Indlela yokubala indawo yephiramidi; isiseko, ngapha ngokupheleleyo?

Ukulungiselela uviwo abafundi zemathematika kufuneka nokumisa ulwazi algebra kunye nejiyometri. Ndingathanda hlanganisa zonke iinkcukacha eyaziwa, ezifana ukubala njani kummandla iphiramidi. Ngaphezu koko, ukuqala ukusuka ezantsi nakwicala lijongene de kummandla komhlaba wonke. Ukuba icala le ijongene meko icacile, njengoko oonxantathu, isiseko usoloko ezahlukeneyo.

Njani ukuba xa kummandla emazantsi iphiramidi?

Kunokuba kakhulu yiyiphi evela unxantathu olithandayo kwi n-gon. Kwaye lo kwisiseko, ngaphandle umahluko kwenani engile, abe nani ezichanekileyo okanye ezingachanekanga. Xa umdla abafundi imisebenzi on ukuhlolwa bafumana imisebenzi kuphela namanani ezichanekileyo siseko. Ngoko ke, siza kuthetha kuphela ngazo.

triangle alinganayo

Oko alinganayo. Lowo onke amaqela bayalingana yaye otyunjwe ileta "a". Kulo mzekelo, indawo isiseko zepiramidi ibalwa yi ifomula:

S = ⁽² * √3 a) / 4.

square

Ifomula ukubala indawo balo elula, kuba "a" - icala kwakhona:

Kwaye S = ^2.

n-gon rhoqo engenasizathu

Asemacaleni polygon isiqu efanayo. Ngokuba inani engile ezisetyenziswayo Latin incwadi n.

S = (n * a ²⁾ / (4 * kubatshaba- (180º / n)) .

Indlela ukungena kubalo ummandla komphezulu osecaleni ngokupheleleyo?

Ekubeni inani isiseko lichanekile, ngoko bonke ubuso iphiramidi iyalingana. Ngasinye apho unxantathu isosceles, kususela imiphetho ecaleni iyalingana. Emva koko, ukuze ubale kummandla icala iphiramidi kufuneka formula eyenziwa wamcacisela inani monomials ezifanayo. Inani amagama ixhomekeke isixa macala kwesiseko.

Ummandla unxantathu isosceles into ayinakubalwa indlela apho isiqingatha imveliso lesiseko liphindaphindwe ukuphakama. Le nokuphakama iphiramidi ebizwa apothem. nesikhundla sayo - "A". Le ndlela yokubala jikelele kummandla kumphezulu osecaleni ngolu hlobo lulandelayo:

S = ½ P * A, apho P - kujikelezo kwisiseko iphiramidi.

Kukho amaxesha xa na yaziwa kwicala lesakhelo, kodwa imiphetho ebezingalindelekanga (a) caba kunye engile njengezona (α). Emva koko uxhomekeke ukusebenzisa le fomyula ilandelayo ukubala ndawo zisecaleni iphiramidi:

S = n / 2 ukuya ² * α isono.

Task № 1

Isimo. Fumana indawo lilonke yephiramidi, ukuba nomzantsi walo unxantathu alinganayo kunye ecaleni-4 cm kwaye ixabiso √3 apothem cm.

Isigqibo. Kufuneka uqale ukubala yomjikelezo isiseko. Ukusukela ukuba oku unxantathu rhoqo, ngoko nje yaziwa, umntu ngoko nangoko ukubala indawo kumphezulu yonke osecaleni :. ½ * 12 * √3 = 6√3 ^cm2 P apothem = 3 * 4 = 12 ^cm.

Ukuze ufumane unxantathu isiseko elahliweyo lixabiso le ndawo (4 ² * √3) / 4 = 4√3 ^cm2.

Ukuze ubone indawo yonke kufuneka phinda amaxabiso amabini kubangela: 6√3 + 4√3 = 10√3 ^cm2.

Phendula. 10√3 ^cm2.

Ingxaki № 2

Isimo. Kukho iphiramidi quadrangular rhoqo. Ubude isiseko ilingana 7 mm, enyeleni osecaleni - 16 mm. Kufuneka wazi indawo yayo umphezulu.

