Diagonal equilateral trapezoid. Mene ne tsakiyar line na trapezoid. Iri trapezoids. Trapeze - shi ..

Trapeze - musamman idan akwai wani mai quadrangle, a cikin abin da daya biyu daga bangarorin ne a layi daya. Kalmar "trapezoid" da aka samu daga kalmar Helenanci τράπεζα, ma'ana "tebur", "tebur". A wannan labarin, za mu duba a iri trapeze kuma da kaddarorin. Har ila yau, mun dubi yadda yin lissafi da mutum abubuwa na geometrical adadi. Alal misali, diagonal na wani equilateral trapezium, tsakiyar line, yankin da sauransu. A kayan dauke a cikin na farko lissafi rare style, t. E. A wani sauƙi m hanya.

Overview

Da farko, bari mu gane abin da mai quadrangle. Wannan adadi ne na musamman idan akwai wani mai polygon da ciwon hudu tarnaƙi, kuma hudu vertices. Biyu vertices na wani quadrilateral, wanda ba m, da ake kira m. A wannan za a iya ce da biyu da ba-m bangarorin biyu. The main iri quadrangles - a parallelogram, murabba'i mai dari, rhombus, square, trapezoid da deltoid.

Saboda haka baya ga trapeze. Kamar yadda muka ce, wannan adadi da bangarorin biyu suke a layi daya. Su kira sansanonin. A wasu biyu (ba a layi daya) - sasanninta. A kayan na daban-daban Nazarin kuma Nazarin sau da yawa sosai za ka iya saduwa da kalubalen da dangantaka da trapezoids wanda bayani sau da yawa na bukatar da dalibi ya sani ba ta rufe shirin. Makaranta Course lissafi ya gabatar da ƴan tare da kusassari dũkiyarsu da kuma diagonals kazalika da tsakãtsaki ce layi na wani isosceles trapezoid. Amma wanin cewa ake magana a kai a lissafi siffar yana da sauran fasali. Amma game da su daga baya ...

iri trapeze

Akwai su da yawa iri na wannan adadi. Duk da haka, mafi sau da yawa m zuwa la'akari biyu na su - isosceles da rectangular.

1. rectangular trapezoid - wani adadi a wanda daya daga cikin bangarorin perpendicular da tushe. Ta na da biyu kusassari ne ko da yaushe daidaita casa'in digiri.

2. isosceles trapezium - wani lissafi da adadi wanda tarnaƙi ne daidai. Saboda haka, da malã'iku a tushe ma ne daidai.

Babban ka'idojin hanyoyin domin karatu da kaddarorin na trapezoid

A ka'idojin sun hada da yin amfani da abin da ake kira aiki m. A gaskiya, babu bukatar ya shiga a cikin wani msar tambayar mana lissafi na sabon Properties na wannan adadi. Suna iya zama bude ko a cikin tsari na samar da nasu daban-daban ayyuka (mafi tsarin). Yana da muhimmanci sosai cewa malamin san abin da ayyuka kana bukatar ka saka a gaban dalibai a kowace lokaci na ilmantarwa tsari. Haka kuma, kowane trapezoid dukiya za a iya wakilta a matsayin mai key aiki a cikin aiki da tsarin.

Doka ta biyu ne da ake kira karkace kungiyar na nazarin "ƙwarai" trapeze Properties. Wannan yakan haifar da wani koma aiwatar da koyon mutum fasali na lissafi adadi. Saboda haka, dalibai sauki a tuna da su. Alal misali, dũkiyar da maki hudu. Yana za a iya tabbatar da matsayin a cikin binciken na kama da baya amfani da vectors. A Daidaita triangles m ga bangarorin da adadi, yana yiwuwa ya tabbatar da yin amfani da ba kawai da kaddarorin triangles tare da daidaita Heights gudanar da bangarorin da abin da karya a kan wani madaidaiciya line, amma kuma ta hanyar yin amfani da dabara S = 1/2 (ab * sinα). Bugu da ƙari kuma, shi ne zai yiwu a yi aiki daga dokokin sines ga rubũtacce trapezium ko dama-angled alwatika da trapezoid aka bayyana a cikin t. D.

Da amfani da "extracurricular" siffofi da wani lissafi da adadi a cikin abun ciki na makaranta hanya - a tasking su fasahar koyarwa. Constant tunani da nazari da kaddarorin da nassi daga cikin sauran damar dalibai don koyi da trapeze zurfi da kuma tabbatar da nasarar da aiki. Saboda haka, za mu ci gaba da nazarin wannan gagarumin adadi.

