Kanjani ukubala indawo iphiramidi: base, ohlangothini negcwele?

Ekulungiseleleni ukuhlolwa e abafundi wezibalo kufanele nokumisa wokwazi algebra futhi geometry. Ngithanda sihlanganise lonke ulwazi eyaziwa, njengokuthi singakunqoba kanjani ukubala indawo iphiramidi. Ngaphezu kwalokho, kusukela phansi futhi uhlangothi ubhekene kwaze kwaba yilapho yonke i-surface area. Uma ohlangothini ubhekene isimo esicacile, njengoba aboncantathu, base ihlukile njalo.

Kanjani ukuba lapho endaweni kumila ocijile?

Kungaba ngempela iyiphi sibalo kusuka unxantathu ngokungenasizathu kuya n-gon. Futhi lokhu base, ngaphandle umehluko inani engeli, kungenzeka isibalo unembile noma azilungile. Esikhathini isithakazelo abafundi imisebenzi ukuhlolwa atholakala kuphela imisebenzi izibalo elungile base. Ngakho-ke, sizosebenzisa kuphela sixoxe ngabo.

unxantathu equilateral

Lokho equilateral. Omunye kubona bonke abathintekayo bayalingana futhi ebekwe incwadi "a". Kulokhu, base endaweni mbhoshongo ibalwa ifomula:

S = ⁽² * √3) / 4.

isikwele

Ifomula ukubala endaweni yalo iyona elula, 'eliyindlu "- side futhi:

Futhi S = ^2.

Ngokungenasizathu n-gon njalo

Ngesikhathi izinhlangothi ipholigoni isiqu efanayo. Ukuze inani engeli esetshenziswa Latin incwadi n.

S = (n * ²⁾ / (4 * TG (180º / n)) .

Indlela ukungena kulesi sibalo endaweni ebusweni ezilingene ngokugcwele?

Njengoba lesi sibalo base ilungile ke bonke ubuso mbhoshongo bayalingana. Ngamunye okuyinto unxantathu isosceles, kusukela emaphethelweni ohlangothini bayalingana. Khona-ke, ukuze ukubala kwendawo uhlangothi mbhoshongo kudingeka ifomula ehlanganisa isamba monomials ezifanayo. Isibalo imigomo kunqunywa inani izinhlangothi isizinda.

Endaweni unxantathu isosceles kufakwa kukhompyutha ngokuhlukanisa ifomula lapho kwesigamu umkhiqizo isizinda iyanda yi ukuphakama. Lokhu ubude mbhoshongo ngokuthi apothem. ukuqokwa - "A". Ifomula jikelele endaweni ebusweni lateral simiswe ngalendlela lelandzelako:

S = ½ P * A, lapho P - azungezwe we base mbhoshongo.

Kunezikhathi lapho kungaziwa ukuthi ohlangothini isizinda, kodwa emaphethelweni emibi (a) flat kanye engela ngesikhathi sokudlondlobala (α). Khona-ke uncika zisebenzisa le ndlela elandelayo ukubala endaweni lateral mbhoshongo:

S = n / 2 ² yesono α.

Umsebenti № 1

Isimo. Thola endaweni Imininingwane mbhoshongo, uma isizinda salo unxantathu equilateral nohlangothi lokuzibona 4 cm begodu inenani √3 apothem cm.

Isinqumo. Kufanele uqale nge sibalo base ipherimitha. Njengoba lena unxantathu njalo-ke P = 3 * 4 = 12 cm apothem Njengoba yaziwa, umuntu angabona ngokushesha ukubala indawo yonke ebusweni lateral :. ½ * 12 * √3 = 6√3 ^cm2.

Ukuze uthole base unxantathu ukubaluleka kwendawo (4 ² * √3) / 4 = 4√3 ^cm2.

Ukuze ubone ukuthi yonke indawo kudingeka Songa amanani amabili okuholela: 6√3 + 4√3 = 10√3 ^cm2.

