Indlela yokuthola azungezwe isikwele ezinhlangothini zalo, phezu kwendawo, futhi kolunye uhlangothi igebe eliphakathi idayagonali yayo futhi ohlangothini isikwele

Ngokuvamile empilweni, abantu badinga ukuthola azungezwe isikwele. kuphakama Le nkinga, isibonelo, ezimweni lapho kudingeka ukubala ubude besikhathi kocingo noma imali edingekayo ukuze uwanamathisele wallpaper izindonga ekamelweni. Nokho, ngo-mBhalo, azungezwe kuphela isixazululo umlamuli imisebenzi esisebenzayo. Kodwa, noma kunjalo, kulesi simo, abantu kudingeka sazi indlela yokuthola azungezwe we isikwele.

Ukuze uqale, Ngithanda ukuchaza ukuthi yini azungezwe. Ipherimitha, eqinisweni, umngcele ethile ukuma Jomethri noma ubude ingqikithi imingcele yayo. Manje chaza inshokutsi yetinhlobo konxantathu. Parallelogram nge engele kwesokudla kufanele uyiswe amabhokisi. Empeleni, isici esiyinhloko eqinisweni engeli yayo, okufanele figure weJiyomethri ezine.

Ngakho, ukuze uthole Ubude umngcele konxantathu, kumele tente ubude zonke izinhlangothi zalo. Njengoba sesibonile, izinhlangothi parallel we isikwele bayalingana, Ngakho-ke, ukwenza lula ukuqonda, kufanele kuqondwe ukuthi azungezwe isikwele ilingana kabili isamba ezinhlangothini zalo ezimbili.

Ngokucaca, lumelela amacala alingane ebhokisini izinhlamvu zamagama Latin "a" futhi "b" ngokulandelana. Ngakho, kubonakala sengathi P (ipherimitha unxande) = a + b + a + b. Lokhu kwesibalo kungenziwa siguqulelwa fomula elandelayo: P = 2 × (a + b).

Kodwa ngokuvamile empilweni, zikhona izikhathi lapho siyazi ubude bohlangothi olulodwa nje, kanti ezinye nezinye izingxenye ebhokisini, noma ngaphandle kwalo. Cabangela ezimbalwa ongakhetha kuzo.

Ngokwesibonelo, kudingeka ukuthola ukuthi yini azungezwe isikwele, inqobo nje ubude kolunye uhlangothi isikwele ayaziwa, kodwa endaweni yayo ezaziwayo. Kudingekile, kusetshenziswa indlela kuhlaziywa indawo isikwele okuyinto ilingana umkhiqizo ezinhlangothini zalo, ukubala ubude besikhathi uhlangothi lwalo lwesibili. Lokhu kalula kwenziwa sokuhlukanisa endaweni ethile nesici esithile. Ukwazi zombili we isikwele ingabalwa kalula, futhi ipherimitha yayo.

Lokhu samuntu kufanelekile lapho kubalwa inani impahla edingekayo ngengxenye kocingo, lapho endaweni eshiwo kumadokhumenti. Umuntu kumelwe amane ngaphezu kwalokho ukukala ingxenye uhlangothi. Kodwa ngezinye izikhathi kudingeka wazi indlela thola azungezwe isikwele, uma wazi omunye ezinhlangothini isikwele kanye idayagonali yayo.

Ngokwemvelo, kathisha isinyathelo sokuqala ukuba uthole ubude ehlangothini lwesibili isikwele. Singasho libalwe theorem kaPythagoras, othi hypotenuse unxantathu kwesokudla, elalizokwakhiwa e isikwele, kuhlanganisa isamba sezikwele kwalezi zinhlangothi zombili. Ngakho-ke, kudingeka ukubala ubude uqonde idayagonali futhi eyaziwa nobude side in isikwele, tfola umehluko phakathi kwabo, futhi kusukela kule umehluko kufanele kube ukuthatha impande square.

I kwaphumela impande square futhi kuyoba engaziwa ohlangothini obuphelele. Futhi indlela yokuthola azungezwe isikwele kungenziwa eligoqiwe eyaziwa ecaleni ubude futhi kuphindaphindwe kabili kubo, wonke umuntu wayeyokwazi kalula babhekane nalesi inqubo.

Ekilasini izibalo nawo acabangela indlela yokuthola azungezwe isikwele esikrinini idayagonali nokukodwa engela oyingozi ezakhiwe idayagonali futhi kolunye uhlangothi isikwele. Lapha sinawo iyisibonelo esihle salokhu ukusetshenziswa kulesi sibalo sine. Kusukela esikoleni sonke siyazi ukuthi sine ye-engeli unxantathu kwesokudla lilingana ratio umlenze eduze futhi hypotenuse. Yingakho ifomula: isono X = cathetus: hypotenuse (diagonals we isikwele).

Itheku kalula liyabonakala by Bradis etafuleni, elaziwayo ifomula sifakwe ngokuphambeneyo - hypotenuse, futhi kalula kukhompyutha, omunye emaceleni isikwele. Manje isinyathelo esilandelayo ukuthola umngane eceleni isikwele yesibili. Kube inures samuntu usebenzisa Pifogora theorem okuxoxwe ngawo ngenhla. Squaring eyaziwa idayagonali nokukhupha esigcawini wasesemukela ohlangothini ezitholakele. Kusukela impendulo empandeni sikwele. Manje izinhlangothi eyaziwa ungakwazi ukubala azungezwe, eligoqiwe ubude bawo kanye ukubuyelela kabili.

Njengoba ungacabanga, lokhu akuyona version oluphelele sezibonelo, eqinisweni, kunabaningi, kodwa evamile kuchaziwe ngenhla.

Ngakho, kungashiwo baphetha ngokuthi kungesemehlweni ubude izinhlangothi ezimbili okuhambisana we isikwele ukuchaza ipherimitha cishe kungenakwenzeka. Nokho, usebenzisa i-arsenal of theorems Jomethri nalabo axioms, kuba njalo kungenzeka ukubala azungezwe isikwele, eligoqiwe nazo zonke izinhlangothi zalo.