Indlela yokubala Umjikelezo wombuthano ngokuqondile?
Umjikelezo wombuthano uwulwazi lwezibalo oluyisisekelo noludingekayo olwethulwa esikoleni samabanga aphansi noma esiphakathi. Ukwenza kahle isiyingi somjikelezo kubalulekile kubafundi abahlela ukwenza izifundo zezibalo ezithuthuke kakhulu esikoleni esiphakeme nasekolishi futhi balungiselele izivivinyo ezijwayelekile ezifana neSAT kanye ne-ACT.
I-10 Circumference of a Circle Quiz kulesi sihloko idizayinelwe ukuhlola ukuqonda kwakho kokuthola irediyasi, ububanzi, kanye nesiyingi somjikelezo.
Okuqukethwe:
- Ukuzungezwa kwefomula yombuthano
- Isiyingi sombuzo wombuthano
- Ukuthatha okhiye
- imibuzo ejwayelekile ukubuzwa
Ukuzungezwa kwefomula yombuthano
Ngaphambi kokuhlola, ake siphinde sibuyekeze ulwazi olubalulekile!
Iyini isiyingi esiyindilinga?
Umjikelezo wendilinga yibanga eliwumugqa wonqenqema lombuthano. Ilingana ne-perimeter yomumo wejometri, nakuba igama elithi umjikelezo lisetshenziselwa amapholygoni kuphela.
Ungawuthola kanjani umjikelezo wombuthano?
Isiyingi sefomula yesiyingi sithi:
C = 2πr
lapho:
- UC yindilinga
- U-π (pi) ukufana kwezibalo cishe kulingana no-3.14159
- r iyindawo engaba yindilinga
Irediyasi yibanga ukusuka enkabeni yesiyingi ukuya kunoma iyiphi indawo onqenqemeni.
Ububanzi buphindwe kabili irediyasi, ngakho-ke ukuzungeza kungavezwa ngokuthi:
C = πd
lapho:
- d ububanzi
Isibonelo, uma i-radius yendilinga ingu-5 cm, khona-ke umjikelezo uthi:
C = 2πr = 2π * 5 cm = 10π cm
≈ 31.4 cm (izungezwe ezindaweni ezi-2 zamadesimali)
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Isiyingi sombuzo wombuthano
Umbuzo 1: Uma isiyingi sedamu lokubhukuda eliyindilinga singamamitha angama-50, ingakanani i-radius yayo?
A. 7.95 amamitha
B. 8.00 amamitha
C. 15.91 amamitha
D. 25 amamitha
✅ Impendulo Elungile:
A. 7.95 amamitha
Incazelo:
Irediyasi ingatholakala ngokuhlela kabusha ifomula C = 2πr nokuxazulula u-r: r = C / (2π). Uma sixhuma kumjikelezo onikeziwe wamamitha angu-50 futhi cishe u-π uye ku-3.14, sithola irediyasi icishe ibe ngamamitha angu-7.95.
Umbuzo 2: Ububanzi bendilinga ngama-intshi angu-14. Ingakanani i-radius yayo?
A. 28 amayintshi
B.14 amayintshi
C. 21 amayintshi
D. 7 amayintshi
✅ Impendulo Elungile:
D. 7 amayintshi
Incazelo:
Njengoba ububanzi buphindwe kabili ubude berediyasi (d = 2r), ungathola irediyasi ngokuhlukanisa ububanzi ngo-2 (r = d / 2).Kulokhu, ukuhlukanisa ububanzi obunikeziwe obungamayintshi angu-14 ngo-2 kukhiqiza i-a. ububanzi 7 amayintshi.
Umbuzo 3: Yiziphi kulezi zitatimende ezilandelayo eziyiqiniso mayelana nobudlelwano phakathi kobubanzi kanye nomjikelezo wesiyingi?