Impendulo. 10√3 ^cm2.

Inkinga № 2

Isimo. Kukhona quadrangular mbhoshongo njalo. Ubude base ilingana 7 mm, onqenqemeni lateral - 16 mm. Udinga ukwazi surface area yayo.

Isinqumo. Kusukela polyhedron - elingunxande futhi elungile, phansi salo isikwele. Ukuzwa isizinda ndawo futhi izinhlangothi lateral akwazi ukubala izinhlayiya mbhoshongo sikwele. Le formula for isikwele kunikezwa ngenhla. Futhi ngazi yonke ohlangothini ubuso unxantathu. Ngakho-ke, ungasebenzisa ifomula Heron sika kuhlaziywa ezindaweni zabo.

Izibalo zokuqala zilula futhi kuholele kule nombolo: 49 mm ^2. Ukubala inani lesibili kudingeka semiperimeter: (7 + 16 * 2): 2 = 19.5 mm. Manje singakwazi ukubala indawo unxantathu isosceles: √ (19,5 * (19,5-7) * (19,5-16) ²⁾ = √2985,9375 = 54.644 mm ^2. Kukhona onxantathu ezine, ngakho lapho kubalwa izinombolo yokugcina kuyodingeka iphindwe 4.

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

Impendulo. 267,576 Inani wayefisa ² mm.

Umsebenti № 3

Isimo. Ngezinye mbhoshongo njalo quadrangular kuyadingeka ukubala ndawo. It is ohlangothini square eyaziwa - 6 cm futhi ukuphakama - 4 cm.

Isinqumo. Indlela elula ukusebenzisa ifomula umkhiqizo azungezwe kanye apothem. Inani lokuqala itholakala nje. Eyesibili kancane nzima.

Sizoba ukukhumbula theorem kaPythagoras futhi ucabangele unxantathu kwesokudla. Lakhiwa ukuphakama mbhoshongo futhi apothem, okungukufa hypotenuse. Umlenze wesibili uhhafu ohlangothini square, njengoba ukuphakama polyhedron awe maphakathi naso.

apothem Somusa (the hypotenuse unxantathu kwesokudla) uyalingana √ ^{(3 2 + 4 2) =} 5 ( cm).

Manje kungenzeka ukubala inani wayefisa: ½ * (4 * 6) * 5 + 6 2 = 96 ( cm ^2).

Impendulo. 96 cm ^2.

Inkinga № 4

Isimo. Dana njalo mbhoshongo olunezinhlangothi. La maqembu kwesisekelo salo elilingana 22 mm, emaphethelweni lateral - 61 mm. Iyini indawo ebusweni lateral zalesi polyhedron?

Isinqumo. Ukucabanga kuwo ziyefana ezichazwe №2 msebenzi. Kuphela mbhoshongo wanikezwa lapho isikwele phansi, futhi manje kuba iheksagoni.

Isinyathelo sokuqala ibalwa base indawo ifomula ngenhla ^{(6 * 22 2) / (} 4 * TG (180º / 6)) = 726 / (tg30º) = 726√3 ^cm2.

Manje udinga ukuthola nengxenye-azungezwe unxantathu isosceles, okuyinto ubuso ohlangothini. (22 + 61 * 2) :. = 72 cm 2 uhlala lwefomula Heron sika ukubala endaweni ngayinye unxantathu, bese kuphinde by eziyisithupha emhlambini futhi lowo kwenzakala base.

Izibalo lwefomula Heron sika: √ (72 * (72-22) * ( 72-61) 2) = √435600 = 660 cm ^2. Izibalo ezonikeza lateral surface area: 660 * 6 = 3960 cm ^2. It uhlala ukuze ungeze zayiqeda ukuthola lonke ebusweni: 5217,47≈5217 cm ^2.

Impendulo. Izizathu - 726√3 cm ^2, ebusweni uhlangothi - 3960 cm ^2, yonke indawo - 5217 cm ^2.