Isigqibo. Ekubeni Ipolihedron - uxande zichanekile, xa nomzantsi walo isikwere. Ukuva indawo isiseko kunye namacala osecaleni bakwazi ukubala iphiramidi square. Ifomula isikwere ezinikwe apha ngentla. Kwaye ndiyazi bonke ubuso icala nxantathu. Ngoko ke, ungakwazi ukusebenzisa ifomula nguHeron kaThixo yokubala kwimimandla yabo.

Ukubala zokuqala ilula kwaye kungakhokelela kule nombolo: 49 mm ^2. Ukubala ixabiso yesibini kufuneka semiperimeter: (7 + 16 * 2): 2 = 19.5 mm. Ngoku sinako ukubala indawo unxantathu isosceles: √ (19,5 * (19,5-7) * (19,5-16) ²⁾ = √2985,9375 = 54.644 mm ^2. Kukho oonxantathu ezine, ngoko xa kubalwa inani lokugqibela kufuneka iphindaphindwe 4.

Wafumana: 49 + 4 * 54,644 = 267,576 ^mm2.

Phendula. 267,576 ixabiso ^olifunayo-2 mm.

Task № 3

Isimo. Xa iphiramidi rhoqo quadrangular kuyimfuneko ukubala ndawo. Yinto eyaziwayo icala kwisikwere - 6 cm kunye nokuphakama - 4 cm.

Isigqibo. Indlela elula ukusebenzisa indlela ukuya imveliso yomjikelezo kunye apothem. Ixabiso lokuqala ifumaneka nje. nzima owesibini kancinane.

Siza ukuba ukukhumbula theorem kaPythagoras kwaye cinga unxantathu tye. Oku akhiwa ukuphakama iphiramidi kunye apothem, yona hypotenuse. Umlenze yesibini isiqingatha ecaleni isikwere, njengendawo ephakamileyo Ipolihedron iwela phakathi kwayo.

apothem ubabalweyo (i hypotenuse kanxantathu ekunene) silingana √ (3 ² + 4 ²⁾ = 5 (cm).

Ngoku ke kunokwenzeka ukuba zibale ixabiso olifunayo: ½ * (4 * 6) * 5 + 6 ² = 96 (cm ^2).

Phendula. 96 cm ^2.

Ingxaki № 4

Isimo. Dana iphiramidi njalo ezinamacala amahlanu. Ezomva isiseko layo alinganayo ukuya kowama-22 mm, imiphetho osecaleni - 61 mm. Yintoni na indawo kumphezulu osecaleni kwalo emacala?

Isigqibo. Ukucamanga kulo ziyafana njengoko kuchaziwe kwi №2 umsebenzi. iphiramidi kuphela Wanikwa ukuba isikwere kwi kwisiseko, kwaye ngoku a iHexagon.

Inyathelo lokuqala ibalwa kummandla isiseko kwifomula ngentla ^{(6 * 22 2) / (} 4 * kubatshaba- (180º / 6)) = 726 / (tg30º) = 726√3 ^cm2.

Ngoku ke kufuneka ukuba ufumane isiqingatha-iperimitha unxantathu isosceles, nto leyo ubuso icala. (22 + 61 + 2) :. = 72 cm 2 ihlala ifomula nguHeron ukuba ukubala ummandla ngamnye unxantathu, uze uphinda-phinde ngo ezintandathu emhlambini kunye nalowo ubuyeleyo ngaphandle kwi siseko.

Ubalo on ifomula nguHeron esithi: √ (72 * (72-22) * ( 72-61) 2) = √435600 = 660 cm ^2. Ubalo eziya kunika indawo umphezulu osecaleni: 660 * 6 = 3960 cm ^2. Kuhleli ukubongeza phezulu ukufumana wonke umphezulu: 5217,47≈5217 cm ^2.

Phendula. Grounds - 726√3 cm ^2, umphezulu ecaleni - 3960 cm ^2, indawo yonke - 5217 cm ^2.