{"id":13259,"date":"2018-01-11T18:01:00","date_gmt":"2018-01-11T18:01:00","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=13259"},"modified":"2022-01-16T23:34:40","modified_gmt":"2022-01-16T23:34:40","slug":"o-triangulo-mar-e-retangulo","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=13259","title":{"rendered":"O tri\u00e2ngulo [MAR] \u00e9 ret\u00e2ngulo"},"content":{"rendered":"<p><ul id='GTTabs_ul_13259' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_13259' class='GTTabs_curr'><a  id=\"13259_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_13259' ><a  id=\"13259_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_13259'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/9V1Pag129-5.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"13260\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=13260\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/9V1Pag129-5.png\" data-orig-size=\"245,220\" 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\/01\/9V1Pag129-5.png\" class=\"alignright size-full wp-image-13260\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/9V1Pag129-5.png\" alt=\"\" width=\"245\" height=\"220\" \/><\/a>O tri\u00e2ngulo [MAR], representado na figura, \u00e9 ret\u00e2ngulo em A e os seus tr\u00eas v\u00e9rtices pertencem \u00e0 circunfer\u00eancia.<\/p>\n<p>Sabendo que \\(\\overparen{MA} = \\overparen{QM}\\) e que \\(M\\widehat RA = 30^\\circ \\), calcula \\(Q\\widehat AR\\).<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_13259' onClick='GTTabs_show(1,13259)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_13259'>\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\/9V1Pag129-5.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"13260\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=13260\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/9V1Pag129-5.png\" data-orig-size=\"245,220\" 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\/01\/9V1Pag129-5.png\" class=\"alignright size-full wp-image-13260\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/9V1Pag129-5.png\" alt=\"\" width=\"245\" height=\"220\" \/><\/a>Como o tri\u00e2ngulo \u00e9 ret\u00e2ngulo e est\u00e1 inscrito na circunfer\u00eancia, ent\u00e3o os arcos MR s\u00e3o semicircunfer\u00eancias.<\/p>\n<p>Ora, \\(\\overparen{MA} = 2 \\times M\\widehat RA = 2 \\times 30^\\circ = 60^\\circ \\).<\/p>\n<p>Logo,\u00a0\\(Q\\widehat AR = \\frac{{\\overparen{QR}}}{2} = \\frac{{\\overparen{MR} &#8211; \\overparen{MQ}}}{2} = \\frac{{180^\\circ &#8211; 60^\\circ }}{2} = 60^\\circ \\).<\/p>\n<p>\u00ad<\/p>\n<p>Ou, se preferir:<\/p>\n<p>\\[Q\\widehat AR = \\frac{{\\overparen{QR}}}{2} = \\frac{{\\overparen{MR} &#8211; \\overparen{MQ}}}{2} = \\frac{{180^\\circ &#8211; \\overparen{MA}}}{2} = \\frac{{180^\\circ &#8211; 2 \\times M\\widehat RA}}{2} = \\frac{{180^\\circ &#8211; 2 \\times 30^\\circ }}{2} = 60^\\circ \\]<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_13259' onClick='GTTabs_show(0,13259)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado O tri\u00e2ngulo [MAR], representado na figura, \u00e9 ret\u00e2ngulo em A e os seus tr\u00eas v\u00e9rtices pertencem \u00e0 circunfer\u00eancia. Sabendo que \\(\\overparen{MA} = \\overparen{QM}\\) e que \\(M\\widehat RA = 30^\\circ \\), calcula&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20449,"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-13259","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":2220,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/9V1Pag129-5_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/13259","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=13259"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/13259\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20449"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=13259"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=13259"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=13259"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=13259"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}