Le ndawo base ezindengezini, kusukela esingunxantathu ukuba zikanhlangothiningi

Okunye amaphrizimu zahlukile esinye kwesinye. Ngesikhathi esifanayo babe kuningi abafana ngakho. Ukuze uthole indawo base ezindengezini, kudingeka ukuthi luhlobo luni kuba.

ithiyori Jikelele

Prism yiliphi polyhedron, ezinhlangothini zazo ngendlela parallelogram. Kulokhu, isizinda kwakho kungenzeka iyiphi polytope - kusuka unxantathu kuya n-gon. Kulokhu-ke base nezibhebhe zihlale ulingana nomunye. Lokho akusebenzi ezinhlangothini - bangathumela ziyahlukahluka kakhulu ngobukhulu bazo.

Ngo ukuxazulula izinkinga abazalwane ababehlangabezana nazo hhayi kuphela endaweni base nezibhebhe. Kungadinga ulwazi ebusweni uhlangothi, okungukuthi, zonke ebusweni ukuthi akuzona elisekela. Gcwalisa ubuso kumele kube inyunyana bonke ubuso ezakha nezibhebhe.

Ngezinye izikhathi ukuphakama livela izinkinga. Kuyinto perpendicular isizinda. Ovundlile we polyhedron liyingxenye exhumanisa iyiphi vertices ezimbili ngazimbili engekho nakunoma ebusweni efanayo.

Kumele kuqashelwe ukuthi le ndawo base a nezibhebhe kwesokudla noma bayichezukisa ezimele ye-engeli phakathi kwawo ubuso lateral. Uma banayo nesimo esifanayo ngesikhathi ubuso phezulu kanye phansi, izindawo zabo bayalingana.

esingunxantathu nezibhebhe

Kuyinto ngasesinqeni se sibalo kokuba vertices ezintathu, okungukuthi unxantathu. Ubizwa ukuba ezahlukene ezaziwayo. Uma unxantathu kuyinto elingunxande, kwanele ukukhumbula ukuthi endaweni ebizwa ngokuthi imilenze kwesigamu umsebenzi.

Inkulumo zezibalo simiswe ngalendlela lelandzelako: S = ½ av.

Ukuze uthole indawo esingunxantathu base nezibhebhe injalo jikelele, ifomula ewusizo Heron futhi eyodwa lapho isandla esithathwayo nengxenye ukuphakama eyenziwa nalokho.

Ifomula sokuqala ukuba bunjesi: S = √ (p (p-kahle) (p-c) (p-c)). semiperimeter (p) elikhona marekhodi, okusho lokhu isamba izinhlangothi ezintathu, zihlukaniswe amabili.

Okwesibili: S = ½ _futhi n * a.

Uma kudingeka zifunde lokungcolisa nezibhebhe esingunxantathu okuyinto efanele, khona-ke unxantathu kuyinto equilateral. Ngoba has ifomula ayo: S = ¼ futhi ² * √3.

nezibhebhe quadrangular

isizinda salo yimiphi quadrangles ezaziwayo. Lokhu kungaba isikwele noma isikwele, rhombus, noma ibhokisi. Esimweni ngasinye, ukuze ukubala indawo base iphrizimu, kuzomele ifomula yabo.

Uma substrate - isikwele, endaweni yayo ichazwa ngokuthi: S = Av, lapho A no B - we isikwele.

Uma kuziwa ku-nezibhebhe quadrangular, base nezibhebhe endaweni efanele ibalwa ifomula ngoba isikwele. Ngoba lokho sika lokho kuvela ukuthi ukuqamba amanga ezansi. Futhi S = ^2.

Esimweni lapho base - kuyinto ibhokisi, kuzomele i equation ezifana: S = a * n _a. Kwenzeka ukuthi ibhokisi eceleni futhi ungomunye emagumbini omabili. Khona-ke, ukubala ukuphakama isidingo sokusebenzisa ifomula ezengeziwe: N _a = b * isono A. Ngaphezu kwalokho, i-engeli A eyakhelene ohlangothini "b" nobude n _futhi okuphambene kule ekhoneni.

