJ UBSBM CFSF%OPLUBTOEBUFóFU m ( D%CA ) = 60° PMEVôVOBHÌ- SF UBSBMalan kaç cm2 dir? BMBOCJSEBJSFIBM- OA B LBTES | 0\"| = r1 OD m DC j % = 30° | OB | = r2 m ( COD ) PMNBLÐ[FSF 5BSBMBMBO= Aa & k -%BJSFEJMJNJOJOBMBO ODC = 6.6 3 - π .108· 30° = 18 3 - 9π r %BJSFIBMLBTOOBMBO= π a r22 - r 2 k 2 360° 1 55 9. Ö 10. 32π 11. 18 3 - 9π 9
11. SINIF 5. MODÜL ¦&.#&37&%\"÷3& www.aydinyayincilik.com.tr ÖRNEK 12 ÖRNEK 14 B #NFSLF[MJ¿FNCFSZBZ\"#$%EJLEËSUHFOJOJ\"WF&OPL- UBMBSOEBLFTJZPS D 6E 6C A OH 4C 63 12 6 3 A 0NFSLF[MJZBSNEBJSFWFSJMNJõUJS#)m\"$ 30° 60° | |%BJSFOJOZBS¿BQDNWF )$ = 4 cm 12 B \"#$ÑÀHFOPMEVôVOBHÌSF, taSBMBMBOLBÀDN2 dir? | | | |\"# =DNWF \"% = 6 3 cm PMEVôVna göre, ta- SBMCÌMHFOJOBMBOLBÀ cm2 dir? | |% |BE| = |AB| = r = 12 m ( ABC ) #&$ÑÀHFOJOJ--ÑÀHFOJEJS = 90° ve AH =CSPMEVôVOEBO Bu durumda m ( A%BE ) = 60° | | | |BH 2 = ²LMJE BH = 8 br 5BSBMBMBO=\" \"#&% -%BJSFEJMJNJOJOBMBO 2 π .10 20.8 5BSBMBMBOMBSUPQMBN= - = 50π - 80 ^ 12 + 6 h.6 3 60° = - π .144· = 54 3 - 24π 22 2 360° ÖRNEK 15 ÖRNEK 13 A NM E S1 D 3a O K 8L 4a O5a S2 S3 BFC \"#$ пHFOJOJO 0 NFSLF[MJ J¿ UFóFU ¿FNCFSJ WFSJMNJõUJS 0NFSLF[MJ¿Fmberin içinde ,-./LBSFTJWFSJMNJõUJS S1 42WF43 J¿JOEFCVMVOEVLMBSEBJSFdilimlerinin alan- MBSOHËTUFSNFLÐ[FSF S1 = S2 = S3 UJS ,BSFOJOCJSLFOBSCSPMEVôVOBHÌSF UBSBMCÌMHFOJO BMBOLBÀCS2 dir? 345 :VLBSEBLJWFSJMFSFHÌSF m ( % ) kaç derecedir? BAC O merkez, O ` [,.]PMEVôVOEBOS= 4 2 SSS Dairenin alanı - A^ KLMN h 32π - 64 123 = = 8π - 16 3 = 4 = 5 41 =4 42 =4 43 =4 5BSBMBMBO= 44 #VCÌMHFMFSJOBMBOMBSNFSLF[BÀÌMÀÑMFSJJMFEPôSVPSBOUM PMEVôVOEBO a =j a = m( % = 180° - 90° = 90° BAC ) 12. Öm 13. Öm 56 14. 54 3 - 24π 15. 90
www.aydinyayincilik.com.tr ¦&.#&37&%\"÷3& 5. MODÜL 11. SINIF ÖRNEK 16 ÖRNEK 18 F E D B 36 6 C4 3 8 30° 43 A C 33 O 33 D B O4 A 4 E [\"#]¿BQMZBSNEBJSFEF [ FC ] [\"#] [&%] [\"#] | | | |0NFSLF[ 0\"#$EJLEËSUHFO 0\" = \"& = 4 br | | | | | |\"# =DN FC = &% = 3 cm #VOBHÌSF UBSBMBMBOkaç br2 dir? :VLBSEBLJ WFSJMFSF HÌSF UBSBM CÌMHFOJO BMBO LBÀ cm2 dir? % = % = 30° & % = 120° | |OB = r =CS 0#$ÑÀHFOJ--ÑÀHFOJEJS m ( FOC ) m ( EOD ) m ( EOF ) | |% 5BSBMBMBO=Ö 120° 3 3 .3 ·2 =Ö+ 9 3 m ( BOC ) = 30°, OC = 4 3 br + 5BSBMBMBO=%BJSFEJMJNJBMBO-\" 0$# 360° 2 30° 4.4 3 16π 2 = 64π · - = - 8 3 br 360° 2 3 ÖRNEK 19 ÖRNEK 17 DKC A C SS S S D S L A F x S S A S AE B O 12 B \"#$%LBSFTJOJOJ¿UFóFU¿FNCFSJJMF#WF%NFSLF[MJ¿FZ- SFL¿FNCFSMFSWFSJMNJõUJS\" \"#$% = 72 br2 [\"0] m [ OB ] [ CB ] 0NFSLF[MJ¿FZSFL¿FNCFSF#OPL Buna göre, UBSBMalan kaç br2 dir? UBTOEBUFóFUUJS ,-&'CJSLFOBSCSPMBOLBSFEJS | |OB =DNWFUBSBMBMBOMBSCJSCJSJOFFöJUPMEVôVOB 5BSBMBMBOMBSUPQMBN=4+ A | |göre, BC LBÀÖDNEJS 4+ A =\" ,-&' = 62 = 36 ¦FZSFLEBJSFEJMJNJOJOBMBO=\" 0#$ 90° 12.x 144π . = & 6π = x 360° 2 16. Ö 9 3 17. 6 57 16π 18. - 8 3 19. 36 3
