{"id":26141,"date":"2023-05-31T08:29:23","date_gmt":"2023-05-31T07:29:23","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=26141"},"modified":"2023-06-01T10:16:25","modified_gmt":"2023-06-01T09:16:25","slug":"a-altura-de-um-triangulo-2","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=26141","title":{"rendered":"A altura de um tri\u00e2ngulo"},"content":{"rendered":"\n<p><ul id='GTTabs_ul_26141' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_26141' class='GTTabs_curr'><a  id=\"26141_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_26141' ><a  id=\"26141_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_26141'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>A express\u00e3o que permite determinar a altura (<em>h<\/em>) de um tri\u00e2ngulo, conhecendo a sua \u00e1rea (<em>A<\/em>) e o comprimento da base (<em>b<\/em>) \u00e9:<\/p>\n<p><strong>[A]<\/strong> \\(h = \\frac{{b \\times A}}{2}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[B]<\/strong> \\(h = \\frac{2}{{b \\times A}}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[C]<\/strong> \\(h = \\frac{{2 \\times A}}{b}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[D]<\/strong> \\(h = \\frac{{2b}}{A}\\)<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_26141' onClick='GTTabs_show(1,26141)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_26141'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p>A express\u00e3o que permite determinar a altura (<em>h<\/em>) de um tri\u00e2ngulo, conhecendo a sua \u00e1rea (<em>A<\/em>) e o comprimento da base (<em>b<\/em>) \u00e9:<\/p>\n<p><strong>[A]<\/strong> \\(h = \\frac{{b \\times A}}{2}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[B]<\/strong> \\(h = \\frac{2}{{b \\times A}}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[C]<\/strong> \\(h = \\frac{{2 \\times A}}{b}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[D]<\/strong> \\(h = \\frac{{2b}}{A}\\)<\/p>\n<\/blockquote>\n<p><br \/>Sabemos que \\(A = \\frac{{b \\times h}}{2}\\).<br \/>Logo, vem: \\[\\begin{array}{*{20}{l}}{A = \\frac{{b \\times h}}{2}}&amp; \\Leftrightarrow &amp;{b \\times h = 2A}\\\\{}&amp; \\Leftrightarrow &amp;{h = \\frac{{2A}}{b}}\\end{array}\\]<\/p>\n<p>Portanto, a op\u00e7\u00e3o correta \u00e9 <strong>[C]<\/strong> \\(h = \\frac{{2 \\times A}}{b}\\).<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_26141' onClick='GTTabs_show(0,26141)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado A express\u00e3o que permite determinar a altura (h) de um tri\u00e2ngulo, conhecendo a sua \u00e1rea (A) e o comprimento da base (b) \u00e9: [A] \\(h = \\frac{{b \\times A}}{2}\\)\u00a0 \u00a0 \u00a0&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":14103,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,714],"tags":[424,192],"series":[],"class_list":["post-26141","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-equacoes-literais-e-sistemas","tag-8-o-ano","tag-equacoes-literais"],"views":62,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat48.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/26141","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=26141"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/26141\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14103"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=26141"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=26141"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=26141"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=26141"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}