Uma kumila iphrizimu kuyinto rhombus ke ukunquma indawo yayo uzodinga indlela efanayo nanjengoba ukuthi we parallelogram (njengoba kunjalo esimweni sakhe esithile). Kodwa umuntu angabona futhi usebenzise ezifana: S = ½ d ₁ d _2. Lapha, d ₁ kanye d ₂ - amabili diagonals we rhombus.

nezibhebhe Pentagonal

Leli cala kuhilela noqhekeko we ipholigoni ku onxantathu kabani izindawo kulula ukufunda. Nakuba-ke kwenzeka ukuthi izibalo kumelwe kube yinombolo ezahlukene vertices.

Njengoba base nezibhebhe - pentagon njalo, zingahlukaniswa zibe ezinhlanu unxantathu equilateral. Khona-ke base nezibhebhe endaweni elilingana endaweni unxantathu (bheka ifomula ngenhla kungaba) iphindwe ezinhlanu.

nezibhebhe avamile olunezinhlangothi

Ngokusho isimiso kuchazwe okwesikhathi nezibhebhe pentagonal, kungenzeka ukuba aphule iheksagoni base 6 equilateral onxantathu. Formula lokungcolisa nezibhebhe ezifana afana odlule. Kuphela ke unxantathu endaweni equilateral kufanele iphindwe eziyisithupha.

Bheka ifomula Ngakho: S = 3/2 futhi ² * √3.

imisebenzi

Inombolo 1. Dana kwesokudla uqonde nezibhebhe unxande. idayagonali elilingana 22 cm, ukuphakama polyhedron Ezazimzungezile - 14 cm Bala nezibhebhe isizinda ndawo futhi bonke ubuso ..

Isinqumo. isizinda nezibhebhe square, kodwa iqembu alwaziwa. Kungenzeka ukuthola ukubaluleka diagonal isikwele (x), okuyinto lihlotshaniswa iphrizimu idayagonali (d) futhi ukuphakama kwalo (n). x ² = d ² - N ^2. Ngakolunye uhlangothi, kulesi sigaba "x" iyona hypotenuse unxantathu eyanqunywa imilenze ayalingana ohlangothini square. Okungukuthi x ² = ² + ^2. Ngakho-ke ematfuba ukuthi ² = (d ² - n ²⁾ / 2.

D esikhundleni inombolo 22, futhi "n" esikhundleni inani layo - 14, kuvela ukuthi ohlangothini square uyalingana 12 cm Manje nje ufunde Footprint: 12 * 12 = 144 cm ^{2 ..}

Ukuze uthole indawo bonke ubuso, kubalulekile ukubeka phansi ukubaluleka kabili base futhi quadruple ohlangothini square. Lesi sakamuva lula ukuthola ifomula ukuze isikwele: uphindaphinde ukuphakama futhi maqondana kumila polyhedron. Okungukuthi 14 no-12, le nombolo kuyoba ulingana 168 cm ^2. Le ndawo ingqikithi ebusweni nezibhebhe kuyinto 960 ^cm2.

Impendulo. Le ndawo base nezibhebhe ilingana 144 cm ^2. Yonke indawo - 960 ^cm2.

Inombolo 2. Dan njalo nezibhebhe esingunxantathu. Ngaphansi kwesithombe senqaba kuyinto unxantathu nohlangothi lokuzibona 6 cm Lokhu idayagonali ohlangothini ubuso 10 cm Bala sikwele: .. A isizinda kanye ebusweni uhlangothi.

Isinqumo. Njengoba iphrizimu ilungile ke isisekelo sayo kuyinto unxantathu equilateral. Ngakho-ke, indawo 6 uyalingana lesikwele, siphindwe ¼ kanye impande skwele 3. ukucabanga elula kunikeza umphumela: 9√3 ^cm2. Le ndawo eyodwa kumila nezibhebhe.

Lonke bhekana ohlangothini ayafana nokujamela calandze emaceleni 6 no-10 cm. Ukuze abale endaweni yabo esanele uphindaphinde izinombolo. Khona-ke ngibandise by ezintathu, ngoba ohlangothini ubhekene e iphrizimu kangaka. Khona-ke ebusweni uhlangothi endaweni nxeba 180 cm ^2.

Impendulo. Square: substrate - 9√3 ^cm2, ohlangothini ebusweni nezibhebhe - 180 cm ^2.