{"id":134524,"date":"2021-11-22T12:54:33","date_gmt":"2021-11-22T14:54:33","guid":{"rendered":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/?post_type=eaja&#038;p=134524"},"modified":"2021-12-22T10:28:50","modified_gmt":"2021-12-22T12:28:50","slug":"matematica-comprimento-da-circunferencia-e-area-do-circulo","status":"publish","type":"eaja","link":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/eaja\/matematica-comprimento-da-circunferencia-e-area-do-circulo\/","title":{"rendered":"Matem\u00e1tica &#8211; Comprimento da Circunfer\u00eancia e \u00c1rea do C\u00edrculo"},"content":{"rendered":"\n<p class=\"has-black-color has-vivid-red-background-color has-text-color has-background\" style=\"font-size:25px\">Ol\u00e1! Esta aula de&nbsp;<strong>Matem\u00e1tica&nbsp;<\/strong>\u00e9 destinada a estudantes da<strong>&nbsp;8\u00aa S\u00e9rie<\/strong>&nbsp;da Eaja. <\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-full\"><img fetchpriority=\"high\" decoding=\"async\" width=\"605\" height=\"605\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/11\/image12-e1637590209243.png\" alt=\"\" class=\"wp-image-134525\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/11\/image12-e1637590209243.png 605w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/11\/image12-e1637590209243-300x300.png 300w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/11\/image12-e1637590209243-150x150.png 150w\" sizes=\"(max-width: 605px) 100vw, 605px\" \/><figcaption> https:\/\/pixabay.com\/pt\/vectors\/globo-mundo-terra-m%c3%a3os-pessoas-304586\/ <\/figcaption><\/figure><\/div>\n\n\n\n<p class=\"has-black-color has-cyan-bluish-gray-background-color has-text-color has-background\" style=\"font-size:25px\">Nesta aula, iremos resolver situa\u00e7\u00f5es-problema envolvendo o c\u00e1lculo das medidas do comprimento da circunfer\u00eancia e a \u00e1rea do c\u00edrculo. <\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-pale-cyan-blue-background-color has-text-color has-background\" style=\"font-size:25px\">Assista \u00e0 videoaula do professor H\u00e9lio sobre essa tem\u00e1tica.<\/p>\n\n\n\n<figure class=\"wp-block-embed aligncenter is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<p class=\"responsive-video-wrap clr\"><iframe title=\"Comprimento da Circunfer\u00eancia e \u00c1rea do C\u00edrculo | Matem\u00e1tica - aula 12 | 8\u00aa s\u00e9rie - Eaja\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/mjTInRGb9GE?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<\/div><figcaption>Comprimento da Circunfer\u00eancia e \u00c1rea do C\u00edrculo | Matem\u00e1tica &#8211; aula 12 | 8\u00aa s\u00e9rie &#8211; Eaja<\/figcaption><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-black-color has-vivid-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Circunfer\u00eancia (Defini\u00e7\u00e3o):<\/strong> \u00e9 a figura geom\u00e9trica formada por todos os pontos do plano que distam igualmente de um ponto fixo fixo desse plano. Esse ponto \u00e9 denominado de <strong>CENTRO DA CIRCUNFER\u00caNCIA<\/strong>. Veja a figura.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/C0Wl-r2PS2lC8UDeUhPE5IMRcEhhhOYCTnT2RfiZB6CFAvZPsD-OY2vFqPOBw2JQzttnHX71aLiW0t-JME8_KWx5GRC_rw91v3Pk5qKPZUS0mcPKU0eI9seba-5MLECz3KoUUwEc6Uc\" alt=\"\" width=\"269\" height=\"246\"\/><figcaption>Imagem dispon\u00edvel em: PNLD Giovanni J\u00fanior, Jos\u00e9 Ruy &#8211; A conquista da matem\u00e1tica: 9o ano p. 127.