Indlela yokubala iSangqa ngokuchanekileyo?
Isangqa sesangqa lulwazi olusisiseko nolufunekayo lwemathematika olwaziswa kwisikolo samabanga aphantsi okanye esiphakathi. Ukuqonda umjikelezo wesangqa kubalulekile kubafundi abaceba ukwenza izifundo zemathematika ezikumgangatho ophezulu kwisikolo samabanga aphakamileyo nakwikholeji kwaye balungiselele iimviwo ezisemgangathweni ezifana neSAT kunye ne-ACT.
I-10 yeSangqa seKwiz yeSangqa kweli nqaku yenzelwe ukuvavanya ukuqonda kwakho ekufumaneni iradiyasi, idayamitha, kunye nesangqa sesangqa.
Isiqulatho:
Isangqa sefomula yesangqa
Phambi kokuba sithathe uvavanyo, makhe siphinde sihlaziye ulwazi olubalulekileyo!
Yintoni isangqa sesangqa?
Umjikelezo wesangqa ngumgama omgca wencam yesangqa. Ilingana nomjikelezo wemilo yejometri, nangona igama lomjikelezo lisetyenziswa kuphela kwiipolygons.
Indlela yokufumana umjikelezo wesangqa?
Umjikelo wefomula yesangqa ngu:
C = 2πr
apho:
- UC ngumjikelo
- π (pi) lutshintsho lwemathematika oluphantse lulingane no-3.14159
- r yiradiyasi yesangqa
Iradiyasi ngumgama ukusuka kumbindi wesangqa ukuya nakweyiphi na indawo emphethweni.
Idayamitha iphinda kabini iradiyasi, ngoko ke i-circumference ingabonakaliswa ngolu hlobo:
C = πd
apho:
- d yidamitha
Ngokomzekelo, ukuba i-radius yesangqa yi-5 cm, i-circumference ithi:
C = 2πr = 2π * 5 cm = 10π cm
≈ 31.4 cm (ingqukuva ukuya kwindawo ezi-2)
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Umjikelezo wemibuzo yesangqa
Umbuzo 1: Ukuba isazinge sedama lokuqubha eziziimitha ezingama-50, ithini iradiyasi?
A. 7.95 yeemitha
B. 8.00 yeemitha
C. 15.91 yeemitha
D. 25 yeemitha
✅ Impendulo echanekileyo:
A. 7.95 yeemitha
Inkcazo:
I-radius inokufunyanwa ngokucwangcisa kwakhona ifomula C = 2πr kunye nokusombulula r: r = C / (2π). Ukuplaga kwisangqa esinikiweyo seemitha ezingama-50 kwaye malunga no-π ukuya ku-3.14, sifumana iradiyasi malunga ne-7.95 yeemitha.
Umbuzo 2: Ubukhulu besangqa zii-intshi ezili-14. Yintoni iradiyasi yayo?
A. 28 intshi
B.14 intshi
C. 21 intshi
D. 7 intshi
✅ Impendulo echanekileyo:
D. 7 intshi
Inkcazo:
Ekubeni i-diameter iphindwe kabini ubude be-radius (d = 2r), ungayifumana i-radius ngokwahlula i-diameter ngo-2 (r = d / 2) . Kule meko, ukwahlula ububanzi obunikiweyo be-intshi ezili-14 nge-2 izivuno. iradius ye-intshi ezi-7.
Umbuzo 3: Yeyiphi kwezi nkcazo zilandelayo eyinyani malunga nobudlelwane phakathi kobubanzi kunye nesangqa sesangqa?
A. Ububanzi sisiqingatha sesangqa.
B. Ububanzi buyafana nesangqa.
C. Idayamitha iphinda kabini isangqa.
D. Idayamitha ngu-π umphinda-phinde kwisangqa.
✅ Impendulo echanekileyo:
A. Ububanzi sisiqingatha sesangqa.
Inkcazo:
Ububanzi bulingana namaxesha ama-2 eradiyasi, ngelixa i-circumference ilingana no-2π amaxesha eradiyasi. Ngoko ke, ububanzi buyisiqingatha se-circumference.
