{"id":13465,"date":"2018-02-04T18:37:16","date_gmt":"2018-02-04T18:37:16","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=13465"},"modified":"2022-01-17T16:24:13","modified_gmt":"2022-01-17T16:24:13","slug":"uma-circunferencia-e-dois-triangulos","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=13465","title":{"rendered":"Uma circunfer\u00eancia e dois tri\u00e2ngulos"},"content":{"rendered":"<p><ul id='GTTabs_ul_13465' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_13465' class='GTTabs_curr'><a  id=\"13465_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_13465' ><a  id=\"13465_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_13465'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"13466\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=13466\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4.png\" data-orig-size=\"350,220\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;,&quot;orientation&quot;:&quot;0&quot;}\" data-image-title=\"Circunfer\u00eancia\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4.png\" class=\"alignright size-medium wp-image-13466\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4-300x189.png\" alt=\"\" width=\"300\" height=\"189\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4-300x189.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4.png 350w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a>Na figura, est\u00e3o representados uma circunfer\u00eancia de centro no ponto O e os tri\u00e2ngulos [ABC] e [CDE].<\/p>\n<p>Sabe-se que:<\/p>\n<ul>\n<li>os pontos A, B e C pertencem \u00e0 circunfer\u00eancia;<\/li>\n<li>[BC] \u00e9 um di\u00e2metro da circunfer\u00eancia;<\/li>\n<li>o tri\u00e2ngulo [CDE] \u00e9 ret\u00e2ngulo em E;<\/li>\n<li>os tri\u00e2ngulos\u00a0[ABC] e [CDE] s\u00e3o semelhantes.<\/li>\n<\/ul>\n<p>A figura n\u00e3o est\u00e1 desenhada \u00e0 escala.<\/p>\n<ol>\n<li>Admite que a amplitude do \u00e2ngulo ACB \u00e9 igual a 36 graus.<br \/>\nQual \u00e9 a amplitude do arco AB?<\/li>\n<li>Admite que \\(\\frac{{\\overline {CD} }}{{\\overline {BC} }} = 0,5\\).<br \/>\nQual \u00e9 o valor do quociente \\(\\frac{{{\\rm{\u00c1rea}}\\;{\\rm{do}}\\;{\\rm{tri\u00e2 ngulo}}\\;\\left[ {CDE} \\right]}}{{{\\rm{\u00c1 rea}}\\;{\\rm{do}}\\;{\\rm{tri\u00e2 ngulo}}\\;\\left[ {ABC} \\right]}}\\)?<\/li>\n<li>Admite que\u00a0 \\(\\overline {AB} = 6\\) cm e \\(\\overline {AC} = 10\\) cm.<br \/>\nDetermina a \u00e1rea do c\u00edrculo de di\u00e2metro [BC].<br \/>\nApresenta o resultado em cm<sup>2<\/sup>, arredondado \u00e0s unidades. Apresenta todos os c\u00e1lculos que efetuares.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_13465' onClick='GTTabs_show(1,13465)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_13465'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ul>\n<li>\n<blockquote>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"13466\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=13466\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4.png\" data-orig-size=\"350,220\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;,&quot;orientation&quot;:&quot;0&quot;}\" data-image-title=\"Circunfer\u00eancia\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4.png\" class=\"alignright size-medium wp-image-13466\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4-300x189.png\" alt=\"\" width=\"300\" height=\"189\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4-300x189.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4.png 350w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a>os pontos A, B e C pertencem \u00e0 circunfer\u00eancia;<\/p>\n<\/blockquote>\n<\/li>\n<li>\n<blockquote>\n<p>[BC] \u00e9 um di\u00e2metro da circunfer\u00eancia;<\/p>\n<\/blockquote>\n<\/li>\n<li>\n<blockquote>\n<p>o tri\u00e2ngulo [CDE] \u00e9 ret\u00e2ngulo em E;<\/p>\n<\/blockquote>\n<\/li>\n<li>\n<blockquote>\n<p>os tri\u00e2ngulos\u00a0[ABC] e [CDE] s\u00e3o semelhantes.<\/p>\n<\/blockquote>\n<\/li>\n<\/ul>\n<ol>\n<li>A amplitude de um \u00e2ngulo inscrito \u00e9 igual a metade da amplitude do arco compreendido entr os seus lados.<br \/>\nLogo, \\(\\overparen{AB} = 2 \\times A\\widehat CB = 2 \\times 36^\\circ = 72^\\circ \\).<br \/>\n\u00ad<\/li>\n<li>Como a raz\u00e3o de semelhan\u00e7a do tri\u00e2ngulo [ABC] para o tri\u00e2ngulo [CDE] \u00e9 \\(r = \\frac{{\\overline {CD} }}{{\\overline {BC} }} = \\frac{1}{2}\\), ent\u00e3o \\(\\frac{{{\\rm{\u00c1rea}}\\;{\\rm{do}}\\;{\\rm{tri\u00e2 ngulo}}\\;\\left[ {CDE} \\right]}}{{{\\rm{\u00c1 rea}}\\;{\\rm{do}}\\;{\\rm{tri\u00e2 ngulo}}\\;\\left[ {ABC} \\right]}} = {r^2} = {\\left( {\\frac{1}{2}} \\right)^2} = \\frac{1}{4}\\).<br \/>\n\u00ad<\/li>\n<li>Aplicando o Teorema de Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo (Porqu\u00ea?) [ABC], vem:\u00a0\\(\\overline {BC} = \\sqrt {{6^2} + {{10}^2}} = \\sqrt {136} = 2\\sqrt {34} \\) cm.<br \/>\nLogo,\u00a0\\({A_\\bigcirc } = \\pi \\times {\\left( {\\sqrt {34} } \\right)^2} = 34\\pi \\approx 107\\) cm<sup>2<\/sup>.<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_13465' onClick='GTTabs_show(0,13465)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Na figura, est\u00e3o representados uma circunfer\u00eancia de centro no ponto O e os tri\u00e2ngulos [ABC] e [CDE]. Sabe-se que: os pontos A, B e C pertencem \u00e0 circunfer\u00eancia; [BC] \u00e9 um&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20505,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,278],"tags":[426,280,188],"series":[],"class_list":["post-13465","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-9--ano","category-aplicando","category-circunferencia-e-poligonos","tag-9-o-ano","tag-angulo-inscrito","tag-circunferencia"],"views":2258,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag149-4_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/13465","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=13465"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/13465\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20505"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=13465"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=13465"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=13465"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=13465"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}