<\/figcaption><\/figure><\/div>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Essa dist\u00e2ncia, constante, \u00e9 chamada de <strong>raio da circunfer\u00eancia<\/strong> (r).<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-light-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Elementos da Circunfer\u00eancia<\/strong><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/nOel0aFc18TDLdGeqvDsu6kSLhk-PRfRgt9vtyxMHhj74evixdzyfAgEanuiCgMBEe-ZSy7G1K1GXVa4ujuODEPbRqpm7Fj_A-L04NOIl_3STp9fkMq9BpxUymcI4ouw9QykW--qOQY\" alt=\"\"\/><figcaption>Imagem dispon\u00edvel em: PNLD Giovanni J\u00fanior, Jos\u00e9 Ruy &#8211; A conquista da matem\u00e1tica: 9o ano p. 127.<\/figcaption><\/figure><\/div>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Podemos notar que o di\u00e2mentro \u00e9 igual ao dobro do raio, isso implica que o raio \u00e9 a metada de di\u00e2mentro.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/BuQlSP2qwEHLovr9A46x_gF9NZyRPbIGu_exDmLa0iwJH6HnS_T0x9T0IlMs4yYYVNmURcsAJ9d4Ktx31OAs5y5Fcvi0vwoUM6IxyGMHq2j9phqLP2IBLyR6ip1PccRdbeeqKdO7x_s\" alt=\"\" width=\"296\" height=\"78\"\/><\/figure><\/div>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/11\/O-valor-do-pi-1-e1637592440180.jpg\" alt=\"\" class=\"wp-image-134527\" width=\"252\" height=\"55\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/11\/O-valor-do-pi-1-e1637592440180.jpg 640w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/11\/O-valor-do-pi-1-e1637592440180-300x66.jpg 300w\" sizes=\"(max-width: 252px) 100vw, 252px\" \/><\/figure><\/div>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">O pi \u00e9 um n\u00famero irracional cujo valor \u00e9 3,14159265358979323846\u2026, uma sequ\u00eancia infinita de d\u00edgitos.<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">O pi resulta da divis\u00e3o entre o per\u00edmetro da circunfer\u00eancia e o seu di\u00e2metro.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/wW2em1HKPSK7ROO3tAru-cmjuQ0URifFM6VtSZEpvgzJ9VOaqUdJFUWJHJQEmH1JTxwO_LpkIUq5WtVf_xs_QJr3LBgMZ_HBDaBeEYlg6KpZuQ4lqVaFJrZZRoEPbMPS3LcpHHL2C_k\" alt=\"\" width=\"232\" height=\"71\"\/><\/figure><\/div>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">O Per\u00edmetro \u00e9 a medida de toda a volta de um c\u00edrculo, tamb\u00e9m chamada de COMPRIMENTO DA CIRCUNFER\u00caNCIA (C). Logo teremos:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/PCMSWzhTSFhv6CCWEdTw27jbMDHN4x-MvOueJDuRa3BHQNVBxaz2kv4X0s7BQcEsiolosFHvbcVzDSsuTw0fvOMB5moQSEsWl6tKP4JVR4K1ha8dfb4nAqITdEMyCy8B9OhZNBk7eso\" alt=\"\" width=\"588\" height=\"63\"\/><\/figure><\/div>\n\n\n\n<p class=\"has-text-align-center has-black-color has-cyan-bluish-gray-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Problemas Resolvidos<\/strong><\/p>\n\n\n\n<ol class=\"has-black-color has-text-color wp-block-list\" style=\"font-size:25px\"><li>Ao percorrer uma dist\u00e2ncia de 6 280 m, H\u00e9lio d\u00e1 20 voltas completas em uma pista circular. Qual \u00e9 o comprimento do raio da pista? (Use: p = 3,14.).