11. SINIF 5. MODÜL ¦&.#&37&%\"÷3& ÖRNEK 22 www.aydinyayincilik.com.tr B ÖRNEK 20 A K R S S O R r A 6 B 3C 3 D [\"%] LпÐL¿FNCFSF$OPLUBTOEBUFóFUUJS0NFS- D 6C LF[MJJLJ¿FNCFSBSBTOEBLBMBOUBSBMBMBOr br2 dir. õekildeki ABCD karesinde A ve C merkezli, 6 cm | | | |AB =CSPMEVôVOBHÌSF \", kaç br dir? ZBSÀBQM ÀFZSFL ÀFNCFSMFSJO BSBTOEB LBMBO UBSBM alan kaç cm2 dir? OC m\"% 0,m\", 90° 6.6 | | | |OC = r, 0, = R, R2Ö- r2Ö=Ö 32 - r2 = 9 S = 36π . - = 9π - 18 j4=Ö- 36 | |0%$ÑÀHFOJOEFQJTBHPS CD = 3 360° 2 0\"$WF0\",ÑÀHFOMFSJOEFQJTBHPS | |92 + r2 = R2 + \", 2 | |81 - 9 = \", 2 6 2 = AK ÖRNEK 21 ÖRNEK 23 C ôFLJMEF CJS LFOBS CS PMBO \"#$% LBSFTJOJO J¿JOF % NFSLF[MJEËSUUFCJS¿FNCFS¿J[JMNJõUJS AB D ai FSS S S A6 O 6B G S S 0NFSLF[MJZBSNçembere [%\"] [ BC ]WF[ CD ]UFóFU DE C | |\"#å= 12 br | | | | | | | |\"' = FD = %& = &$ PMEVôVOBHÌSF UBSBMBMBO Buna göre, tarBMBMBOMBSUPQMBNLBÀCJSJNLBSFEJS kaç br2 dir? [DO]æWF[CO]BÀPSUBZPMEVôVOEBO m ( D%OC ) = 90° bulunur. ( 0$#ÑÀHFOJOJOBôSMLNFSLF[J S = A^ ADC h = 12 6 Bu durumda a + i =EJS 5BSBMBMBO=¦FZSFLEBJSFOJOBMBO-4 5BSBMBMBOMBSUPQMBN= 36π · 90° = 9π 144π = - 4.12 = 36π - 48 360° 4 20. 6 2 21. Ö 58 22. Öm23. Öm
www.aydinyayincilik.com.tr ¦&.#&37&%\"÷3& 5. MODÜL 11. SINIF ÖRNEK 24 ÖRNEK 26 D B C \"#$%LBSFTJOJOJ¿UFóFU ôFLJMEF\"#$EJLпHF- A A ¿FNCFSJ WF ¿FWSFM ¿FN- OJOJO LFOBSMBSOB O1 CFSJWFSJMNJõUJS A S2 O2 03 NFSLF[MJ ZBSN B O2 çembeSMFS¿J[JMNJõUJr. B S1 O1 B O3 C S3 A A A BB ,BSFOJOCJSLFOBSCSPMEVôVOBHÌSF UBSBMBMBOMBS S1 42 43CVMVOEVLMBSUBSBM bölgeleSJOBMBOMBSPMNBL UPQMBNLBÀCS2 dir? Ð[FSF 41 = 10r br2 42 = 18r br2 PMEVóVOBHËSF 43 kaç br2 dir? ,ÑÀÑLEBJSFOJOZBSÀBQ= 4, #ÑZÑLEBJSFOJOZBSÀBQ= 4 2 | | | | | |O A=r = r, = ve 2 = 2 + 2 4A =,BSFOJOBMBO-,ÑÀÑLEBJSFOJOBMBO 1 4B =#ÑZÑLEBJSFOJOBMBO-,BSFOJOBMBO 1 , OA 2 OB r r r r ÷LJEFOLMFNPSUBLÀÌ[ÑMÑSTF 2 13 2A + 2B =Ö 2 3 3 2 2 2 π .r π .r π .r 123 S= ,S = ,S= 12 22 32 #VSBEBO 42 =41 +43 jÖ=Ö+43 j43 Ö ÖRNEK 25 D 4C \"#$% EJLEËSUHFOJ [\"%] ÖRNEK 27 8–4 2 LFOBS EPóSVMUVTVOEB #JS NPUPTJLMFUJO ZBSN EBJSF õFLMJOEFLJ ËO DBN \" NFS- CJSEVWBSCVMVOBOCJOB- LF[OPLUBTFUSBGOEBMJLB¿JMFEËOFCJMFOCJSTJMFDFL OOÐTUUFOHËSÐOÐõÐEÐS JMF UFNJ[MFONFLUFEJS 4JMFDFL \" OPLUBTOB FO B[ DN FO¿PLDNV[BLMLUBLJOPLUBMBSUFNJ[MFZFCJMNFLUFEJS AB | DC | = 4 m | |CB = - 4 2 m \"#$%CJOBTOO$LÌöFTJOFNV[VOMVôVnda birJQ- MF CBôMBONö JOFôJO PUMBZBCJMFDFôJ NBLTJNVN BMBO kaç metrekaredir? 30° 8 ÷OFôJO PUMBZBCJMF- DFôJ UPQMBN BMBO UBSBONöUS D 60° C 210° 115500°° 30° 4 A 8–4 2 4JMFDFôJOUFNJ[MFZFCJMEJôJBMBOÌODBNO 2 ÑPMEV- A 4 B 3 45° ôVOBHÌSF ÌODBNZÑ[FZJOJOZBSÀBQLBÀDNEJS 42 42 Ö 2 - 82 150 = πr 2 2 · 4.4 3 210° 45° 4.4 360 2 3 j r = 8 10 cm 5PQMBNBMBO= 2 + 64π · + 32π · + 360° 360° 2 124π = +8 3+8 3 24. Ö 124π 59 26. Ö 27. 8 10 25. + 8 3 + 8 3