Abubuwa da kaddarorin wani isosceles trapezoid

Kamar yadda muka gani, a cikin wannan lissafi adadi tarnaƙi ne daidai. Amma duk da haka shi ne da aka sani a matsayin mai dama trapezoid. Kuma abin da yake da shi don haka na ƙwarai kuma ya sa samu da sunan? A musamman fasali na wannan adadi da dangantaka da cewa tana da ba kawai daidaita bangarorin da kusassari a tushe, amma kuma diagonally. Bugu da kari, Naira Miliyan Xari da kusassari da wani isosceles trapezoid ne daidai to 360 digiri. Amma dai ba duka! Kawai a kusa da isosceles za a iya bayyana ta a da'irar duk aka sani trapezoids. Wannan shi ne saboda cewa da Naira Miliyan Xari da m kusassari a cikin wannan adadi ne 180 digiri, kuma kawai a karkashin wannan yanayin za a iya bayyana a matsayin mai da'ira a kewayen quadrangle. Wadannan Properties na da lissafi adadi ne cewa da nisa daga saman da tushe ga aikin shirya na adawa da kololuwa a kan layi cewa yana dauke da wannan tushe zai zama daidai da midline.

Yanzu bari mu dubi yadda za a sami sasanninta na wani isosceles trapezoid. La'akari da wani bayani ga wannan matsala, bayar da cewa size daga cikin jam'iyyun da aka sani adadi.

yanke shawara

Shi ne m don nuna a fakaice da quadrangle haruffa A, B, C, D, inda BS kuma BP - mai tushe. A wani isosceles trapezoid tarnaƙi ne daidai. Muna zaton cewa su size ne daidai X kuma Y girma ne sansanonin da kuma Z (karami, kuma mafi girma, bi da bi). Domin da lissafi na kwana na bukatar ku ciyar a cikin tsawo H. A sakamakon haka ne dama-angled alwatika ABN inda AB - da hypotenuse, kuma BN kuma AN - kafafu. Lissafi da size da kafar AN: debewa daga fi girma tushe kadan, da kuma sakamakon da aka raba ta 2. Rubuta dabara: (ZY) / 2 = F. Yanzu, yin lissafi da m kwana na alwatika amfani aiki cos. Mun samu da wadannan shigarwa: cos (β) = X / F. Yanzu lissafi da kwana: β = arcos (X / F). Bugu da ari, da sanin daya kusurwa, za mu iya sanin da na biyu, to yin wannan na farko ilmin lissafi aiki: 180 - β. All kusassari suna tsare.

Akwai kuma na biyu bayani daga wannan matsala. A farkon da aka tsallake daga kusurwa a tsawo na kafa N. calculates tamanin da BN. Mun san cewa square na hypotenuse wata dama alwatika ne daidai da Naira Miliyan Xari da murabba'ai na sauran bangarorin biyu. Mun samu: BN = √ (x2 F2). Next, mun yi amfani da trigonometric aiki KU. A sakamakon haka ne: β = arctg (BN / F). A m kwana da aka samu. Next, mun ayyana wani obtuse kwana kamar yadda a cikin farko hanya.

Dũkiyar diagonals na wani isosceles trapezoid

Da farko, mun rubuta da hudu dokoki. Idan diagonal cikin wani isosceles trapezoid ne perpendicular, sa'an nan:

- tsawo daga cikin adadi ne daidai da Naira Miliyan Xari da kwasfansu, raba ta biyu.

- rufinta, kuma tsakiyar line ne daidai.

- yanki na trapezoid ne daidaita da square na tsawo (cibiyar line zuwa rabin sansanonin).

- square daga cikin diagonal na wani square ne daidai da rabin Naira Miliyan Xari biyu da square sansanonin ko midline (tsawo).

Yanzu dubi dabara fassara da diagonal wani equilateral trapezoid. Wannan yanki na bayanai za a iya zuwa kashi hudu sassa:

1. Formula diagonal tsawon hanyar da gefen.

Muna zaton cewa A ne - wani m tushe, B - Top, C - daidaita tarnaƙi, D - diagonal. A wannan yanayin, da tsawon za a iya ƙaddara kamar haka:

D = √ (C 2 + A * B).

2. Formula ga diagonal tsawon na cosine.

Muna zaton cewa A ne - wani m tushe, B - Top, C - daidaita tarnaƙi, D - diagonal, α (a cikin ƙananan tushe) da kuma β (babba tushe) - trapezoid sasanninta. Mun samu da wadannan dabara, ta hanyar da daya zai iya lissafi da tsawon na diagonal:

- D = √ (A2 + S2-2A * C * cosα).

