Polygon rhoqo. Inani macala yipolygon

Unxantathu, isikwere, namacala - la manani zaziwa phantse wonke. Kodwa apha leyo yipolygon, akazi bonke. Kodwa noko kunjalo iimilo zejiyometri. A polygon rhoqo ekuthiwa oye iiengile ngokulinganayo phakathi kwabo kunye icala. La manani baninzi, kodwa bonke babe iimpawu ezifanayo, kwaye kusebenza kubo indlela efanayo.

Iimpawu yeepoligoni rhoqo

Nawuphi polygon rhoqo, nokuba isikwere okanye octagon, ibe abhaliwe kwisangqa. Le propati esisiseko lidla asetyenziswe ekwakheni amanani. Ukongeza, isangqa ibe abhaliwe buyimilo kunye. Inani lamanqaku zoqhagamshelwano lilingana nenani emacaleni aso. Kwakhona kubalulekile ukuba circle abhaliwe yipolygon uya kuba naye iziko eqhelekileyo. La manani zejiyometri kuxhomekeke theorems omnye. Naliphi na iqela echanekileyo n-gon edibene owela ngaphakathi kwama-wesangqa ngeenxa R. Ngoko ke, oko ibalwe kusetyenziswa le ndlela ilandelayo: a = 2R ∙ ° sin180. Esebenzisa engaba wesangqa inokufumaneka kuphela amaqela kodwa kwibala buyimilo.

Indlela ukufumana inani macala yipolygon

Nayiphi rhoqo n-gon sakhiwa inani amacandelo elingana enye kwenye, apho, xa zidibene, zenza umgca elivaliweyo. Kulo mzekelo, zonke iimilo engile exonxiweyo anexabiso efanayo. Iimilo yahlulwe alula nanzima. Iqela lokuqala kuquka unxantathu kunye isikwere. yeepoligoni Complex abe inani elikhulu lamacala. Kwakhona ziquka inani inkwenkwezi-ezimile. Kumacala polygon entsonkothileyo qho amfumana inscribing nabo kwisangqa. Nantsi ke ubungqina. Zoba polygon rhoqo ezinenani engenasizathu lamacala n. Chaza isangqa ngeenxa zonke kuye. Cela radius R. Ngoku thelekela ukuba ezinye bobethwa n-gon. Ukuba ingongoma ezimbombeni zaso kulala kwisangqa kunye ulingana omnye komnye, ngoko ke isandla ingafunyanwa yi ifomula: a = 2R ∙ sinα: 2.

Ukufumana inani lamacala lo nxantathu rhoqo sibhalwe

unxantathu alinganayo - yinto yipolygon. Formula iya kusetyenziswa ngendlela efanayo naleyo yembutho, kunye n-gon. Triangle ziya kuthathwa njengezisemthethweni ukuba unalo enye ecaleni ubude yenxenye. Ezi angles bayalingana 60⁰. Ukwakha unxantathu zinamacala kwangaphambili ubude. Ukwazi median nokuphakama kwalo, unako ukufumana ixabiso emacaleni aso. Kuba oku ukusebenzisa indlela lokufumana ifomula ngokusebenzisa = x: cosα, apho x - median okanye ubude. Ekubeni onke amaqela unxantathu bayalingana, sinokucela = b = c. Ngoko qiniseka ukuba le ngxelo ilandelayo = b = c = x: cosα. Ngokufanayo, singakwazi ukufumana ixabiso amaqela unxantathu alinganayo, kodwa uya x ukuphakama elinikiweyo. Kulo mzekelo, kuthelekelelwe ukuba ngqongqo ngokwesiseko ngamanani. Ngoko ke, ukwazi ukuphakama x, ukufumana icala unxantathu isosceles usebenzisa ifomula A = B = x: cosα. Emva kokufumanisa ukuba amaxabiso a nga ibalwe ukusuka ubude siseko. Sisebenzisa theorem bakaPythagoras. Sifuna a isiseko isiqingatha ixabiso c: 2 = √ (x: cosα) ^ 2 - (x 2) = √x ^ 2 (1 - cos ^ 2α): cos ^ 2α = x ∙ tgα. Ke c = 2xtgα. Nantso indlela elula unako ukufumana naliphi na inani macala polygon sibhalwe.

