Ummandla unxantathu alinganayo

Phakathi amanani zejiyometri, nekuxoxwa ngazo kwicandelo geometry, sesona obufunyenwe ngesisombululo iingxaki ezahlukeneyo kunye nxantathu. It is a nani zejometri lisekwa imigca emithathu. Baya ngaxa lithile musa phambana kwaye zifana. Kuyenzeka ukunika inkcazo ezahlukeneyo: nxantathu yi egopheni polygonal closed eliquka iiyunithi ezintathu apho ekuqaleni kwayo kunye nesiphelo oxhulumene kwi ngenqaku elinye. Ukuba omathathu macala zinexabiso elilinganayo, ngoko ke unxantathu alinganayo, okanye, njengoko bathi, i alinganayo.

Sazi njani ukuqinisekisa indawo unxantathu alinganayo? Ukusombulula ezi ngxaki kuyimfuneko ukwazi ezinye iimpawu amanani zejometri. Okokuqala, kule hlobo unxantathu zonke engile bayalingana. Okwesibini, ukuphakama elehla ukusuka phezulu ukuya isiseko, kukuthi median nokuphakama. Oku kubonisa ukuba ukuphakama njengezona nxantathu ihlukane engile ezimbini ngokulinganayo, kwaye icala - ibe amacandelo amabini alinganayo. Ekubeni unxantathu alinganayo libunjwa ezimbini oonxantathu ekunene-engile egqithe, xa kujongwa amaxabiso ezinqwenelekayo kufuneka asebenzise theorem kaPythagoras.

indawo Ibala kanxantathu kungenziwa ngeendlela ezahlukeneyo, ngokuxhomekeke amanani ezaziwayo.

1. Cinga unxantathu alinganayo kunye kwicala b eyaziwa kunye nokuphakama h. indawo unxantathu kulo mzekelo iya kulingana omnye-isiqingatha kwicala imveliso kunye nokuphakama. Xa ifomula bekuya ikhangeleke ngolu hlobo:

S = 1/2 * h * b

Xa amagama, indawo triangle alinganayo ilingana mnye-isiqingatha ecaleni layo umsebenzi kunye nokuphakama.

2. Ukuba uyazi kuphela icala ixabiso, ngaphambi kokuba afune ndawo, kuyimfuneko ukuba ukubala ukuphakama kwayo. Kuba oku cinga isiqingatha unxantathu, leyo ukuphakama omnye umlenze, lo hypotenuse - nganeno unxantathu, kunye nomlenze yesibini - isiqingatha emacaleni nxantathu ngokweempawu yayo. Zonke evela theorem efanayo kaPythagoras thina ezichaza ukuphakama nxantathu. Njengokuba kubhaliwe ndisazi, isikweri hypotenuse ingqamana udibaniso lwezikweri imilenze. Xa siqwalasela kwisiqingatha nxantathu, kule meko Uhlangothi hypotenuse, kwicala wesiqingatha - emlenzeni, nokuphakama - yesibini.

(B / 2) ² + h2 = b², kungoko

h² = b²- (b / 2) ². Nantsi zifana:

h² = 3b² / 4,

h = √3b² / 4,

h = b / 2√3.

Njengoko ubona, ukuphakama mzobo ethathelwa ingqalelo lilingana imveliso isiqingatha ubuso bakhe ingcambu ka emithathu.

Umjikelezi ifomula ubone: S = 1/2 * b * b / 2√3 = b² / 4√3.

Oko kukuthi, indawo unxantathu alinganayo ilingana imveliso kwicala yesine isikwere kunye ingcambu ka emithathu.

3. Kukho eminye imisebenzi apho kufuneka ukuba lijonge indawo unxantathu alinganayo ubude ethile. Ke kulula kunakuqala. Sele bazisa kwimeko yangaphambili, ukuba h² = 3 b² / 4. Ngaphezu koko kuyimfuneko apha lokusirhoxisa kwicala kwaye endaweni kungena ifomula ndawo. Kuya kukhangela ngolu hlobo:

b² = 4/3 * h², yiyo b = 2H / √3. Esikhundleni ifomula ukuba isikwere, siya nakufumana nto noko;

S = 1/2 * h * 2H / √3, kungoko S = h² / √3.

Kukho iingxaki xa kuyimfuneko ukufumana indawo unxantathu alinganayo kunye embindini wesangqa sibhalwe okanye nezisekelwe. Ngokuba lo ukubala, kukwakho ifomula ezithile zezi zilandelayo: r = √3 * b / 6, R = √3 * b / 3.

Act sele eziqhelekileyo kuthi lo mgaqo. Nge radius eyaziwa, lisenza kwicala Indlela kunye nokubala ngokuthi kufakwe ixabiso ezaziwayo kwirediyasi. Ixabiso wafumana endaweni kwifomula sele eyaziwa yokubala indawo nxantathu ekunene ukwenza izibalo kwaye ufumane ixabiso efunekayo.

Njengoko ubona, ukuze bacombulule iingxaki efanayo, kufuneka wazi kuphela iimpawu unxantathu alinganayo kunye theorem kaPythagoras, kwaye, kunye, kwaye embindini wesangqa sibhalwe. Ngokubamba isicombululo ulwazi iingxaki ezinjalo abayi bazobe nzima kakhulu.