- D = √ (A2 + S2-2A * C * cosβ).

- D = √ (B2 + S2-2V * C * cosβ).

- D = √ (B2 + S2-2V * C * cosα).

3. Formula diagonal tsawon wani isosceles trapezoid.

Muna zaton cewa A ne - wani m tushe, B - babba, D - diagonal, M - tsakiyar layin H - tsawo, P - yanki na trapezoid, α da β - da kwana tsakanin diagonals. Ƙayyade da tsawon wadannan dabarbari:

- D = √ (M2 + N2).

- D = √ (H 2 + (A + B) 2/4).

- D = √ (N (A + B) / sinα) = √ (2n / sinα) = √ (2m * N / sinα).

Domin wannan harka, da daidaitakar: sinα = sinβ.

4. Formula diagonal tsawon hanyar da tarnaƙi, kuma tsawo.

Muna zaton cewa A ne - wani m tushe, B - Top, C - tarnaƙi, D - diagonal, H - tsawo, α - kwana tare da m tushe.

Ƙayyade da tsawon wadannan dabarbari:

- D = √ (H 2 + (A-P * ctgα) 2);

- D = √ (H 2 + (B + F * ctgα) 2);

- D = √ (A2 + S2-2A * √ (C2-H2)).

Abubuwa da kaddarorin wani rectangular trapezium

Bari mu dubi abin da ake sha'awar a cikin wannan geometrical adadi. Kamar yadda muka ce, muna da rectangular trapezoid biyu dama kusassari.

Bayan da na gargajiya definition, akwai wasu. Alal misali, a rectangular trapezoid - a trapezoid a wadda gefe daya ne perpendicular da tushe. Ko siffar da ciwon a gefen kusassari. A irin wannan trapezoids tsawo ne gefen cewa shi ne perpendicular zuwa sansanonin. A tsakiyar layin - wani sashi cewa haɗu da midpoints na bangarorin biyu. Dũkiyar ce kashi ne cewa shi ne a layi daya zuwa sansanonin da kuma daidaita rabin Miliyan Xari.

Yanzu bari la'akari da asali dabarbari cewa ayyana lissafi siffofi. Don yin wannan, muna zaton cewa A da B - tushe. C (perpendicular da tushe) da kuma D - tarnaƙi na rectangular trapezium, M - tsakiyar line, α - m kwana, P - yanki.

1. A gefe perpendicular zuwa sansanonin, wani adadi daidaita da tsawo (C = N), da kuma daidai da tsawon na biyu gefen A kuma ba tare da na kwana α a wata mafi girma tushe (C = A * sinα). Haka kuma, shi ne daidaita da samfurin na tangent na m kwana α da bambanci a sansanonin: C = (A-B) * tgα.

2. A gefe D (ba perpendicular da tushe) daidaita da quotient na bambanci A da B da cosine (α) ko wani m kwana da masu zaman kansu tsawo Figures H, kuma ba tare da m kwana: A = (A-B) / cos α = C / sinα.

3. A gefe cewa shi ne perpendicular zuwa sansanonin, shi ne daidaita da square tushen daga cikin square da bambanci D - na biyu gefen - da kuma wani square tushe bambance-bambance:

C = √ (q2 (A-B) 2).

4. Side A rectangular trapezoid ne daidaita da square tushen wata square Naira Miliyan Xari da wani square gefen kuma C sansanonin lissafi siffar bambanci: D = √ (C 2 + (A-B) 2).

5. A gefe C ne, sunã daidaita da quotient na square biyu Naira Miliyan Xari da sansanin: C = P / M = 2P / (A + B).

6. A wurin da aka ayyana da samfurin M (da cibiyar layi na rectangular trapezoid) a tsawo ko a kaikaice shugabanci perpendicular zuwa sansanonin: P = M * N = M * C.

7. Position C ne quotient na sau biyu square siffar da samfurin ba tare da m kwana da kuma Naira Miliyan Xari da sansanin: C = P / M * sinα = 2P / ((A + B) * sinα).

8. Formula gefen wani rectangular trapezium ta hanyar da diagonal, da kuma kwana tsakanin su:

- sinα = sinβ.

- C = (D1 * D2 / (A + B)) * sinα = (D1 * D2 / (A + B)) * sinβ,

inda D1 da D2 - diagonal na trapezoid. α da β - da kwana tsakanin su.