TEST - 23 %BJSFOJO¦FWSFTJWF\"MBO 1. %ËSUFõEBJSFEJMJNJOEFO 4. O 0 NFSLF[MJ EBJSF diliminde ZBQMNõ UBTBSN LÐQF E õFLJMEFLJHJCJEJS A | |OF =CS | |F FB = 3 br Bir daire diliminin NFSLF[BÀT ZBS- % ÀBQ DNPMEVôVOB B EF = 6r br HÌSF CVLÑQFOJOÀFW- SF V[VOMVôV LBÀ CJ- & rimdir? :VLBSEBLJWFSJMFSF göre, | AB | kaç birimdir? \" r # r $ r % r & r \" Õ # Õ+ $ Õ+ 36 % Õ & Õ+ 36 5. K B 0 NFSLF[MJ EBJSF- de A m ( B%AC ) = 45° A % O ( 2. BC = 4r cm m (AKB) = 60° 45° BC :VLBSEBWFSJMFOMFSFHÌSF EBJSFOJOZBSÀBQLBÀ 5BSBMCÌMHFOJOÀFWSFTJ Ö+ CSPMEVôVOB cm dir? HÌSF EBJSFOJOZBSÀBQLBÀCJSJNEJS \" # $ % & \" # $ % & 6. \"#$EJLпHFOJOJOJ¿UFóFU¿FNCFSJOJOUFóFUEFóNF 3. D C \"#$% LBSFTJOJO J¿JOEF OPLUBMBS% & 'OPLUBMBSES \"WF#NFSLF[MJ¿FZSFL A E EBJSFMFSWFSJMNJõUJS 5 E D 12 AB BF C % &% | | | |\"& =CS &$ = 12 br PMEVôVOBHÌSF ÀFNCF- DE = 5π birim PMEVóVOBHËSF AC + BE UPQ- MBNLBÀÖCJSJNEJS rin çevresi kaç birimdir? \" # $ % & \" Õ # Õ $ Õ % Õ & Õ 1. C 2. B 3. B 60 4. E 5. A 6. E
%BJSFOJO¦FWSFTJWF\"MBO TEST - 24 1. 15 C ôFLJMEF 0 NFSLF[MJ 4. :BS¿BQ CJSJN PMBO . NFSLF[MJ ¿FNCFS [\"#] O 8 A [\"#] ¿BQM¿FNCFS¿J- ¿BQM ZBSN ¿FNCFSF UFóFU PMBDBL õFLJMEF , LPOV- NVOEBO-LPOVNVOBHFUJSJMJZPS [JMNJõUJS M | |B \"$ =DN | BC | = 8 cm ) KA BL :VLBSEBLJWFSJMFSFgöre, |ACB | kaç cm dir? \" r # 17r $ 17r | |\"# = 24 br PMEVôVOB HÌSF . OPLUBTOO BM- 2 3 EôUPQMBNZPMLBÀCJSJNEJS & 17r 7 \" Õ # Õ $ Õ % Õ & Õ % 17r 5. 4 A 2. O1WF02NFSLF[MJZBSNEBJSFMFSWFSJMNJõUJS A O1 O2 B B | |AB = DN PMEVôVOB HÌSF UBSBM CÌMHFOJO ¥FNCFSZBZõFLMJOEFLJTVLBZESBóOBZFSFEJLWF N V[VOMVóVOEBLJ CJS NFSEJWFO JMF ¿LMNBLUBES çevresi kaç cm dir? ,BZESBóOV¿MBS\"OPLUBTOEBNFSEJWFOF #OPL- UBTOEBZFSFUFóFUPMBDBLõFLJMEFTBCJUMFONJõUJS \" Õ+ # Õ $ Õ+ 6 #VOBHÌSF LBZESBLUBOLBZBOCJSJLBZESBLÑ[F- SJOEFLBÀNFUSFZPMBMBDBLUS % Õ+ & Õ+ 6 \" Õ # Õ $ Õ % Õ & Õ 3. ¥FWSFTJCJSJNPMBO\"#$пHFOJOJOUÐNLËõFMF- 6. [\"#] ¿BQM ZBSN ¿FNCFSJO J¿JOF [\"$] [CD] [DB] SJOEFOZBS¿BQMBSCJSJNPMBOEBJSFEJMJNMFSJLFTJMJQ ¿BQMZBSN¿FNCFSMFS¿J[JMNJõUJS BUMZPS A 24 6 BC AC D B #VOBHÌSF LBMBOöFLMJOÀFWSFTJLBÀCJSJNPMVS %&% AC = 2 br CD = 4 br DB = 6 br \" +Õ # +Õ $ +Õ & Buna göre, AB kaç birimdir? % & Õ \" # $ % & 1. B 2. D 3. C 61 4. C 5. A 6. C
