{"id":23260,"date":"2022-11-20T17:04:49","date_gmt":"2022-11-20T17:04:49","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=23260"},"modified":"2022-11-24T19:07:02","modified_gmt":"2022-11-24T19:07:02","slug":"a-area-de-um-triangulo-equilatero","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=23260","title":{"rendered":"A \u00e1rea de um tri\u00e2ngulo equil\u00e1tero"},"content":{"rendered":"\n<p><ul id='GTTabs_ul_23260' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_23260' class='GTTabs_curr'><a  id=\"23260_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_23260' ><a  id=\"23260_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_23260'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag057-9.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"23261\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=23261\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag057-9.png\" data-orig-size=\"250,200\" 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=\"8_Pag057-9\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag057-9.png\" class=\"alignright wp-image-23261 size-full\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag057-9.png\" alt=\"\" width=\"250\" height=\"200\" \/><\/a>Determina um valor arredondado \u00e0s d\u00e9cimas da \u00e1rea do tri\u00e2ngulo equil\u00e1tero da figura.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_23260' onClick='GTTabs_show(1,23260)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_23260'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag057-9b.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"23267\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=23267\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag057-9b.png\" data-orig-size=\"250,200\" 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=\"8_Pag057-9b\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag057-9b.png\" class=\"alignright wp-image-23267 size-full\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag057-9b.png\" alt=\"\" width=\"250\" height=\"200\" \/><\/a>Aplicando o Teorema de Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo [BCM], temos:<\/p>\n<p>\\[\\overline {CM} = \\sqrt {{{\\overline {BC} }^2} &#8211; {{\\overline {BM} }^2}} = \\sqrt {{{4,5}^2} &#8211; {{2,25}^2}} = \\sqrt {20,25 &#8211; 5,0625} = \\sqrt {15,1875} \\]<\/p>\n<p>Portanto, a altura do tri\u00e2ngulo \u00e9, aproximadamente, \\(h\\simeq 3,9\\) cm.<\/p>\n<p>A \u00e1rea do tri\u00e2ngulo, aproximada \u00e0s d\u00e9cimas, \u00e9: \\(A=\\frac{4,5\\times \\sqrt{15,1875}}{2}\\simeq 8,8\\) cm<sup>2<\/sup>.<\/p>\n<p>\u00a0<\/p>\n<h6>Os valores exatos da altura e da \u00e1rea do tri\u00e2ngulo<\/h6>\n<p>Aplicando o Teorema de Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo [BCM], temos:<\/p>\n<p>\\[\\overline {CM} = \\sqrt {{{\\overline {BC} }^2} &#8211; {{\\overline {BM} }^2}} = \\sqrt {{{\\left( {\\frac{9}{2}} \\right)}^2} &#8211; {{\\left( {\\frac{9}{4}} \\right)}^2}} = \\sqrt {\\frac{{{9^2}}}{4} &#8211; \\frac{{{9^2}}}{{16}}} = \\sqrt {\\frac{{4 \\times {9^2}}}{{16}} &#8211; \\frac{{{9^2}}}{{16}}} = \\sqrt {3 \\times \\frac{{{9^2}}}{{16}}} = \\frac{9}{4}\\sqrt 3 \\]<\/p>\n<p>Portanto, a altura do tri\u00e2ngulo \u00e9 \\(h = \\frac{9}{4}\\sqrt 3 \\) cm.<\/p>\n<p>A \u00e1rea do tri\u00e2ngulo, em cent\u00edmetros quadrados, \u00e9:<\/p>\n<p>\\[A = \\frac{{\\frac{9}{2} \\times \\frac{9}{4}\\sqrt 3 }}{2} = \\frac{1}{2} \\times \\frac{{81}}{8}\\sqrt 3 = \\frac{{81}}{{16}}\\sqrt 3 \\]<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_23260' onClick='GTTabs_show(0,23260)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Determina um valor arredondado \u00e0s d\u00e9cimas da \u00e1rea do tri\u00e2ngulo equil\u00e1tero da figura. Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":23262,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,682],"tags":[424,67,118],"series":[],"class_list":["post-23260","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-teorema-de-pitagoras","tag-8-o-ano","tag-geometria","tag-teorema-de-pitagoras"],"views":319,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag057-9_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/23260","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=23260"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/23260\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/23262"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=23260"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=23260"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=23260"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=23260"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}