<\/li><\/ol>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resolu\u00e7\u00e3o<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Se 20 voltas equivalem a 6280 metros, ent\u00e3o 1 volta d\u00e1 6280:20=314 metros. Utilizando a express\u00e3o do comprimento teremos:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/LeXsCRcP61wdY6r2NIn8I-3f5QyYp3vy8Lvw1x8KV-i_m1skENSZzLnt3fVXrB2Bw5v61QCoEtnlZFyUIMNr6DY7KEt0tKOkFQilmTOwUMEbScQDKMGNktpARfNu8fwG3obqhrki13w\" alt=\"\" width=\"368\" height=\"76\"\/><\/figure><\/div>\n\n\n\n<ol start=\"2\" class=\"has-black-color has-text-color wp-block-list\" style=\"font-size:25px\"><li>(Enem-2014) Um homem, determinado a melhorar sua sa\u00fade, resolveu andar diariamente numa pra\u00e7a circular que h\u00e1 em frente \u00e0 sua casa. Todos os dias ele d\u00e1 exatamente 15 voltas em torno da pra\u00e7a, que tem 50 m de raio. Use 3 como aproxima\u00e7\u00e3o para pi. Qual \u00e9 a dist\u00e2ncia percorrida por esse homem em sua caminhada di\u00e1ria?<\/li><\/ol>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resolu\u00e7\u00e3o<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Calculando a dist\u00e2ncia percorrida ao dar 1 volta:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/M1DZ_cN7I_-has2Duu5ehwsSMAGh6R3znrB5QD_BsiOBFVSxy1YXThTnncmNfBQD5CE3xsF8l70eRfJrKJ-GNDgko2vy5XhKPoXBEFYfmk5MgQVcFPwnhQ9GG-ukp0VwzGb0j1veKYo\" alt=\"\" width=\"394\" height=\"41\"\/><\/figure><\/div>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Se 1 volta corresponde a 300 metros, 15 voltas corresponder\u00e3o a 300.15=4500 metros o que equivale a 4,5km<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Portanto a dist\u00e2ncia percorrida por esse homem \u00e9 igual a 4,5km.<\/p>\n\n\n\n<p class=\"has-black-color has-vivid-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>C\u00edrculo (Defini\u00e7\u00e3o):<\/strong> tamb\u00e9m chamado de disco, \u00e9 a reuni\u00e3o dos pontos de uma circunfer\u00eancia e seus pontos internos. A diferen\u00e7a fundamental entre c\u00edrculo e circunfer\u00eancia \u00e9 que c\u00edrculo \u00e9 toda a regi\u00e3o interna da circunfer\u00eancia.<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-light-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>\u00c1rea do C\u00edrculo<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Considerando que o c\u00edrculo surge \u00e0 medida que o n\u00famero de lados de um pol\u00edgono regular (medida dos lados iguais) vai aumentando, a \u00e1rea do c\u00edrculo \u00e9 igual a \u00e1rea desse pol\u00edgono regular.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/mz6y68eTv1E4Dg8OmW5bBTI6njqs19RZf6nra8AMkSSJywgQTqcbpQquXdcqi6z0muiExuAP8KYrNNSAXJUEjVaM-xKnJ7G8XfWgUMIA7wX13EsJmAT4cuZ59YiXwnj2_Z3JCHs-m5Y\" alt=\"\" width=\"480\" height=\"102\"\/><figcaption>Imagem dispon\u00edvel em: PNLD Giovanni J\u00fanior, Jos\u00e9 Ruy &#8211; A conquista da matem\u00e1tica: 9o ano p. 232.<\/figcaption><\/figure><\/div>\n\n\n\n<p style=\"font-size:25px\">A \u00e1rea do pol\u00edgono regular \u00e9 dada pelo produto entre o semiper\u00edmetro e o ap\u00f3tema. Para o c\u00edrculo, o semiper\u00edmetro \u00e9 dado pela metade do comprimento da circunfer\u00eancia e o ap\u00f3tema \u00e9 igual ao raio do c\u00edrculo. Logo teremos<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/hDJ4jJj9YugwiGbmvUKuM2564oVG-DE-1bMDZ1oDf29W0gIf12JsXs7GmmNHYboTmA_OHJD7r2sfVkkxfTVvRz72T5t8e9Jmv7n2UM93UONwKoFmLs_jWtVyxTs_tPrfcPzAzrSNSxw\" alt=\"\" width=\"315\" height=\"40\"\/><\/figure><\/div>\n\n\n\n<p