TEST - 25 %BJSFOJO¦FWSFTJWF\"MBO 1. A :BS¿BQ DN PMBO õF- 4. [\"$ 0NFSLF[MJ¿FNCFSF$OPLUBTOEBUFóFUUJS LJMEFLJEBJSFEF 45° |\"#| = |\"$| A 1B % O m ( BAC ) = BC 3 C :VLBSEBLJ WFSJMFSF HÌSF UBSBM CÌMHFOJO BMBO | | | | \" # 0 EPóSVTBM \"# = DN \"$ = 3 cm kaç cm2 dir? PMEVó VOB HËSF UBSBM CÌMHFOJO BMBO LBÀ DN2 \" + 4r # 8 2 + 4r $ 8 2 + 8r dir? % 16 + 8r & 16 2 + 8r \" 3 - r # 3 - r $ 3 - r 3 2 3 2 6 % 3 - r & 3 - r 4 6 4 12 2. 5BCBOEBJSFõFLMJOEFPMBOCBSVUG¿MBSOEBOUB- 5. :BS¿BQMBS PSBO 1 2 OFTJ õFLJMEFLJ HJCJ ÐTU ÐTUF ZBUSMBSBL TBLMBOZPS A '¿MBSOLBZNBNBTJ¿JOFUSBGHFSHJOCJSJQMFJLJLF[ B PMBO BZO NFSLF[MJ TBSMZPS dairelerden kпÐóÐ [\"#]ZFUFóFUUJS | |\"# =DNPMEVóVOBHËSF UBSBMCÌMH FOJOBMB- OLBÀDN2 dir? \" r - 12 3 # r - 8 3 $ 16r - 6 3 % r - 8 3 'ÀMBSEBOCJSJOJOUBCBOZBSÀBQDNPMEVôV- & r - 6 3 OBHÌSF LVMMBOMBOJQJOV[VOMVôVen az kaç cm dir? \" Õ # Õ 6. :BS¿BQMBS DN $ Õ % Õ PMBO Fõ ¿FNCFSMFS õFLJMEFLJ HJCJ CJSCJ & Õ SJOFUFóFUUJSMFS 3. 4BCJUCJSOPLUBZBCJSJQMFCBóMPMBOLPZVOTBCJUCJS I[MBTBBUCPZVODBPUMBNõGBLBUEPZNBNõUS ,PZVOVOEPZNBTJÀJOBZOI[MBTBBUEBIBPU- #VOBHÌSF UBSBMCÌMHFOJOBMBOLBÀcm2 dir? MBNBTHFSFLUJôJOFHÌSF JQV[VOMVôV en az kaç \" - 2r # - 4r $ - 4r LBUOBÀLBSMNBMES \" # $ 3 % 2 & 3 % - 8r & - 16r 2 1. B 2. E 3. D 62 4. C 5. A 6. E
%BJSFOJO¦FWSFTJWF\"MBO TEST - 26 1. C [\"#]¿BQMZBSNEB- 4. C [\"#] ¿BQM ZB- ireye [ BC ] #OPLUa- 30° SNEBirede TOEBUFóFUUJS A D % = 30° m ( CAB ) \" % $EPóSVTBM | |B \"# = 6 cm | \"%| = | DC| = 2 cm :VLBSEBLJ WFSJMFSF HÌSF UBSBM CÌMHFOJO BMBO AB kaç cm2 dir? :VLBSEBLJWFSJMFSFHÌSF UBSBMCÌMHFMFSJOBMBOMB- \" 3π - 3 # 3π - 3 3 $ 3π - 7 3 SUPQMBNLBÀDN2 dir? 24 \" # $ % & % 3π - 2 3 93 & 3π - 4 5. D C \"#$% LBSFTJOJO J¿J- 2. \"#$% LBSFTJOJO ¿FW- OF \" NFSLF[MJ [\"#] B SFM¿FNCFSJ¿J[JMNJõUJS A ZBS¿BQMWF#NFS- ) ABC = 4π cm LF[MJ[#\"]ZBS¿BQM A çeyrek çemberler çi- [JMNJõUJS B DC | |AB =CSPMEVôVOBHÌSF UBSBMBMBOLBÀCS2 dir? :VLBSEBLJ WFSJMFSF HÌSF UBSBM CÌMHFOJO BMBO \" 32π # 32π - 16 3 kaç cm2 dir? 3 2 \" r # r $ r - 4 $ 64π - 32 3 % 64π - 16 3 3 3 & 64π - 16 3 % r - & r - 2 3. 0NFSLF[MJ¿FNCFSF 6. öQFLFMJOEFLJUFTUUF¿Ë[FNFEJóJCJSTPSVOVOGPUPóSB- [\"#]UFóFU GO¿FLJQBSLBEBõOBHËOEFSNFLJTUJZPS O 0\"#FõLFOBSпHFO r 4BZGBEBLJ IFS TPSV TBZGBOO MJL LTNO ¿FNCFSJOZBS¿BQCS LBQMBNBLUBES AB r .BLJOBOO NFSDFL ¿BQ CS JLFO UÐN TBZGB LBESBKBTóNBLUBES :VLBSEBLJ WFSJMFSF HÌSF UBSBM BMBO LBÀ CJSJN- karedir? r öQFLTBEFDF¿Ë[FNFEJóJCJSTPSVZVLBESBKBTó- ESZPS \" 12 3 - 6π # 12 3 - 2π $ 24 3 - 6π ÷QFL NBLJOBOOTBZGBEÑ[MFNJOFPMBOV[BLMô- % 24 3 - 2π & 12 3 - 4π O EFôJöUJSNFEFO LBESBKB UFL TPSVZV TôESE- ôOB HÌSF JTUFEJôJ TPSVZV ÀFLNFL JÀJO NFSDFL ÀBQOLBÀCJSJNPMBSBLBZBSMBNöUS \" 2 # 6 5 $ % 3 5 & 1. B 2. D 3. A 63 4. E 5. D 6. B