Umbuzo 4: Itheyibhile ekufuneka sihlale kuyo inomlinganiselo weemitha ezi-6.28. Kufuneka sifumane i-diameter yetafile.
A. 1 iyadi
B. 2 yeeyadi
C. iiyadi ezi-3
D. 4 yeeyadi
✅ Impendulo echanekileyo:
B. 2 yeeyadi
Inkcazo:
Umjikelezo wesangqa ubalwa ngokuphinda-phinda i-diameter ngo-pi (π). Kule meko, i-circumference inikezelwa njenge-6.28 yards. Ukufumana i-diameter, kufuneka sahlule i-circumference nge-pi. Ukwahlula iiyadi ezi-6.28 nge-pi kusinika malunga neeyadi ezi-2. Ngoko ke, ububanzi betafile buyi-2 yeeyadi.
Umbuzo 5: Isitiya esisisangqa sinesangqa seemitha ezingama-36. Ingaba ingakanani iradiyasi yegadi?
A. 3.14 yeemitha
B. 6 yeemitha
C. 9 yeemitha
D. 18 yeemitha
✅ Impendulo echanekileyo:
C. 9 yeemitha
Inkcazo:
Ukufumana i-radius, sebenzisa ifomula yomjikelezo: C = 2πr. Lungisa kwakhona ifomula yokusombulula iradiyasi: r = C / (2π). Ukuplaga kumjikelezo onikiweyo weemitha ezingama-36 kwaye usebenzisa ixabiso eliqikelelweyo lika-π njenge-3.14, ufumana r = 36 / (2 * 3.14) ≈ 9 yeemitha.
Umbuzo 6: Idama lokuqubha elinesetyhula linemitha ezisi-8. Uqikelelwa umgama ongakanani indadi ewuhambayo ukujikeleza idama xa igqibezela umjikelo omnye?
A. 16 yeemitha
B. 25 yeemitha
C. 50 yeemitha
D. 100 yeemitha
✅ Impendulo echanekileyo:
C. 50 yeemitha
Inkcazo:
Ukufumana umgama ohanjwa yindadi ukujikeleza idama kumjikelo omnye, usebenzisa ifomula yesangqa (C = 2πr). Kule meko, yi-2 * 3.14 * 8 yeemitha ≈ 50.24 yeemitha, ezimalunga neemitha ezingama-50.
Umbuzo wesi-7: Xa lilinganisa ihula hoop eklasini, iqela C lafumanisa ukuba linemitha ye-intshi ezisi-7. Yintoni isangqa sehula hoop?
A. 39.6 intshi
B. 37.6 intshi
C. 47.6 intshi
D. 49.6 intshi
✅ Impendulo echanekileyo:
C. 47.6 intshi
Inkcazo:
Umjikelezo wesangqa unokufumaneka usebenzisa i-formula C = 2πr, apho i-r yi-radius yesangqa. Kule meko, i-radius ye-hula hoop inikezelwa njenge-intshi ezi-7. Ukuplaga eli xabiso kwifomula, sifumana C = 2π(7) = 14π intshi. Ngokumalunga no-π ukuya ku-3.14, singabala umjikelezo njengo-14(3.14) = 43.96 intshi. Ukusondezwa kweyona ndawo ikufutshane yeshumi, i-circumference yi-47.6 intshi, ehambelana nempendulo enikiweyo.
Umbuzo 8: Isiqingatha sesangqa sineemitha ezili-10. Yintoni iperimeter yayo?
A. 20 yeemitha
B. 15 yeemitha
C. 31.42 yeemitha
D. 62.84 yeemitha
✅ Impendulo echanekileyo:
C. 31.42 yeemitha
Inkcazo: Ukufumana iperimeter yesiqingatha sesemicircle, bala isiqingatha sesangqa esipheleleyo kunye neradius yeemitha ezili-10.
Umbuzo 9: Iqela lebhola yomnyazi lidlala ngebhola ene-radius ye-5.6 intshi. Uthini umjikelo webhola yomnyazi ngamnye?