class=\"has-text-align-center has-black-color has-cyan-bluish-gray-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Problemas Resolvidos<\/strong><\/p>\n\n\n\n<ol class=\"has-black-color has-text-color wp-block-list\" style=\"font-size:25px\"><li>Uma folha de papel\u00e3o tem a forma circular de raio 21 cm. Qual \u00e9, em cm<sup>2<\/sup>, a \u00e1rea ocupada por essa folha? (Usar: p = 3,14)<\/li><\/ol>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resolu\u00e7\u00e3o<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Utilizando a f\u00f3rmula da \u00e1rea do c\u00edrculo, teremos:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/IBJ8vmABlWeuz4D_RNra66YGMKtAeCLLL4tzxQERv6eruvqO-r5OMVf8lWq3v629L1EZEGTdrDLABXF-HKPOOM44O0xxOQkf2Ga7eMvzL4j4mpMgXy2AV32j7Z22CWjloxnfa2qKTr8\" alt=\"\" width=\"502\" height=\"34\"\/><\/figure><\/div>\n\n\n\n<ol start=\"2\" class=\"has-black-color has-text-color wp-block-list\" style=\"font-size:25px\"><li>Um jardineiro cultiva suas plantas em um canteiro cuja forma \u00e9 a da figura a seguir, em que uma parte \u00e9 uma semicircunfer\u00eancia. Para cobrir todo o canteiro, ele calculou que precisaria comprar uma lona com 170 m\u00b2 de \u00e1rea. Voc\u00ea pode afirmar que a \u00e1rea da lona \u00e9 suficiente para cobrir esse canteiro? Use pi = 3,14<\/li><\/ol>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/dJBEe4MswUiGJJxW87KYA7iUdf21jGAEywN-i6PRFeomILiq4iufvQMLAAhfG5BfZrMgXVjHyTMPXlnW1u8-v2Iosfr4x9NqqQkr15hqLTRYtiM4Z79r1yIKPwGoR7bqZQEeHJGFokw\" alt=\"\" width=\"250\" height=\"321\"\/><figcaption>Imagem dispon\u00edvel em: PNLD Giovanni J\u00fanior, Jos\u00e9 Ruy &#8211; A conquista da matem\u00e1tica: 9o ano p. 233.<\/figcaption><\/figure><\/div>\n\n\n\n<p class=\"has-text-align-left has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resolu\u00e7\u00e3o&nbsp;<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">O canteiro \u00e9 formado por um ret\u00e2ngulo e um semic\u00edrculo, vamos calcular a \u00e1rea de cada um e depois som\u00e1-las.<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">\u00c1rea do ret\u00e2ngulo = 10 . 10 = 100 m<sup>2<\/sup><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">\u00c1rea do semic\u00edrculo = (3,14. 5<sup>2<\/sup>):2 = (3,14 . 25);2 = 78,5 : 2 = 39,25 m<sup>2<\/sup><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">\u00c1rea do canteiro = 100 + 39,25 = 139,25m<sup>2<\/sup><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Como a \u00e1rea do canteiro foi menor do que a \u00e1rea da lona, podemos afirmar que a \u00e1rea da lona \u00e9 suficiente para cobrir o canteiro.<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-pale-cyan-blue-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Problemas Propostos<\/strong><\/p>\n\n\n\n<ol class=\"has-black-color has-text-color wp-block-list\" style=\"font-size:25px\"><li>Considerando que uma pizza tradicional grande possui 35 cm de raio e uma pizza tradicional pequena apresenta 25 cm, determine a diferen\u00e7a entre a \u00e1rea das duas pizzas.<\/li><li>Determine a medida do raio de uma pra\u00e7a circular que possui 9420 m de comprimento (Use \u03c0 = 3,14.).