Yuav ua li cas mus nrhiav tau cov hypotenuse ntawm ib txoj cai daim duab peb sab

Ntawm cov heev heev suav tau rau lub xam los ntawm ntau yam ntau ntawm txawv geometric duab, yog nrhiav lub hypotenuse ntawm daim duab peb sab. Nco qab hais tias ib tug daim duab peb sab yog hu ua ib tug polyhedron muaj peb cov ces kaum. Hauv qab no yog ib tug ob peb txoj kev sib txawv los mus laij lub hypotenuse ntawm lub duab peb ceg yuav muab.

Chiv, cia saib yuav ua li cas nrhiav tau cov hypotenuse ntawm ib txoj cai daim duab peb sab. Rau cov neeg xeb, hu ua lub duab peb sab uas muaj ib lub ntawm 90 degrees. sab ntawm lub duab peb sab, nyob rau lub opposite sab ntawm txoj cai lub yog hu ua tus hypotenuse. Nyob rau hauv tas li ntawd, nws yog ib lub coos sab ntawm lub duab peb sab. Nyob rau qhov ntev ntawm lub hypotenuse paub ntau yog xam raws li nram no:

• Paub ntev ntawm ob txhais ceg. Hypotenuse nyob rau hauv cov ntaub ntawv no yog muab los xam siv cov Pythagorean theorem, uas nyeem hais tias raws li nram no: square ntawm lub hypotenuse sib npaug rau hauv lub sum ntawm cov plaub fab ntawm lub ob sab. Yog hais tias peb xav txog ib txoj cai-angled daim duab peb sab BKF, qhov twg BK thiab KF ob txhais ceg thiab FB - lub hypotenuse, lub FB2 = BK2 + KF2. Nws hais ntxiv hais tias nyob rau hauv kev xam xyuas qhov ntev ntawm lub hypotenuse yuav tsum tau tsa qho nyob rau hauv txhua tus ntawm cov square qhov tseem ceeb ntawm qhov lwm ob sab. Ces ntxiv txog cov xov tooj thiab hais tias npaum li cas los ntawm cov square hauv paus.

Xav txog qhov no piv txwv: Dan daim duab peb sab uas muaj ib lub cai lub. Ib ceg yog 3 cm, 4 cm lwm. Nrhiav cov hypotenuse. Cov tshuaj no yog raws li nram no.

FB2 = BK2 + KF2 = (3cm) 2+ (4 cm) 2 = + 9sm2 16sm2 = 25 cm2. Peb extract lub square hauv paus thiab tau FB = 5cm.

• Paub cathetus (BK) thiab lub kaum sab xis uas nyob ib sab mus rau nws, uas tas hypotenuse thiab hais tias tus ceg. Yuav ua li cas mus nrhiav tau cov hypotenuse ntawm daim duab peb sab? Peb txhais lub npe hu lub α. Raws li rau cov cuab yeej ntawm ib lub duab peb sab, uas hais tias tus piv ntawm ceg ntev rau qhov ntev ntawm lub hypotenuse yog sib npaug zos rau cov cosine ntawm lub kaum sab xis ntawm lub hypotenuse thiab cov ceg. Xav no daim duab peb sab yuav sau li: FB = BK * cos (α).
• Paub cathetus (KF) thiab tib lub kaum sab xis α, tsuas tam sim no nws tau raug tawm tsam. Yuav ua li cas mus nrhiav tau cov hypotenuse nyob rau hauv cov ntaub ntawv no? Cia peb tag nrho rau tib lub thaj chaw ntawm ib txoj cai daim duab peb sab thiab peb kawm tau hais tias tus piv ntawm ceg ntev rau qhov ntev ntawm lub hypotenuse yog sib npaug zos rau cov sine ntawm lub kaum sab xis ntawm lub opposing sab. Hais tias yog, FB = KF * kev txhaum (α).

Xav txog cov piv txwv nram qab. Muab tag nrho cov cai-angled daim duab peb sab nrog hypotenuse BKF FB. Cia lub kaum sab xis F sib npaug 30 degrees, qhov thib ob lub B yog 60 degrees. Lwm lub npe hu cathetus BK, qhov ntev ntawm uas sau raws nkaus Ii 8 cm laij tau qhov yam nqi li sai tau .:

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

• Paub vajvoog voos kheej-kheej (R), piav txog ib tug daim duab peb sab uas muaj ib lub cai lub. Yuav ua li cas mus nrhiav tau cov hypotenuse nyob rau hauv qhov kev xam ntawm xws li ib tug teeb meem? Los ntawm cov thaj chaw ntawm lub vajvoog circumscribing daim duab peb sab uas muaj ib lub cai kaum yog lub npe hu, xws li hais tias ntawm qhov chaw ntawm lub vajvoog coincides nrog lub point ntawm lub hypotenuse faib nws nyob rau hauv ib nrab. Nyob rau hauv cov lus yooj yim - lub voos kheej-kheej sau raws nkaus Ii mus rau ib nrab ntawm lub hypotenuse. Li no, lub hypotenuse yog sib npaug zos rau ob zaug lub voos kheej-kheej. FB = 2 * R. Yog hais tias muab ib tug zoo xws li cov teeb meem, uas yog tsis paub voos kheej-kheej, thiab lub nruab nrab, koj yuav tsum xyuam xim rau cov khoom teejtug uas yog lub vajvoog circumscribed txog daim duab peb sab uas muaj ib lub cai kaum, uas hais tias cov voos kheej-kheej yog sib npaug zos rau theem nrab twv mus rau lub hypotenuse. Siv tag nrho cov khoom, qhov teeb meem yog solved nyob rau hauv tib txoj kev.

Yog hais tias cov nqe lus nug yog ua li cas nrhiav tus hypotenuse ntawm ib tug ceskaum txoj cai daim duab peb sab, nws yog tsim nyog los hu tag nrho rau tib lub Pythagorean theorem. Tab sis, ua ntej ntawm tag nrho cov nco ntsoov hais tias lub ceskaum daim duab peb sab yog ib tug daim duab peb sab uas muaj ob sib npaug zos sab. Nyob rau hauv cov ntaub ntawv ntawm ib txoj cai daim duab peb sab sib npaug zos sab yog cov ob txhais ceg. Muaj FB2 = BK2 + KF2, tab sis raws li BK = KF peb tau raws li nram no: FB2 = 2 BK2, FB = BK√2

Raws li koj tau pom, paub txog cov kev Pythagorean theorem thiab cov khoom ntawm ib txoj cai daim duab peb sab, kom daws tau qhov teeb meem rau cov uas koj yuav tsum xam qhov ntev ntawm lub hypotenuse, nws yog heev yooj yim. Yog hais tias tag nrho cov khoom ntawm ib qho nyuaj rau nco ntsoov, kawm npaj ua qauv, hloov paub qhov tseem ceeb nyob rau hauv uas nws yuav tau los xam yuav tsum tau ntev ntawm lub hypotenuse.