Ukubalwa emacaleni square abhaliwe kwisangqa

Njengazo naziphi na ezinye yipolygon square sibhalwe kungoku nje inamacala namaqela alinganayo engile. Ukuba isebenzisa indlela efanayo naleyo kanxantathu. Bala icala isikwere singavuselelwa yexabiso idiagonal. Xiya ndlela zolwimi ngokunzulu. Yinto eyaziwayo ukuba idiagonal bisects ekujikeni. Ekuqaleni ixabiso layo iqondo 90. Ngoko ke, lo ezimbini emva ukulaba unxantathu uxande. angles emazantsi iya kulingana degrees 45. Ngako oko, kwicala ngalinye isikwere uyalingana, oko kukuthi: a = b = c = d = e e√2 ∙ cosα = 2, apho e - ke idiagonal isikwere okanye isiseko kuyilwa emva kolwahlulo kanxantathu yoxande. Le asiyiyo yodwa indlela lokufumana engontsini isikwere. Sizibhale inani kwisangqa. Ukwazi embindini wesangqa R, sifumana ulwalathiso lwe ngcambu. We ukubala oko ngolu hlobo lulandelayo A4 = R√2. Le radii yeepolygons rhoqo ebalwa ifomula R = a: 2tg (360 ^o: 2n), apho a - ubude icala.

Indlela yokubala kujikelezo n-gon

Kujikelezo n-gon kudityaniswe zonke macala ayo. Kulula ukubala. Kufuneka wazi imilinganiselo onke amaqela. Kuba ezinye iintlobo weepholigoni, kukho iifomyula ezikhethekileyo. Amvumela ukuba ufumane kwibala eninzi ngokukhawuleza. Yinto eyaziwayo ukuba nayiphi yipolygon buphindeke alinganayo. Ngoko ke, ukuze ukubala iperimitha yayo, izimamva ezinokuthelekiswa ukwazi ubuncinane omnye kubo. Ifomula liya kuxhomekeka kwinani engontsini imilo. Ngokubanzi, kukhangeleka ngolu hlobo: R = i, apho a - ixabiso ecaleni, kunye n - inani engile. Umzekelo, ukuba ufumane yomjikelezo octagon rhoqo kunye ecaleni 3 cm, kufuneka uphinda-phinde ngo-8, oko kukuthi, P = 3 ∙ 8 = 24 cm Ngokuba namacala a ecaleni kwe-5 cm ibalwa ngolu hlobo lulandelayo :. P = 5 ∙ 6 = 30 cm kunye njalo. polygon ngalinye.

Ukufumana yomjikelezo parallelogram, isikwere kunye nedayimani

Kuxhomekeka zingaphi iindonga wenza yipolygon, ukubala iperimitha yayo. Oku kube lula kakhulu lo msebenzi. Eneneni, xa kuthelekiswa nezinye iinyama, kule meko akuyomfuneko ukuba ukukhangela konke esandleni sakhe, sekwanele omnye. Kumgaqo kwenziwa kujikelezo Ikwadrilatherali, oko kukuthi, isikwere kunye nekalikedo. Nangona ukuba amanani ezahlukeneyo, ifomula apho omnye P = 4a, apho a - icala. Nanku umzekelo. Ukuba iqela isikwere okanye irhombus 6 cm, sifumana iperimitha lulandelayo: P = 4 ∙ 6 = 24 cm V parallelogram ezi ephambana kuphela .. Ngoko ke, umjikelezo yayo usebenzisa enye indlela. Ngoko ke, kufuneka sazi ubude kunye nobubanzi ngokomzekeliso. Emva koko sisebenzisa P ifomula = (a + b) ∙ 2. parallelogram yakhe macala onke alinganayo kunye engile phakathi kwabo, ebizwa nekalikedo.

Ukufumana lokubiyela unxantathu alinganayo kunye uxande

Umhlaba ojikeleze ekunene triangle alinganayo ingafunyanwa kwi kwifomula P = 3a, apho a - ubude icala. Ukuba ayaziwa, ingafunyanwa ngokusebenzisa udibaniso. Xa unxantathu ilungelo elingana nexabiso kukho emacaleni nje emibini. Isiseko ingafunyanwa ngokusebenzisa theorem kaPythagoras. Emva uya kuyazi amaxabiso zonke zontathu amaqela, esibala lokubiyela. Izakufunyanwa usebenzisa ifomula R = a + b + c, apho kwaye b - amacala alingane, yaye - kwisiseko. Khumbula ukuba kukho unxantathu alinganayo, a = b = a, ngoko a + b = 2a, ngoko P = 2a + c. Umzekelo, icala unxantathu isosceles ilingana-4 cm, fumana, noseko lwalo kunye ongaphandle kwebala. Ukubala inani hypotenuse kaPythagoras kunye √a = ² + ² = √16 + 16 = √32 = 5,65 cm. Ngoku ukubala iiperimeter P = 2 ∙ 4 + 5.65 = 13.65 cm.