A. 11.2 intshi
B. 17.6 intshi
C. 22.4 intshi
D. 35.2 intshi
✅ Impendulo echanekileyo:
C. 22.4 intshi
Inkcazo:
Ungasebenzisa ifomula yomjikelezo wesangqa, eyi-C = 2πr. Iradiyasi enikiweyo yi-5.6 intshi. Xhuma eli xabiso kwifomula, sinayo C = 2π * 5.6 intshi. C ≈ 2 * 3.14 * 5.6 intshi. C ≈ 11.2 * 5.6 intshi. C ≈ 22.4 intshi. Ke, umjikelo webhola yomnyazi ngamnye umalunga nee-intshi ezingama-22.4. Oku kubonisa umgama ojikeleze ibhola yomnyazi.
Umbuzo 10: USarah nabahlobo bakhe ababini bebesakha itafile yepikiniki esesangqa ukulungiselela indibano yabo. Babesazi ukuba ukuze bonke bahlale ngokukhululeka bejikeleze itafile, kwakufuneka kumjikelo oziimitha ezili-18. Yeyiphi idayimitha emayibe nayo itafile yepikiniki ukufikelela kwisangqa esichanekileyo?
A. 3 iinyawo
B. 6 iinyawo
C. iinyawo ezili-9
D. 12 iinyawo
✅ Impendulo echanekileyo:
B. 6 iinyawo
Inkcazo:
Ukufumana i-radius, yahlula i-circumference ngo-2π, sinayo r = C / (2π) r = 18 iinyawo / (2 * 3.14) r ≈ 18 iinyawo / 6.28 r ≈ 2.87 iinyawo (ijikelezwe kwikhulu elikufutshane).
Ngoku, ukufumana i-diameter, ngokulula kabini i-radius: I-Diameter = 2 * I-Radius Diameter ≈ 2 * 2.87 iinyawo Ububanzi ≈ 5.74 iinyawo. Ke, itafile yepikiniki kufuneka ibe nobubanzi obumalunga neemitha ezi-5.74
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Imibuzo ebuzwa qho
Yintoni 2πr yesangqa?
2πr yifomula yesangqa sesangqa. Kule fomula:
- "2" imele ukuba uthatha kabini ubude beradiyasi. Isangqa ngumgama ojikeleze isangqa, ngoko ke kufuneka ujikeleze isangqa kube kanye kwaye kwakhona, yiyo loo nto siphindaphinda ngo-2.
- "π" (pi) yimathematika engaguqukiyo malunga ne-3.14159. Isetyenziswa kuba imele ubudlelwane phakathi kwesangqa kunye nobubanzi besangqa.
- "r" imele iradiyasi yesangqa, engumgama ukusuka kumbindi wesangqa ukuya kuyo nayiphi na indawo kwisangqa saso.
Kutheni isangqa ngu-2πr?
Ifomula yesangqa sesangqa, C = 2πr, ivela kwinkcazo ye-pi (π) kunye neempawu zejometri zesangqa. I-Pi (π) imele umlinganiselo wesangqa kwidayamitha yaso. Xa uphinda-phinda iradiyasi (r) ngo 2π, ubala ngokuyimfuneko umgama ojikeleze isangqa, eyinkcazelo yesangqa.
Ngaba isangqa si-3.14 ngokuphinda-phindwe ngeradiyasi?
Hayi, umjikelezo awukho ncam u-3.14 umphinda-phinde kwiradiyasi. Ubudlelwane phakathi komjikelezo kunye ne-radius yesangqa inikwe ngefomula C = 2πr. Ngelixa u-π (pi) emalunga ne-3.14159, i-circumference ngu-2 umphinda-phinde ngo-p kwiradiyasi. Ngoko ke, i-circumference ingaphezulu ko-3.14 umphinda-phinde ngeradiyasi; ngu 2 umphinda-phinde ngo π umphinda-phinde kwiradiyasi.
Ref: I-Omni Caculator | UProf
