{"id":14438,"date":"2018-04-13T18:11:01","date_gmt":"2018-04-13T17:11:01","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=14438"},"modified":"2022-01-07T21:14:23","modified_gmt":"2022-01-07T21:14:23","slug":"uma-caixa-aberta","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=14438","title":{"rendered":"Uma caixa aberta"},"content":{"rendered":"<p><ul id='GTTabs_ul_14438' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_14438' class='GTTabs_curr'><a  id=\"14438_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_14438' ><a  id=\"14438_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_14438'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"14439\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=14439\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11.png\" data-orig-size=\"406,363\" 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=\"Caixa aberta\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11.png\" class=\"alignright size-medium wp-image-14439\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11-300x268.png\" alt=\"\" width=\"300\" height=\"268\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11-300x268.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11.png 406w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a>De um quadrado de cart\u00e3o, de lado <em>x<\/em> cm, foi cortado, em cada canto, um quadradinho com 2 cm de lado, como mostra a figura.<\/p>\n<ol>\n<li>Calcula o valor de <em>x<\/em>, sabendo que a figura resultante tem \u00e1rea 65 cm<sup>2<\/sup>.<\/li>\n<li>Depois de cortado o cart\u00e3o, constru\u00edmos uma caixa aberta.<br \/>\nDetermina o valor de <em>x<\/em>, de modo que o volume da caixa seja 50 cm<sup>3<\/sup>.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_14438' onClick='GTTabs_show(1,14438)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_14438'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"14439\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=14439\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11.png\" data-orig-size=\"406,363\" 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=\"Caixa aberta\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11.png\" class=\"alignright size-medium wp-image-14439\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11-300x268.png\" alt=\"\" width=\"300\" height=\"268\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11-300x268.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11.png 406w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/>Comecemos por exprimir a \u00e1rea da figura resultante em fun\u00e7\u00e3o de <em>x<\/em>:<br \/>\n\\[\\begin{array}{*{20}{l}}{{A_{FR}}}&amp; = &amp;{{A_{Cart\u00e3o}} &#8211; 4 \\times {A_{Quadradinho}}}\\\\{}&amp; = &amp;{{x^2} &#8211; 4 \\times {2^2}}\\\\{}&amp; = &amp;{{x^2} &#8211; 16}\\end{array}\\]<br \/>\nOra, como se sabe que a \u00e1rea da figura resultante \u00e9 65 cm<sup>2<\/sup>, ent\u00e3o vem:\u00a0\\[{x^2} &#8211; 16 = 65\\]<br \/>\nNo entanto, esta equa\u00e7\u00e3o possui duas solu\u00e7\u00f5es e uma delas n\u00e3o \u00e9 solu\u00e7\u00e3o do problema, como veremos de seguida:<br \/>\n\\[\\begin{array}{*{20}{l}}{{x^2} &#8211; 16 = 65}&amp; \\Leftrightarrow &amp;{{x^2} = 81}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; 9}&amp; \\vee &amp;{x = 9}\\end{array}}\\end{array}\\]<br \/>\nComo\u00a0\\(x &gt; 4\\) (<strong>Porqu\u00ea?<\/strong>), ent\u00e3o a solu\u00e7\u00e3o do problema \u00e9\u00a0\\({x = 9}\\).<\/p>\n<p><strong>Alternativa<\/strong>:<br \/>\nAo repararmos que ter\u00e1 de ser\u00a0\\(x &gt; 4\\), poder\u00edamos ter resolvido da seguinte forma:\\[\\begin{array}{*{20}{l}}{{A_{FR}} = 65}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{{x^2} &#8211; 16 = 65}&amp; \\wedge &amp;{x &gt; 4}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{{x^2} = 81}&amp; \\wedge &amp;{x &gt; 4}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{\\left( {\\begin{array}{*{20}{c}}{x = &#8211; 9}&amp; \\vee &amp;{x = 9}\\end{array}} \\right)}&amp; \\wedge &amp;{x &gt; 4}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{x = 9}\\end{array}\\]<\/p>\n<p>\u00ad<\/p>\n<\/li>\n<li>Ora, a base da caixa \u00e9 um quadrado cujo lado mede\u00a0\\(\\left( {x &#8211; 4} \\right)\\) cm e a altura da caixa \u00e9 2 cm.\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/caixa2.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"11697\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=11697\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/caixa2.png\" data-orig-size=\"523,244\" 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;}\" data-image-title=\"Caixa\" data-image-description=\"\" data-image-caption=\"&lt;p&gt;Caixa&lt;\/p&gt;\n\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/caixa2.png\" class=\"alignright wp-image-11697 size-medium\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/caixa2-300x139.png\" alt=\"\" width=\"300\" height=\"139\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/caixa2-300x139.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/caixa2-150x69.png 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/caixa2-400x186.png 400w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/caixa2.png 523w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/p>\n<p>Assim, temos:<br \/>\n\\[\\begin{array}{*{20}{l}}{{V_{Caixa}} = 50}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{{{\\left( {x &#8211; 4} \\right)}^2} \\times 2 = 50}&amp; \\wedge &amp;{x &gt; 4}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{{{\\left( {x &#8211; 4} \\right)}^2} = 25}&amp; \\wedge &amp;{x &gt; 4}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{\\left( {\\begin{array}{*{20}{c}}{x &#8211; 4 = &#8211; 5}&amp; \\vee &amp;{x &#8211; 4 = 5}\\end{array}} \\right)}&amp; \\wedge &amp;{x &gt; 4}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{\\left( {\\begin{array}{*{20}{c}}{x = &#8211; 1}&amp; \\vee &amp;{x = 9}\\end{array}} \\right)}&amp; \\wedge &amp;{x &gt; 4}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{x = 9}\\end{array}\\]<br \/>\nPortanto, para que o volume da caixa seja 50 cm<sup>3<\/sup>, ter\u00e1 de ser\u00a0\\({x = 9}\\).<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_14438' onClick='GTTabs_show(0,14438)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado De um quadrado de cart\u00e3o, de lado x cm, foi cortado, em cada canto, um quadradinho com 2 cm de lado, como mostra a figura. Calcula o valor de x, sabendo&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":14440,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,303],"tags":[426,306,109],"series":[],"class_list":["post-14438","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-9--ano","category-aplicando","category-equacoes-do-2-o-grau","tag-9-o-ano","tag-equacao-do-2-o-grau","tag-volume"],"views":2621,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-11a.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14438","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=14438"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14438\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14440"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=14438"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=14438"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=14438"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=14438"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}