<\/li><li><strong>(UEM-PR)&nbsp;<\/strong>Uma pista de atletismo tem a forma circular e seu di\u00e2metro mede 80 m. Um atleta treinando nessa pista deseja correr 10 km diariamente. Determine o n\u00famero m\u00ednimo de voltas completas que ele deve dar nessa pista a cada dia.<\/li><li><strong>(UESPI)&nbsp;<\/strong>Um trabalhador gasta 3 horas para limpar um terreno circular de 6 metros de raio. Se o terreno tivesse 12 metros de raio, quanto tempo o trabalhador gastaria para limpar tal terreno?<\/li><li>Deseja-se pregar uma fita decorativa ao redor da tampa de um pote redondo. Se o di\u00e2metro da tampa mede 12 cm, qual o comprimento m\u00ednimo que a fita deve ter para dar a volta completa na tampa?<\/li><li>A roda de um \u00f4nibus tem 90 cm de raio. Que dist\u00e2ncia o \u00f4nibus ter\u00e1 percorrido quando a roda der 120 voltas?&nbsp;<\/li><\/ol>\n\n\n\n<p class=\"has-black-color has-cyan-bluish-gray-background-color has-text-color has-background\" style=\"font-size:25px\">Ficamos por aqui, n\u00e3o esque\u00e7am de acessar os v\u00eddeos do canal do Prof. H\u00e9lio para aprender um pouco mais. Link: <a href=\"https:\/\/www.youtube.com\/channel\/UCKmf3-WVHXY61WgWrz-OO4Q\">(4) Professor Helio Roberto da Rocha &#8211; YouTube<\/a>&nbsp;<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<figure class=\"wp-block-table\"><table><tbody><tr><td><br><strong>Objetivos de Aprendizagem e Desenvolvimento:<\/strong><\/td><td>(EAJAMA0822) Resolver situa\u00e7\u00f5es-problema envolvendo o c\u00e1lculo das medidas do comprimento da circunfer\u00eancia e \u00e1rea do c\u00edrculo.<\/td><\/tr><tr><td><strong>Refer\u00eancias<\/strong>:<\/td><td>GIOVANNI J\u00daNIOR, Jos\u00e9 Ruy &#8211; A conquista da matem\u00e1tica: 9o ano: ensino fundamental: anos finais \/ Jos\u00e9 Ruy Giovanni J\u00fanior, Benedicto Castrucci. \u2014 4. ed. \u2014 S\u00e3o Paulo: FTD, 2018.&nbsp;SOUZA, Joamir Roberto de: Matem\u00e1tica realidade &amp; tecnologia: 9\u00ba ano: ensino fundamental: anos finais \/ Joamir Roberto de Souza. \u2013 1. ed. \u2013 S\u00e3o Paulo: FTD, 2018.<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"has-cyan-bluish-gray-background-color has-background\" style=\"font-size:25px\"> Professor, essa aula segue a Matriz Estruturante para a Eaja 2021. Foi elaborada no ano de 2020, com a suspens\u00e3o das aulas presenciais devido a pandemia da Covid-19 e segue as orienta\u00e7\u00f5es de flexibiliza\u00e7\u00e3o curricular para o bi\u00eanio 2020\/2021 (Of\u00edcio Circular 149\/2020 Dirped). <\/p>\n","protected":false},"author":42,"featured_media":134525,"template":"","meta":{"_acf_changed":false,"ocean_post_layout":"","ocean_both_sidebars_style":"","ocean_both_sidebars_content_width":0,"ocean_both_sidebars_sidebars_width":0,"ocean_sidebar":"","ocean_second_sidebar":"","ocean_disable_margins":"enable","ocean_add_body_class":"","ocean_shortcode_before_top_bar":"","ocean_shortcode_after_top_bar":"","ocean_shortcode_before_header":"","ocean_shortcode_after_header":"","ocean_has_shortcode":"","ocean_shortcode_after_title":"","ocean_shortcode_before_footer_widgets":"","ocean_shortcode_after_footer_widgets":"","ocean_shortcode_before_footer_bottom":"","ocean_shortcode_after_footer_bottom":"","ocean_display_top_bar":"default","ocean_display_header":"default","ocean_header_style":"","ocean_center_header