TEST - 27 0NFSLF[MJ¿FNCFS 4. D %BJSFOJO¦FWSFTJWF\"MBO 1. A | |BD å= 16 br C \"#$%LBSFTJOJOJ¿JOEF O D m ( A%DC ) = 30° [\"%] [ BC ] [\"#]¿BQ- B | |\"%å= |DC| M ZBSN ¿FNCFSMFS WF- SJMNJõUJS C AB :VLBSEBLJ WFSJMFSF HÌSF UBSBM BMBOMBS UPQMBN | |\"# =CSPMEVóVOBHËSF UBSBMBMBOLBÀCS2 dir? kaç r br2 dir? \" # $ % & 40 3 \" r - # r - $ r - 32 5. 0WF.NFSLF[MJFõEBJSFMFSJOZBS¿BQMBSDNEJS % r - & + 36r 6 2. ôFLJMEF\"#$FõLFOBSпHFOEJS% & (CVMVO- OM EVLMBSLFOBSMBSOPSUBOPLUBMBSPMNBLÐ[FSF\" # $NFSLF[MJ¿FNCFSZBZMBS¿J[JMNJõUJS A DE ÷LJEBJSFOJOLFTJöJNMFSJPMBOUBSBMBMBOLBÀDN2 dir? FB GC \" 18r - 12 3 # 20r - 18 3 $ r - 12 3 % 24r - 18 3 &öLFOBSÑÀHFOJOBMBO 12 3 br2 PMEVôVOBgö- & 30r - 18 3 SF UBSBMBMBOLBÀCJSJNLBSFEJS \" 8 3 # 10 3 C 12 3 % & 3. 0 NFSLF[MJ ¿FNCFSMFS- 6. [BC] 0NFSLF[MJ¿FNCFSZBZOB#OPLUBTOEBUF- EF óFUUJS A S1 O m % = 120° A 120° ( COD ) O C S2 36° |OB| = 2 |BD| D B D CB S1WF42 CVMVOEVLMBSCËMHFMFSJOBMBOMBSOHËTUFS- % = 36° % = BC EJóJOFHËSF S1 PSBOLBÀUS m ( AOC ) AB S2 5BSBM CÌMHFOJO BMBO Ö DN2 PMEVôVOB HÌSF ÀFNCFSJOZBSÀBQLBÀDNEJS \" 8 # 5 % 8 & 9 9 8 $ 58 \" # $ % & 1. A 2. C 3. D 64 4. B 5. D 6. A
Çember ve Daire 0 NFSLF[MJ ¿FZSFL KARMA TEST - 1 ¿FNCFSWFSJMNJõUJS 1. B 4. \"#$% EJLEËSUHFOJOF 5 OPLUBTOEB UFóFU ZBSN BC m DC ¿FNCFSWFSJMNJõUJS C |OD| = |DC| D TC x F O DA xE % AB :VLBSEBLJWFSJMFSFHÌSF m ( ADC ) = x kaç de- | | | |[DB]LËõFHFO &' = &#PMEVóVOBHËSF recedir? m ( A%ED ) = x kaç derecedir? \" # $ % & \" # $ % & 5. A 2. B T 5)m OC x H D x | |0) = 4 br | |\"$ = 8 br B C O4 H A8 C #\" WF #$ ¿FNCFSF TSBTZMB \" WF $ OPLUBMBSOEB UFóFU %)m\"$ m % = m & m ( A%BC ) = 80° (AD) ( DC) 0NFSLF[MJ¿FZSFL¿FNCFSEF5UFóFUEFóNFOPLUB- PMEVóVOBHËSF % = x kaç derecedir? m ( ADH ) | |TPMEVóVOBHËSF OB = x kaç birimdir? \" # $ % & \" # $ % & A 6. \"#$пHFO 3. C D x %&#$ 45° |\"#| = |\"$| 43 DNE | |%& = 6 br | BC | = 12 br AH 12 B KM O | |% %)m\"# m ( BDC ) = 45° %) = 4 3 br | |)# = 12 br BL C | |[AB]ÀBQMÀFNCFSEFWFSJMFOMFSFHÌSF BC = x 0 NFSLF[MJ ÀFNCFSEF , - . / UFôFU EFôNF OPLUBMBSPMEVôVOBHÌSF \"#$ÑÀHFOJOJOÀFWSF- kaç birimdir? si kaç birimdir? \" 8 2 # $ 4 3 \" # $ % & % 4 2 & 1. D 2. C 3. A 65 4. C 5. E 6. B
KARMA TEST - 2 Çember ve Daire 1. 4. C % = 30° m ( AOC ) D T 10 2 | |\"# = 8 br A BC A BO [\"#]¿BQMZBSN¿FNCFSEF%UFóFUEFóNFOPLUBT [\"#] ¿BQM ZBSN EBJSF 0 NFSLF[MJ EBJSF EJMJNJOF \" | | | | | |\" # $ EPóSVTBM CD = 10 2 CS \"# = BC ol- OPLUBTOEB 0$EPóSVTVOB5OPLUBTOEBUFóFUUJS | |dVóVOBHËSF AC kaç birimdir? :VLBSEB WFSJMFOMFSF HÌSF UBSBM BMBO LBÀ CS2 \" # $ % & dir? \" r # r $ r % r & r 2. ôFLJMEFLËõFMFSJ¿FNCFSÐ[FSJOEFPMBO\"#$%EJL- EËSUHFOJWFSJMNJõUJS D I IC \"% = 6 birim I I\"# = 12 birim 6 5. BE F [\"'] a [%&] = {K} A B C A K \"#%$ x m ( B%AF ) = 32° 12 :VLBSEBLJWFSJMFSF HÌSF UBSBMCÌMHFMFSJOBMBOMBS D UPQMBNLBÀCJSJNLBSFEJS %%% \" 27r # 45r $ r % r & r m (BE) = m (EF) = m (FC)PMEVóVOBHËSF 22 m ( D%KF ) = x kaç derecedir? \" # $ % & 3. A 6. B \"#$% x 2 A |\"#| = |$&| L C FD H 44° 5 P m % ) = 44° ( DCE B4 K C \"#$пHFO 1)m\"# 1-m\"$ 1,m#$ E | | | | | | | | | |1) = 1, = 1- \") =CS CL =CS :VLBSEBLJ WFSJMFSF HÌSF % = x kaç de- | BK | = 4 br m ( ABE ) :VLBSEBLJWFSJMFSFHÌSF \"#$ÑÀHFOJOJOÀFWSF- recedir? si kaç birimdir? \" # $ % & \" # $ % & 1. C 2. B 3. E 66 4. C 5. A 6. D
Çember ve Daire KARMA TEST - 3 1. \"õBóEBLJ BENMBS J[MFOFSFL CJS HFPNFUSJL ¿J[JN 4. A 0NFSLF[ ZBQMZPS 9 [\"%]¿BQ 68 r [\"#] m [ BC ]PMBDBLõFLJMEF\"#$EJLпHFOJ¿J- B C [\")] m [BC] [JMJZPS OH | |r \" NFSLF[MJ ZBS¿BQ \"# PMBO ¿FNCFS ZBZOO | |\"# =CS [\"$]OLFTUJóJOPLUB%JõBSFUMFOJZPS r [DC]¿BQMZBSN¿FNCFS[BC]LFOBSO&OPLUB- | |\")= 6 br TOEBLFTFDFLõFLJMEF¿J[JMJZPS | |D \"$ = 8 br | | | | | |AB = 6 br, DC = CS PMEVôVOB HÌSF BE :VLBSEBLJWFSJMFSFHÌSF ÀFNCFSJOZBSÀBQLBÀ birimdir? kaç birimdir? \" # $ % & \" # $ % & 2. %Ñ[MFNEF 01 NFSLF[MJ DN ZBSÀBQM MJL 5. [#\"] m [\"$] PMBO \"#$ EJL пHFOJ ¿J[JMJZPS %BIB TPOSB \" NFSLF[MJ [\"#] ZBS¿BQM ¿FNCFS ¿J[JMJZPS NFSLF[ BÀ JMF HÌTUFSJMFO EBJSF EJMJNJOJO BMBO Çember [BC]WF[\"$]OTSBTZMB&WF%OPLUBMB- O2NFSLF[MJDNZBSÀBQMEBJSFEFLBÀEFSFDF- SOEBLFTJZPS MJLNFSLF[BÀJMFHÌTUFSJMJS % \" # $ % & | | | |BE = EC PMEVôVOBHÌSF m(ED) kaç derece- dir? \" # $ % & 3. 0 NFSLF[MJ ZBSN ¿FNCFS WF ) NFSLF[MJ ¿FZSFL 6. A 45° ¿FNCFSWFSJMNJõUJS C 30° E A HO BD C B m ( B%AD ) % m ( DAC ) \"#$пHFO = 30°, = 45° | | | | | |$& = &) =CS 0) = 3 br 3 | \"#| = |\"$| Buna göre, UBSBM BMBOMBSO GBSLMBSOO NVUMBL DC EFôFSJLBÀCJSJNLBSFEJS :VLBSEBLJ WFSJMFSF HÌSF PSBO BöBôEB- LJMFSEFOIBOHJTJEJS BD \" r # 7r $ r % 9r & r 2 2 \" 3 3 # 3 2 $ % 6 & 3 1. C 2. E 3. B 67 4. B 5. B 6. B
KARMA TEST - 4 Çember ve Daire 1. 0NFSLF[MJ¿FNCFS\"#$пHFOJOJOEõUFóFU¿FN- 4. :BS¿BQCSPMBOEBJSFõFLMJOEFLJLºóU[\"#]¿BQ beridir. CPZVODBLBUMBOZPS D E A BA O B A O B CF &MEFFEJMFOLºóUEBIBTPOSB\"WF#OPLUBMBS¿BL- ( õBDBLõFLJMEFZFOJEFOLBUMBOZPS4POEVSVNEBFM- EF FEJMFO LºóUUBO NBLBT ZBSENZMB ZBS¿BQ CS #%WF#'UFóFUEPóSVMBS m (DEF) = 260° PMBOZBSN¿FNCFSMFSõFLJMEFLJHJCJLFTJMJQBUMZPS % = 55° PMEVôVOB HÌSF m ( % ) kaç A OA O m ( OCF ) BAC derecedir? \" # $ % & 2. \"0#EJLпHFOWF0NFSLF[MJ¿FZSFLEBJSFWFSJMNJõ- KK UJS 4POEVSVNEBLBMBOL»ôUUBNBNFOBÀMEôOEB FMEFFEJMFOöFLMJOBMBOLBÀCJSJNLBSFEJS A \" Õ # Õ $ Õ % Õ & Õ O CB 5. ôFLJMEFBSBMBSOEBLJV[BLMLSCJSJNPMBOEWFLQB- I I\"0 = 6 br WFUBSBMBMBOMBSCJSCJSJOFFõJUPMEVôV- ralel EPóSVMBS BSBTOEB 0 NFSLF[MJ ZBSN EBJSF I Ina göre, OB kaç br dir? \"#$%LBSFTJWF134FõLFOBSпHFOJWFSJMNJõUJS \" r # r $ r % r & r O D CP d S1 S2 S3 r T A BR S k 3. 