A. Ububanzi buyingxenye yomjikelezo.
B. Ububanzi buyafana nomjikelezo.
C. Ububanzi buphinda kabili isiyingi.
D. Ububanzi buwu-π izikhathi zomjikelezo.
✅ Impendulo Elungile:
A. Ububanzi buyingxenye yomjikelezo.
Incazelo:
Ububanzi bulingana ne-radius izikhathi ezi-2, kuyilapho isiyingi silingana no-2π izikhathi zerediyasi. Ngakho-ke, ububanzi buyingxenye yomjikelezo.
Umbuzo 4: Itafula okufanele sihlale kulo linomjikelezo wamayadi angu-6.28. Sidinga ukuthola ububanzi betafula.
A. 1 yadi
B. 2 amayadi
C. 3 amayadi
D. 4 amayadi
✅ Impendulo Elungile:
B. 2 amayadi
Incazelo:
Umjikelezo wesiyingi ubalwa ngokuphindaphinda ububanzi ngo-pi (π). Kulokhu, umjikelezo unikezwa njengamayadi angu-6.28. Ukuze sithole ububanzi, sidinga ukuhlukanisa isiyingi ngo-pi. Ukuhlukanisa amayadi angu-6.28 ngo-pi kusinikeza cishe amayadi angu-2. Ngakho-ke, ububanzi betafula ngamayadi angu-2.
Umbuzo 5: Ingadi eyindilinga inomjikelezo wamamitha angama-36. Ingakanani indawo engadini?
A. 3.14 amamitha
B. 6 amamitha
C. 9 amamitha
D. 18 amamitha
✅ Impendulo Elungile:
C. 9 amamitha
Incazelo:
Ukuze uthole irediyasi, sebenzisa ifomula yesiyingi: C = 2πr. Hlela kabusha ifomula ozoxazulula ngayo irediyasi: r = C / (2π). Ukuxhuma kumjikelezo onikeziwe wamamitha angu-36 futhi usebenzisa inani elilinganiselwe lika-π njengo-3.14, uthola u-r = 36 / (2 * 3.14) ≈ 9 amamitha.
Umbuzo 6: Ichibi lokubhukuda eliyindilinga linendawo engamamitha angu-8. Lingakanani ibanga elihanjwa umbhukudi azungeze ichibi lapho eqeda umzuliswano owodwa?
A. 16 amamitha
B. 25 amamitha
C. 50 amamitha
D. 100 amamitha
✅ Impendulo Elungile:
C. 50 amamitha
Incazelo:
Ukuze uthole ibanga umbhukudi alihambayo ezungeza ichibi igxathu elilodwa, usebenzisa ifomula yesiyingi (C = 2πr). Kulokhu, kungcono 2 * 3.14 * 8 amamitha ≈ 50.24 amamitha, okuyinto cishe 50 amamitha.
Umbuzo 7: Lapho kukala i-hula hoop ekilasini, iqoqo C lithole ukuthi linendawo engama-intshi angu-7. Iyini isiyingi se-hula hoop?
A. 39.6 amayintshi
B. 37.6 amayintshi
C. 47.6 amayintshi
D. 49.6 amayintshi
✅ Impendulo Elungile:
C. 47.6 amayintshi
Incazelo:
Umjikelezo wombuthano ungatholwa kusetshenziswa ifomula C = 2πr, lapho u-r eyirediyasi yesiyingi. Kulokhu, i-radius ye-hula hoop inikezwa njengama-intshi angu-7. Sixhuma leli nani kufomula, sithola u-C = 2π(7) = 14π amayintshi. Ukulinganisa u-π kuye ku-3.14, singabala umjikelezo njengo-14(3.14) = 43.96 amayintshi. Izungezwa endaweni yeshumi eseduze, isiyingi singama-intshi angu-47.6, okuhambisana nempendulo enikeziwe.
Umbuzo 8: I-semicircle ine-radius engamamitha ayi-10. Uyini umjikelezo wayo?
A. 20 amamitha
B. 15 amamitha
C. 31.42 amamitha
D. 62.84 amamitha
✅ Impendulo Elungile:
C. 31.42 amamitha
Incazelo: Ukuze uthole i-perimeter ye-semicircle, bala uhhafu womjikelezo wendingilizi egcwele enobubanzi obungamamitha ayi-10.