Indlela yokufumana eliyi yipolygon

A polygon rhoqo ufumaneka ebomini bethu yonke imihla, umzekelo, i-njengesiqhelo isikwere, unxantathu, octagon. Kubonakala ngathi akukho nto kulula kunokuba kwakhiwe le piece wena. Kodwa loo nto kanye xa efika kuqala. Ukwenzela ukwakha nayiphi n-eYangon, kuyimfuneko ukwazi ixabiso engile yayo. Kodwa ngaba njani? izazinzulu Nkqu yamandulo ziye bezama ukwakha yeepoligoni rhoqo. Babegqibe ukuze angqamane ibe isangqa. Kwaye ke phezu kwayo uphawula imfuno kwindawo, izihlanganisi nabo imigca ethe ngqo. le ngxaki isonjululwe ukwakha iimilo ezilula. Iifomula kunye theorems zifunyenwe. Umzekelo, i-Euclid emsebenzini wakhe odumileyo "Ekhaya" ukuba isisombululo yeengxaki inxaxheba 3-, 4-, 5-, 6- no-15-gons. Wafumana iindlela ukwakha ukufumana engile. Makhe sibone indlela yokuyenza le-15-gon. Okokuqala, kufuneka Ungabala udibaniso engile zayo. Kuyimfuneko ukusebenzisa S ifomula = 180⁰ (n-2). Ngoko ke, sinikwe 15-eYangon, ngenxa yoko, inani n ngu 15. Ukusetyenziswa idata eyaziwa kwaye afumane S ifomula = 180⁰ (15 - 2) = 180⁰ x 13 = 2340⁰. Siyifumene udibaniso lwamaxabiso wonke engile ngaphakathi buyimilo-15 macala. Ngoku ke kufuneka ukuba ufumane ixabiso ngamnye kubo. Zonke angles 15 ukwenza izibalo 2340⁰: 15 = 156⁰. Ngenxa yoko, angle yangaphakathi ngasinye 156⁰, ngoku nomlawuli kunye nekhampasi nga akhe echanekileyo 15-gon. Kodwa kuthekani ezintsonkothileyo ngakumbi n-eYangon? Kwiinkulungwane ezininzi izazinzulu baye bazama ukulungisa le ngxaki. Kwafunyanwa kuphela ngenkulungwane ye-18 ngu Carl Fridrihom Gaussom. Wakwazi ukwakha-square 65537. Ukususela ngoko ke ingxaki ngokusemthethweni zicingeleka isonjululwe ngokupheleleyo.

Ukubalwa angela n-gon kwilayini eziqala kwisangqa

Kakade ke, kukho iindlela ezininzi lokufumana engile zepolygons. Amaninzi ukuba zibalwa ngokwamazinga. Kodwa sinako ukuyiveza kwilayini eziqala kwisangqa. Indlela ukukwenza oko? Maluqhutywe ngolu hlobo. Okokuqala, sifumanisa inani macala yipolygon, uze emva koko uthabathe zoko 2. Ngenxa yoko, sifumana ixabiso: n - 2. Qhamani umahluko yafunyanwa inani n ( "pi" = 3.14). Ngoku besizakwahlula ukuba imveliso ngenani ezimbombeni kwi n-gon. Cinga nje ngomzekelo ka zokubala idatha pyatnadtsatiugolnika efanayo. Ngenxa yoko, inani n ilingana no 15. asebenzise S indlela = n (n - 2): n = 3,14 (15 - 2): 15 = 3,14 ∙ 13: 15 = 2.72. Oku, Kakade ke, hayi indlela kuphela ukubala engile eziqala. Uyakwazi ukwahlula ngokulula ubungakanani-engile ngamaqondo ngenani 57,3. Ngapha koko, izidanga ezininzi kangaka kuyalingana yelayini omnye.

Ukubalwa angles ku grads

Ukongeza izidanga kunye eziqala, nokuphakama yipolygon, unga zama ukufumana ixabiso ngokwamazinga. Oku kwenziwa ngolu hlobo lulandelayo. Thina nothabatha kwinani lilonke engile inombolo 2, ebabela umahluko waba inani macala yipolygon. Found isiphumo lithi liphindaphindwe 200. Ngendlela, le yunithi yomlinganiselo ka-engile njengoko grads, esetyenziswa ngenkankulu.

Ukubalwa engile umphandle n-gon

Nawuphi polygon rhoqo, ukongeza basekhaya, sinako ukubala kwakhona kwikona engaphandle. ixabiso layo iyafana kwezinye amanani. Ngoko ke, ukufumana engile sangaphandle yipolygon, kufuneka wazi ukubaluleka lwangaphakathi. Ngapha koko, siyazi ukuba isiphumo sale engile ezimbini isoloko degrees 180. Ngoko ke, ubalo kwenziwa ngolu hlobo lulandelayo: 180⁰ thabatha kwikona embilwini. Sifumana umahluko. Kuya kuba ixabiso engile ezikufutshane kuyo. Umzekelo, kwikona engaphakathi sigcawu 90, ngoko ke imbonakalo iya kuba 180⁰ - 90⁰ = 90⁰. Njengoko sibona, kulula ukufumana. angle lwangaphandle ingathatha ixabiso ukusuka + 180⁰ ukusa, ngokulandelelanayo, -180⁰.