9. Formula gefen ta hanyar wani kwana a ƙananan tushe da kuma wasu: A = (A-B) / cosα = C / sinα = H / sinα.

Tun da trapezoid da dama kusassari ne wani batu na trapezoid, da sauran dabarbari cewa sanin wadannan Figures, zai hadu da rectangular.

Kadarorin incircle

Idan yanayin da aka ce cewa a cikin wani rectangular trapezoid rubũtacce da'irar, sa'an nan za ka iya amfani da wadannan Properties:

- da adadin tushe ne Naira Miliyan Xari da bangarorin.

- nesa daga saman rectangular siffar da maki na tangency na rubũtacce da'irar ne ko da yaushe daidaita;

- tsawo daga cikin trapezoid ne daidai da gefe, perpendicular zuwa sansanonin, kuma shi ne daidai ga diamita daga cikin da'irar .

- da'irar cibiyar ne batu a wadda rarraba bisectors na kusassari .

- idan kaikaice gefen batu na lamba ne zuwa kashi tsawo N da M, sa'an nan da radius daga cikin da'irar ne, sunã daidaita da square tushen da samfurin na wadannan segments.

- quadrangle kafa ta maki na lamba, saman trapezoid da cibiyar da rubũtacce da'irar - shi ne mai square, wanda gefen shi ne daidaita da radius.

- yanki na adadi ne da samfurin na dalilin da samfurin na rabin-ware Naira Miliyan Xari da kwasfansu a rufinta.

similar trapeze

Wannan topic ne da amfani sosai ga karatu da kaddarorin lissafi Figures. Alal misali, diagonal tsaga cikin hudu triangles trapezoid, kuma suna dab da tushe daga cikin kamar, kuma ga bangarorin - na daidaita. Wannan bayani za a iya kira a dũkiyar triangles, wanda shi ne karye trapeze ta diagonals. A kashi na farko na wannan sanarwa da aka tabbatar ta hanyar alamar kama daga kusurwoyin nan biyu. Don tabbatar da kashi na biyu shi ne mafi alhẽri a yi amfani da hanyar kayyade a kasa.

A hujja

Yarda cewa adadi ABSD (AD kuma BC - tushen da trapezoid) ne karye diagonals HP da AC. A batu na mahada - O. Mun samu hudu triangles: AOC - a ƙananan tushe, BOS - babba tushe, Abo da sod a tarnaƙi. Triangles sod da biofeedback da na kowa tsawo a wannan lõkacin, idan segments na BO da OD ne da kwasfansu guda. Mun samu cewa bambanci na da yankunan (P) daidaita da bambanci daga wadannan segments: PBOS / PSOD = BO / ML = K. Saboda haka, PSOD = PBOS / K. Hakazalika, triangles AOB da biofeedback da na kowa tsawo. Yarda a gare su da tushe segments SB kuma OA. Mun samu PBOS / PAOB = CO / OA = K kuma PAOB = PBOS / K. Daga wannan shi ya bi cewa PSOD = PAOB.

Don ƙarfafa abu dalibai suna karfafa samu alaka tsakanin yankunan triangles samu, wanda ya karye trapeze ta diagonals, shata gaba aiki. An sani cewa triangles BOS da ADP yankunan su ne daidai, shi wajibi ne a sami fannin wani trapezoid. Tun PSOD = PAOB, sa'an nan PABSD PBOS + = PAOD + 2 * PSOD. Daga cikin kama daga triangles BOS da ANM haka cewa BO / OD = √ (PBOS / PAOD). Saboda haka, PBOS / PSOD = BO / OD = √ (PBOS / PAOD). Samun PSOD = √ (* PBOS PAOD). Sa'an nan PABSD PBOS + = PAOD + 2 * √ (PAOD PBOS *) = (+ √PBOS √PAOD) 2.

Properties kama

Ci gaba wajen samar da wannan batu, yana yiwuwa ya tabbatar, da kuma sauran ban sha'awa fasali na trapezoids. Saboda haka, tare da taimakon da kama zai iya tabbatar da dukiya da rabi, wanda ya wuce ta nufi kafa ta ratsawa na diagonals na lissafi adadi, a layi daya zuwa ga ƙasa. Domin wannan mun shirya da wadannan matsala: wajibi ne a samu tsawon RK kashi cewa ya wuce ta nufi O. Daga cikin kama daga triangles ADP da SPU bi cewa AO / OS = AD / BS. Daga cikin kama daga triangles ADP da Asb haka cewa AB / AC = PO / AD = BS / (BP + BS). Wannan ya nuna cewa BS * PO = AD / (AD + BC). Hakazalika, daga kama da triangles MLC da Abr haka cewa OK * BP = BS / (BP + BS). Wannan ya nuna cewa OC da kuma RC = RC = 2 * BS * AD / (AD + BC). Kashi wucewa ta cikin mahada batu na diagonals layi daya zuwa ga tushe da kuma a haɗa da bangarorin biyu, da rarrabawa aya ta tsãge a cikin rabin. Its tsawon - ne masu jituwa nufin na dalilin Figures.

