Indlela ukufumana hypotenuse kanxantathu ilungelo

Phakathi izibalo amaninzi okuba ukubalwa omyinge ezahlukeneyo ezahlukeneyo iimilo zejiyometri, ufumene i hypotenuse lo nxantathu. Khumbula ukuba unxantathu kuthiwa Ipolihedron ukuba engile ezintathu. Ngezantsi ezimbalwa ngeendlela ezahlukeneyo ukubala hypotenuse le oonxantathu ziya kunikwa.

Ekuqaleni, makhe sibone ukuba ukufumana indlela hypotenuse kanxantathu tye. Kwabo ngekopi, ebizwa unxantathu eziziingxande ukuba i engile degrees 90. ecaleni unxantathu, kubekwe phezu kwelinye icala akwiengile elungileyo kuthiwa hypotenuse. Ngaphezu koko, akukho icala mde lo nxantathu. Kuxhomekeka ubude ubuninzi hypotenuse eyaziwa ibalwa ngolu hlobo lulandelayo:

• Eyaziwa ubude emilenzeni. Hypotenuse kule meko ibalwa ngokusebenzisa theorem kaPythagoras, leyo ifundeka ngolu hlobo lulandelayo: isikweri hypotenuse ilingana udibaniso lwezikweri elinye macala mabini. Xa siqwalasela unxantathu BKF ekunene-engile egqithe, apho BK kunye kf imilenze kunye FB - le hypotenuse, lo FB2 = BK2 + KF2. Oku kulandela ukuba ibale ubude hypotenuse kufuneka kuvuswa yanjalo nganye amaxabiso square ezinye macala mabini. Ke dibanisa la amanani nokuba bethatyathwe ngenxa ingcambu.

Nanku umzekelo: Dan unxantathu kunye esinekona sasekunene. Ngomlenze omnye yi-3 cm, 4 cm omnye. Fumana hypotenuse. Isisombululo ngolu hlobo.

FB2 = BK2 + KF2 = (3cm) 2+ (4 cm) 2 = + 9sm2 16sm2 = 25 cm2. Esilikhuphayo i ingcambu kunye uthole FB = 5cm.

• Eyaziwa cathetus (BK) kunye engile ezikufutshane kuyo, nto leyo yenza ukuba hypotenuse nokuba emlenzeni. Indlela ukufumana hypotenuse lo nxantathu? Thina isho α eyaziwa ekujikeni. Ngokutsho kwipropati kanxantathu buxande, nto leyo ithi umlinganiselo ubude umlenze ubude hypotenuse lilingana i cosine ye-engile phakathi hypotenuse kunye nomlenze. Ukuqwalasela lo nxantathu angabhalwa ngolu hlobo: FB = BK * cos (α).
• Eyaziwa cathetus (kf) kunye angle α efanayo, kodwa ngoku sele kwenziwe alwayo. Indlela yokufumana le hypotenuse kule meko? Makhe sonke ukuba iimpawu ezifanayo kanxantathu ekunene kwaye sifunda ukuba umlinganiselo ubude umlenze ubude hypotenuse lilingana i sine ye-engile lecala eliphikisayo. Oko kukuthi, FB = kf * isono (α).

Khangela lo mzekelo ulandelayo. Banikwa bonke unxantathu omnye ngasekunene-angled kunye hypotenuse BKF FB. Vumela le angle F ilingana degrees 30, le angle B yesibini degrees 60. Enye cathetus eyaziwa BK, ubude kufanekisela 8 cm ukubala ixabiso olifunayo kangangoko kunokwenzeka .:

FB = BK / cos60 = 8 cm.
FB = BK / sin30 = 8 cm.

• Eyaziwa circle radius (R), echazwe malunga unxantathu kunye esinekona sasekunene. Indlela ukufumana hypotenuse xa kuqwalaselwa ingxaki enjalo? Ukususela iimpawu wesangqa circumscribing unxantathu kunye esinekona sasekunene yaziwa, kangangokuba embindini wesangqa ingqamana ingongoma hypotenuse hlula sibe ngamacala amabini. Ngamagama alula - embindini ingqamana kwisiqingatha hypotenuse. Ngenxa yoko, le hypotenuse ilingana kabini embindini. FB = 2 * R. Ukuba ukukhanya akunike ingxaki efanayo, nto leyo eyaziwa radius, kunye udibaniso, kufuneka sinikele ingqalelo kwipropati wesangqa nezisekelwe malunga unxantathu kunye esinekona sasekunene, ukuba uthi kwirediyasi lilingana udibaniso batsaleleke hypotenuse. Ukusebenzisa zonke ezi zakhiwo, ingxaki isonjululwe ngendlela efanayo.

Ukuba umbuzo kukufumana indlela hypotenuse ye unxantathu isosceles ekunene, kuyimfuneko ukuba uqhagamshelane yonke le theorem efanayo kaPythagoras. Kodwa ke, qala zonke khumbula ukuba unxantathu isosceles kukho unxantathu buphindeke kabini ngokulinganayo. Kwimeko unxantathu ilungelo amacala alingane ke imilenze. Yiba FB2 = BK2 + KF2, kodwa njengoko BK = kf thina zilandelayo: FB2 = 2 BK2, FB = BK√2

Njengoko ubona, ukwazi theorem kaPythagoras kunye iimpawu unxantathu ilungelo, ukusombulula ingxaki leyo kufuneka ukubala ubude hypotenuse, kulula kakhulu. Ukuba zonke iimpawu nzima ukuze ukhumbule, funda iifomyula esele zenziwe, wokusebenzisa amaxabiso abaziwayo apho kuya kuba lula ukubala ubude efunekayo le hypotenuse.