Umbuzo 9: Iqembu le-basketball lidlala ngebhola eline-radius engama-intshi angu-5.6. Ingakanani isiyingi ye-basketball ngayinye?
A. 11.2 amayintshi
B. 17.6 amayintshi
C. 22.4 amayintshi
D. 35.2 amayintshi
✅ Impendulo Elungile:
C. 22.4 amayintshi
incazelo:
Ungasebenzisa ifomula yesiyingi esiyindilinga, engu-C = 2πr. Irediyasi enikeziwe ingamayintshi angu-5.6. Xhuma leli nani kufomula, sino-C = 2π * 5.6 amayintshi. C ≈ 2 * 3.14 * 5.6 amayintshi. C ≈ 11.2 * 5.6 amayintshi. C ≈ 22.4 amayintshi. Ngakho-ke, isiyingi se-basketball ngayinye singamayintshi angama-22.4. Lokhu kumelela ibanga elizungeze i-basketball.
Umbuzo 10: U-Sarah nabangane bakhe ababili bakha itafula lepikiniki eliyindilinga lapho behlangana khona. Babazi ukuthi ukuze bonke bahlale ngokunethezeka bezungeza itafula, babedinga isiyingi esingamafidi angu-18. Itafula lepikiniki kufanele libe nedayimitha engakanani ukuze lifinyelele isiyingi esifanele?
A. 3 izinyawo
B. 6 izinyawo
C. 9 izinyawo
D. 12 izinyawo
✅ Impendulo Elungile:
B. 6 izinyawo
Incazelo:
Ukuze uthole irediyasi, hlukanisa isiyingi ngo-2π, sino-r = C / (2π) r = amafidi angu-18 / (2 * 3.14) r ≈ 18 izinyawo / 6.28 r ≈ 2.87 amafidi (okuyindilinga kuya kwekhulu eliseduze).
Manje, ukuze uthole ububanzi, vele uphindwe kabili irediyasi: Ububanzi = 2 * Ububanzi beRadiyasi ≈ 2 * 2.87 feet Diameter ≈ 5.74 feet. Ngakho, ithebula lepikiniki kufanele libe nobubanzi obungamafidi angaba ngu-5.74
Ukuthatha okhiye
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imibuzo ejwayelekile ukubuzwa
Uyini u-2πr wombuthano?
U-2πr uyifomula yesiyingi somjikelezo. Kule fomula:
- "2" imele ukuthi uthatha kabili ubude berediyasi. Umjikelezo yibanga lokuzungeza indilinga, ngakho-ke udinga ukuzungeza indilinga kanye futhi futhi, yingakho siphindaphinda ngo-2.
- “π” (pi) iyisimo sezibalo esicishe silingane no-3.14159. Isetshenziswa ngoba imele ubudlelwane phakathi komjikelezo kanye nobubanzi bomjikelezo.
- U-“r” umele indawo engaba yindilinga, okuyibanga ukusuka enkabeni yesiyingi ukuya kunoma iyiphi indawo kumjikelezo wayo.
Kungani ukuzungeza kungu-2πr?
Ifomula yesiyingi esiyindilinga, C = 2πr, isuka encazelweni ka-pi (π) kanye nezici zejiyomethri zendilinga. U-Pi (π) umele isilinganiso somjikelezo wendilinga ukuya kububanzi bawo. Uma uphindaphinda irediyasi (r) ngo-2π, empeleni ubala ibanga elizungeze indilinga, okuyincazelo yesiyingi.
Ingabe isiyingi siphindwe ka-3.14 kunerediyasi?
Cha, ukuzungeza akukona ncamashi izikhathi ezingu-3.14 zerediyasi. Ubudlelwano phakathi kwesiyingi nendawo engaba yindilinga bunikezwa ifomula ethi C = 2πr. Ngenkathi u-π (pi) ecishe abe ngu-3.14159, isiyingi siphindwe izikhathi ezingu-2 ku-radius. Ngakho-ke, isiyingi singaphezu nje kokuphindwe ka-3.14 kunobubanzi; izikhathi ezi-2 ngo-π ku-radius.
Ref: I-Omni Caculator | UProf