La'akari da wadannan halaye na wani trapezoid, wanda ake ce da dukiyar da maki hudu. batu na rarrabawa da diagonals (D), na mahada na ci gaba da bangarorin (E), kazalika da tsakiyar sansanonin (T da kuma G) ko da yaushe karya a kan wannan layin. Abu ne mai sauki a tabbatar da kama hanya. A sakamakon triangles ne irin wannan BES da kuma AED, da kuma kowane ciki har da wani tsakãtsaki ce ET da DLY raba koli kwana E a daidai sassa. Saboda haka, batu E, T da kuma F ne collinear. Hakazalika, a kan wannan layin aka shirya cikin sharuddan T, yã, kuma G. Wannan haka daga kama da triangles BOS da ANM. Saboda haka zamu iya gane cewa duk hudu sharuddan - E, T, ya kuma F - zai karya a kan wani madaidaiciya line.

Amfani da irin wannan trapezoids, za a iya miƙa wa dalibai samun da tsawon da rabi (LF), wanda ya raba da siffa cikin biyu kamar. Wannan yanke dole ne a layi daya zuwa sansanonin. Tun lokacin da ya karbi trapezoid ALFD LBSF da kuma irin wannan, da BS / LF = LF / AD. Wannan ya nuna cewa LF = √ (BS * BP). Mun kammala da cewa, kashi cewa ya raba zuwa biyu trapezium kamar, yana da tsawon daidaita da lissafi nufin na tsawo daga cikin sansanonin gane.

Ka yi la'akari da abubuwan da suka dace daidai. A gininsa yana da wani ɓangaren da ya rarraba trapezoid a cikin siffa biyu. Muna ɗauka cewa sashen trapzoid na ABSD ya raba ta wani ɓangare na EH cikin nau'i biyu. An tsayi tsawo daga ramin B, wanda aka raba ta kashi EH cikin sassa biyu - B1 da B2. Muna samun: PABSD / 2 = (BS + EH) * B1 / 2 = (AD + EH) * B2 / 2 da PABSD = (BS + AD) * (B1 + B2) / 2. Gaba kuma, muna tsara tsarin wanda shine farko (BS + EH) * B1 = (AD + EH) * B2 da na biyu (BS + EH) * B1 = (BS + AD) * (B1 + B2) / 2. Saboda haka ya bi B2 / B1 = (BS + EH) / (AD + EH) da BS + EH = ((BS + AD) / 2) * (1 + B2 / B1). Mun sami cewa tsawon rabon rarraba trapezoid zuwa sassa guda biyu daidai yake da tsayin daka mai tsawo: √ ((BS2 + AD2) / 2).

Ƙarin kama da juna

Ta haka ne, mun tabbatar da cewa:

1. Rashin haɗin da ke haɗuwa a tsakiyar ɓangaren ƙananan gefe shi ne na layi da na BS kuma yana daidai da ma'anar ƙididdiga na BS da AD (tsawon ma'auni na trapezium).

2. Layin da yake wucewa ta hanyar O na haɗuwa na diagonals a layi daidai da AD da BS zai daidaita da daidaitaccen jitu na lambobin AD da BS (2 * BS * AD / (BS + AD).

3. Sashin rarraba trapezoid a cikin irin wannan yana da tsawon tsaka-tsakin geometric na BS da AD.

4. Ra'ayin da ke rarraba adadi a sassa biyu daidai yana da tsawon ma'auni na ƙananan lambobin AD da BS.

Don ƙarfafa kayan da kuma gane danganta tsakanin sassa da aka bincika, ɗalibin ya buƙaci gina su don takamaiman trapezoid. Zai iya nuna layin tsakiya da sashi wanda ke wucewa ta hanyar ma'ana O - haɗuwa da diagonals na adadi - a layi daya zuwa ga asali. Amma ina za ta uku da ta huɗu? Wannan amsar za ta jagoranci ɗalibi don gano ma'anar da ake so a tsakanin dabi'un ƙira.