0NFSLF[MJ¿FZSFL¿FNCFS¿J[JMJZPS%BIBTPOSB\" 41 42WF43JÀJOEFCVMVOEVLMBSCÌMHFMFSJOBMBO- MBSOHÌTUFSdiklerineHÌSF BöBôEBLJTSBMBNB- WF#LËõFMFSJJLJGBSLMZBS¿BQÐ[FSJOEF $WF%LË- MBSEBOIBOHJTJEPôSVEVS õFMFSJ¿FNCFSZBZÐ[FSJOEFPMBO\"#$%LBSFTJ¿J[J- liyor. \" S1 < S2 < S3 # S1 < S3 < S2 #VöBSUMBSBVZHVOLBSFOJOBMBOCS2PMEVôV- OBHÌSF ÀFNCFSJOZBSÀBQLBÀCJSJNEJS $ S2 < S < S % S3 < S < S \" # 5 2 $ 5 3 3 1 2 1 % & & S3 < S1 < S2 1. E 2. A 3. B 68 4. C 5. D
Çember ve Daire YAZILI SORULARI 1. ôFLJMEFLJ 0 NFS- 3. C D ôFLJMEFLJ 0 NFS- LF[MJ¿FNCFSEF A LF[MJ [\"#] ¿BQM 140° 6O6 ¿FNCFSEF a O |\"$| = |CD| We A B |\"0| = 6 cm B m % ) = 140° | |CD =DNWF ( ACD x3 | |BD =DN 5 2H 1D C | |:VLBSEBLJWFSJMFSFgöre, OD = x kaç cm dir? :VLBSEBLJWFSJMFSFHÌSF % = a kaç de- m ( CAB ) | | | |OH m$#BMOSTB BH = 3, DH = 2 | |0)#ÑÀHFOJOEFQJTBHPS OH = 3 3 recedir? | |0)%ÑÀHFOJOEFQJTBHPS OD = 31 ) m (ABD) = 280° ¦FWSFFÀ & m (DB) = 100° [AB]ÀBQ && m (AC) = m (CD) = 40° ) Bu durumda 2a = m (CDB) = 140° a 2. B 4. C a D D 7 100° 40° A F E C A 52 B :VLBSEBLJ¿FNCFSEF [#&] a [CD] = {F} [\"#]¿BQM¿FNCFS m ( % ) = m ( % ) ABD DBC % % | | | |\"# = 5 2 CS DB = 7 br m ( BAC ) BFC = 40° m ( ) = 100° PMEVôVOBHÌSF | |:VLBSEBLJWFSJMFSFHÌSF DC kaç birimdir? % = a kaç derecedir? m ( ABE ) &% m ( % ) = 90° m (BC) - m (DE) = 80° %öBÀ ADC &% | |\"%#ÑÀHFOJOEF1JTBHPS AD = 1 m (BC) + m (DE) = 200° ÷ÀBÀ |AD|= |DC| = 1 % Bu durumda m (DE) = 60° j 2a j a 1. 31 2. 30 69 3. 70 4. 1
YAZILI SORULARI Çember ve Daire 5. C 7. #JSJNLBSFMFSFBZSMNõYBZSUMBSOEBLJLBSF- A OJOJ¿JOF0NFSLF[MJEBJSF¿J[JMNJõUJS O BD O 0NFSLF[MJJLJ¿FNCFSZBZWFSJMNJõUJS #VOBHÌSF UBSBMBMBOMBSOUPQMBNLBÀCJSJNLB- redir? | | | || | % 5BSBM BMBOMBS UBöOSTB UPQMBNMBS ÀFZSFL EBJSFZF FöJU OB = 3 OD \"# + AB =CSPMEVóVOBHË- olur. 25π | |& 4 SF CD + CD UPQMBNLBç birimdir? & AB AB 3 = = & 5 CD CD 18 3 = & 5 CD + CD & | |CD + CD = 6. 8. D 6 L C ôFLJMEF\"#$%LBSF- A TJOJOJ¿UFóFU¿FNCF- #JSJNLBSFMFSFBZSMNöEÑ[MFNEFWFSJMFOÀFN- S SJJMF%NFSLF[MJ¿FZ- CFSZBZMBSZMBPMVöUVSVMNVöCÌMHFOJOÀFWSFTJOJ S K SFLEBJSF¿J[JMNJõUJS bulunuz. E :BSÀBQCSPMBOJLJÀFZSFL ZBSÀBQCSPMBOJLJÀFZSFL S ZBSÀBQCSPMBOJLJÀFZSFL S6 ZBSÀBQCSPMBOCJSZBSNÀFNCFSMFSJOÀFWSFMFSJUPQ- A MBN 10π 4π 6π 8π A F6B + + + = 14π I IDL = 6 cm PMEVôVOBHÌSF UBSBMCÌMHFOJOBMBO 2 222 kaç cm2 dir? A+S= 6.6 = 18 2 2A +4= 36 cm2 5. 30 6. Ö 70 25π 8. 36 7. 4