_left_menu":"","ocean_custom_header_template":"","ocean_custom_logo":0,"ocean_custom_retina_logo":0,"ocean_custom_logo_max_width":0,"ocean_custom_logo_tablet_max_width":0,"ocean_custom_logo_mobile_max_width":0,"ocean_custom_logo_max_height":0,"ocean_custom_logo_tablet_max_height":0,"ocean_custom_logo_mobile_max_height":0,"ocean_header_custom_menu":"","ocean_menu_typo_font_family":"","ocean_menu_typo_font_subset":"","ocean_menu_typo_font_size":0,"ocean_menu_typo_font_size_tablet":0,"ocean_menu_typo_font_size_mobile":0,"ocean_menu_typo_font_size_unit":"px","ocean_menu_typo_font_weight":"","ocean_menu_typo_font_weight_tablet":"","ocean_menu_typo_font_weight_mobile":"","ocean_menu_typo_transform":"","ocean_menu_typo_transform_tablet":"","ocean_menu_typo_transform_mobile":"","ocean_menu_typo_line_height":0,"ocean_menu_typo_line_height_tablet":0,"ocean_menu_typo_line_height_mobile":0,"ocean_menu_typo_line_height_unit":"","ocean_menu_typo_spacing":0,"ocean_menu_typo_spacing_tablet":0,"ocean_menu_typo_spacing_mobile":0,"ocean_menu_typo_spacing_unit":"","ocean_menu_link_color":"","ocean_menu_link_color_hover":"","ocean_menu_link_color_active":"","ocean_menu_link_background":"","ocean_menu_link_hover_background":"","ocean_menu_link_active_background":"","ocean_menu_social_links_bg":"","ocean_menu_social_hover_links_bg":"","ocean_menu_social_links_color":"","ocean_menu_social_hover_links_color":"","ocean_disable_title":"default","ocean_disable_heading":"default","ocean_post_title":"","ocean_post_subheading":"","ocean_post_title_style":"","ocean_post_title_background_color":"","ocean_post_title_background":0,"ocean_post_title_bg_image_position":"","ocean_post_title_bg_image_attachment":"","ocean_post_title_bg_image_repeat":"","ocean_post_title_bg_image_size":"","ocean_post_title_height":0,"ocean_post_title_bg_overlay":0.5,"ocean_post_title_bg_overlay_color":"","ocean_disable_breadcrumbs":"default","ocean_breadcrumbs_color":"","ocean_breadcrumbs_separator_color":"","ocean_breadcrumbs_links_color":"","ocean_breadcrumbs_links_hover_color":"","ocean_display_footer_widgets":"default","ocean_display_footer_bottom":"default","ocean_custom_footer_template":""},"eaja_categoria":[69],"serie":[100],"eaja_componente":[78],"class_list":["post-134524","eaja","type-eaja","status-publish","has-post-thumbnail","hentry","eaja_categoria-2o-segmento-7a-e-8a-serie","serie-8a-serie","eaja_componente-matematica","entry","has-media"],"acf":[],"_links":{"self":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/eaja\/134524","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/eaja"}],"about":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/types\/eaja"}],"author":[{"embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/users\/42"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/media\/134525"}],"wp:attachment":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/media?parent=134524"}],"wp:term":[{"taxonomy":"eaja_categoria","embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/eaja_categoria?post=134524"},{"taxonomy":"serie","embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/serie?post=134524"},{"taxonomy":"eaja_componente","embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/eaja_componente?post=134524"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}