{"id":13208,"date":"2018-01-10T11:40:39","date_gmt":"2018-01-10T11:40:39","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=13208"},"modified":"2022-01-16T22:33:35","modified_gmt":"2022-01-16T22:33:35","slug":"observa-a-figura-2","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=13208","title":{"rendered":"Observa a figura"},"content":{"rendered":"<p><ul id='GTTabs_ul_13208' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_13208' class='GTTabs_curr'><a  id=\"13208_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_13208' ><a  id=\"13208_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_13208'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/segcirculo.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"13209\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=13209\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/segcirculo.png\" data-orig-size=\"255,270\" 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=\"Segmento de c\u00edrculo\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/segcirculo.png\" class=\"alignright wp-image-13209\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/segcirculo.png\" alt=\"\" width=\"220\" height=\"233\" \/><\/a>Determina a \u00e1rea do segmento de c\u00edrculo menor determinado pela corda [AB].<\/p>\n<p>Apresenta o valor da \u00e1rea arredondado \u00e0s unidades.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_13208' onClick='GTTabs_show(1,13208)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_13208'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/segcirculo.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"13209\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=13209\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/segcirculo.png\" data-orig-size=\"255,270\" 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=\"Segmento de c\u00edrculo\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/segcirculo.png\" class=\"alignright wp-image-13209\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/segcirculo.png\" alt=\"\" width=\"220\" height=\"233\" \/><\/a>De acordo com as indica\u00e7\u00f5es da figura, conclui-se que o tri\u00e2ngulo [ABO] \u00e9 equil\u00e1tero e, consequentemente, \u00e9 equi\u00e2ngulo. Logo,\u00a0\\(A\\widehat OB = 60^\\circ \\).<\/p>\n<p>Determinemos agora a altura do tri\u00e2ngulo equil\u00e1tero [ABO].<br \/>\nSendo M o ponto m\u00e9dio do segmento de reta [AB] e por aplica\u00e7\u00e3o do Teorema de Pit\u00e1goras, vem:<\/p>\n<p>\\[\\overline {OM} = \\sqrt {{{\\overline {OA} }^2} &#8211; {{\\overline {AM} }^2}} = \\sqrt {{{10}^2} &#8211; {5^2}} = \\sqrt {75} = 5\\sqrt 3 \\]<\/p>\n<p>A \u00e1rea do setor circular correspondente ao arco AB \u00e9 dado por:<\/p>\n<p>\\[{A_S} = \\frac{{60^\\circ }}{{360^\\circ }} \\times \\pi \\times {10^2} = \\frac{{50\\pi }}{3}\\]<\/p>\n<p>Assim, obtemos o seguinte valor (em cm<sup>2<\/sup>) para a \u00e1rea do segmento de c\u00edrculo menor determinado pela corda [AB]:<\/p>\n<p>\\[{A_{SegC\u00edrculo}} = {A_S} &#8211; {A_{\\left[ {ABO} \\right]}} = \\frac{{50\\pi }}{3} &#8211; \\frac{{10 \\times 5\\sqrt 3 }}{2} = \\frac{{50\\pi }}{3} &#8211; 25\\sqrt 3 \\approx 9\\]<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_13208' onClick='GTTabs_show(0,13208)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Determina a \u00e1rea do segmento de c\u00edrculo menor determinado pela corda [AB]. Apresenta o valor da \u00e1rea arredondado \u00e0s unidades. Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":20435,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,278],"tags":[426,457,458,284],"series":[],"class_list":["post-13208","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-arco-de-circunferencia","tag-segmento-de-circulo","tag-setor-circular"],"views":2895,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/9V1Pag119-8_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/13208","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=13208"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/13208\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20435"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=13208"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=13208"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=13208"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=13208"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}