Çember ve Daire <(1m1(6m/6258/$5 1. 3. #JS TBBU LVMFTJOEF CVMVOBO TBBUJO ZFSF FO V[BL \"ZSUMBSWFDNPMBOEJLEËSUHFOõFLMJOEFLJCJS OPLUBTOO ZFSF V[BLMó NFUSFEJS 4BBUJO ZB- QF¿FUF ZÐ[FZJOF BLBO NÐSFLLFQ QF¿FUF Ð[FSJOEF S¿BQ NFUSF PMVQ BLSFCJOJO V[VOMVóV CJMJONF- EBJSFTFMCJSõFLJMEFEBóMNBLUBES NFLUFEJS .ÑSFLLFCJOEBôMNBI[TBOJZFEF DNPMEV- #VOBHÌSF TBBUJMFBSBTOEBLJIFS- IBOHJCJS[BNBOEBBLSFCJOVÀOPLUBTOOZFSF ôVOB HÌSF TBOJZF TPOSB QFÀFUFOJO NÑSFL- V[BLMôNFUSFDJOTJOEFOBöBôEBLJMFSEFOIBO- LFQTJ[LTNOOZÑ[FZBMBOLBÀDN2 dir? gisi olabilir? \" mÕ # mÕ $ mÕ \" 23 # 21 $ 17 % 13 & 11 % mÕ & -Õ 4. ¥FWSFTJCSPMBOFõLFOBSEËSUHFOõFLMJOEFLJLB- óU[\"$]LËõFHFOJCPZVODBLBUMBOZPS 2. ¶TU ZÐ[FZJOJO BZSUMBS WF NFUSF PMBO EJLEËSU- DC C HFO õFLMJOEFLJ NBTBZB EBJSF õFLMJOEFLJ NBTB ËS- UÐTÐËSUÐMNÐõUÐS A BA B %BIBTPOSBPMVõBOJLJLBUMLºóEO\" # $LËõFMF- SJOEFOCSZBS¿BQMEBJSFEJMJNMFSJLFTJMJQBUMZPS C 1 1 .BTB ËSUÐTÐ UÐN NBTBZ LBQMBNõ WF NBTBOO 1 EËSULFOBSOEBOBõBóZBEPóSVõFLJMEFLJHJCJLËõF- 1 MFSBSBTOEBEBJSFLFTNFMFSJPMVõUVSBSBLTBSLNõUS A 1 1B #VOB HÌSF NBTB ÌSUÑTÑOÑO TBSLBO LTNMBS- ,BMBO JLJ LBUM L»ôU ZFOJEFO BÀMEôOEB FMEF OOBMBOMBSUPQMBNLBÀNFUSFLBSFEJS FEJMFOöFLMJOÀFWSFTJLBÀCJSJNEJS \" Õ- # Õ- $ Õ- 12 \" -Õ # +Õ $ +Õ % Õ- & Õ- 6 % +Õ & +Õ 1. C 2. A 71 3. D 4. E
<(1m1(6m/6258/$5 Çember ve Daire 1. -VOBQBSLBHJEFO.FMJT¿FNCFSIBMLBBUNBPZVOV 3. A ôFLJMEFLJ IBNTUFS ¿BSL PZOBZBDBLUS .FMJT ZBS¿BQ CS PMBO ¿FNCFSJ BU- L B TÐTMFOFSFL Fõ BSBMLM \" UóOEB TUBOEEBLJ EJLEËSUHFOMFSEFO CJSJ ¿FNCFSJO J¿JOEFLBMSTBPZVOVLB[BOBDBLUS # $ % & ' , -OPLUB- 5 4 7 K C MBSOBCJSFSBNQVMZFSMFõ- 8 9 8 UJSJMNJõUJS )BNTUFS ¿BSL 5 7 8 F D EBLJLBEB UVS I[ZMB 9 6 6 E PL ZËOÐOEF EËOEÐSFCJM- NFLUFEJS #VOBHÌSF TOIJÀEVSNBEBOZÑSÑZFOIBNT- UFSEVSEVôVBOEBÀBSLOTPOHÌSÑOUÑTÑBöBô- EBLJMFSEFOIBOHJTJPMVS 4UBOEEBLJEJLEÌSUHFOMFSJOBZSUV[VOMVLMBSöF- A) B B) K C) F LJMEFLJ HJCJ PMEVôVOB HÌSF .FMJThJO PZVOV LB- C L [BONBLJÀJOIFEFGBMBCJMFDFôJEJLEÌSUHFOTBZT A F E K BöBôEBLJMFSEFOIBOHJTJEJS L DE AD L \" # $ % & KE DB CA F C B D) E) D E CE DF B FC K AK BL L A 4. 5FMEFOZBQMNõ¿FWSFTJÕDNPMBO\"#$FõLFOBS пHFOJOF õFLJM WFSJMFSFL ËODF CJS LBSFZF BSEOEBO 2. EBCJS¿FNCFSFEËOÐõUÐSÐMÐZPS A A1 A2 C1 O B C B1 B2 C2 ,BSF õFLMJOEFLJ CJS CBI¿FOJO BóSML NFSLF[JOF ¶¿HFOJOLËõFMFSJ\" # $OPLUBMBSLBSFOJOÐ[FSJOEF EFOL HFMFO OPLUBEB CJS TVMBNB TJTUFNJ WBSES #V \"1 #1 $1 EBJSFOJOÐ[FSJOEFJTF\"2 #2 $2OPLUB- TJTUFN FO GB[MB NFUSF V[BLMóB LBEBS TVMBZBCJM- MBSOBEËOÐõNFLUFEJS. NFLUFEJS#VCBI¿FOJOEËSULËõFTJOFJQV[VOMVLMBS BZOPMBOEËSULPZVOCBóMBONõUS #VOBHÌSF LBSFOJOJÀJOEFPMVöBO\"1B1C1ÑÀHF- OJOJO BMBO 41, çemberin içinde oluöBn ABC 222 #BIÀFOJOBMBON2PMEVôVOBHÌSF LPZVOMB- ÑÀHFOJOJOBMBO42 olmakÑ[FSe, S1 PSBOBöB- SOTMBONBEBOPUMBZBCJMNFMFSJJÀJOLVMMBOMBDBL ôEBLJMFSEFOIBOHJTJEJS S2 UPQMBN JQ V[VOMVôV FO GB[MB LBÀ NFUSF PMNBM- ES 27 3 93 $ 4π2 \" # 27 3 4 2 23 & \" # $ % & 4 3π2 % 3π2 3 1. D 2. D 72 3. B 4. C
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