Jigon da ke haɗuwa da matsakaicin sakonni na trapezoid

Yi la'akari da dukiyar da ke cikin wannan adadi. Muna ɗauka cewa sashin MN yana da layi tare da asali da rarraba diagonals a rabi. Za a kira maki da tsinkayyi na W da W. Wannan sashi zai daidaita da bambancin rabi na asali. Bari mu bincika wannan dalla-dalla. MS shine tsakiyar layin mai kwakwalwa ABC, yana daidai da BS / 2. MN ita ce tsakiyar zangon ABD, yana daidai da AD / 2. Sa'an nan kuma mu sami M, = MN-MN, kuma saboda haka, M, = A / 2-BC / 2 = (AD + BC) / 2.

Cibiyar nauyi

Bari mu dubi yadda za'a rarraba wannan nauyin don adadi na lissafi. Don wannan, wajibi ne don mika kwasfofin a cikin wasu hanyoyi. Mene ne wannan yake nufi? Dole ne a ƙara zuwa babba babba ƙananan - a kowane gefe, alal misali, zuwa dama. Kuma ƙasa ta kara ta tsawon tsawon hagu. Sa'an nan kuma haɗa su tare da diagonal. Maganin tsinkayar wannan sashi tare da tsakiyar tsakiyar adadi shi ne tsakiyar karfin trapezoid.

An rubuta kuma an bayyana trapeziums

Bari mu lissafa siffofin irin wannan adadi:

1. Za'a iya rubuta trapezoid a cikin zagaye kawai idan yana da isosceles.

2. A cikin zagaye wanda zai iya kwatanta trapezoid, idan dai jimlar tsayin dakalansu daidai yake da ƙididdigar tsaka-tsakin sassan.

Sakamakon da'irar da aka rubuta:

1. Tsawancin trapezium wanda aka kwatanta shi ne ko yaushe daidai da guda biyu.

2. A gefe na gefen gefen gefen da aka bayyana a tsakiyar tsakiyar kewaya a kusurwar dama.

Hanya na farko shine bayyane, kuma don tabbatar da na biyu an buƙatar tabbatar da cewa kusurwar SOD ɗin tsaye ne, wanda, a gaskiya ma, ba ya da wahala sosai. Amma sanin wannan dukiya zai ba mu damar amfani da alƙalan haɗin gwiwar a yayin warware matsalar.

Yanzu bari mu yi la'akari da sakamakon da aka samu na trapezoid, wanda aka rubuta a cikin da'irar. Mun sami cewa tsawo shi ne ma'anar siffar tushe na adadi: H = 2R = √ (BS * AD). Yin aiki na hanyar warware matsalar matsalolin trapezoids (ka'idar riƙe da ma'auni biyu), ɗalibin dole ne ya warware aikin da ke gaba. Muna ɗauka cewa BT shine girman wani adadi na ABSD. Dole ne a sami sassan AT da TD. Yin amfani da ma'anar da aka bayyana a sama, wannan ba zai yi wuya a yi ba.

Yanzu bari mu kwatanta yadda za mu gano radius na da'ira ta amfani da yankin trapezium wanda aka bayyana. Mun rage girman daga saman B zuwa tushe na hawan jini. Tun lokacin da aka rubuta layin a cikin trapezoid, to BS + AD = 2AB ko AB = (BS + AD) / 2. Daga alwashi ABN mun sami sinal = BN / AB = 2 * BN / (BS + AD). PABSD = (BS + AD) * BN / 2, BN = 2R. Mun sami PABSD = (BS + AD) * R, ya bi cewa R = PABSD / (BS + AD).

.

Dukkan takardu na layin tsakiya na trapezium

Yanzu lokaci ya yi zuwa zuwa kashi na ƙarshe na wannan nau'in siffar halitta. Bari mu ga yadda tsakiyar tsakiyar trapezoid (M) shine:

1. Ta hanyar asali: M = (A + B) / 2.

2. Ta hanyar tsawo, tushe da kusurwa:

• M = A-H * (ctgα + ctgβ) / 2;

• M = B + H * (ctgα + ctgβ) / 2.

3. Ta hanyar tsawo, zane-zane da kusurwa tsakanin su. Alal misali, D1 da D2 suna diagonals na trapezoid; Α, β ne kusurwa tsakanin su:

M = D1 * D2 * sinal / 2H = D1 * D2 * sinβ / 2H.

4. Ta wurin yankin da tsawo: M = P / H.