{"id":14448,"date":"2018-04-13T19:26:42","date_gmt":"2018-04-13T18:26:42","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=14448"},"modified":"2022-01-07T21:16:12","modified_gmt":"2022-01-07T21:16:12","slug":"um-jardim-retangular","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=14448","title":{"rendered":"Um jardim retangular"},"content":{"rendered":"<p><ul id='GTTabs_ul_14448' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_14448' class='GTTabs_curr'><a  id=\"14448_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_14448' ><a  id=\"14448_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_14448'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Um jardim retangular tem 6 metros de comprimento e 4 metros de largura.<br \/>\nEste jardim foi aumentado de modo a ficar com uma \u00e1rea de 143 m<sup>2<\/sup>.<br \/>\nO acrescento a cada lado foi igual.<\/p>\n<ol>\n<li>Quantos metros foram acrescentados ao comprimento e \u00e0 largura deste jardim?<\/li>\n<li>Um rolo com 45 m de rede chega para vedar o novo jardim?<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_14448' onClick='GTTabs_show(1,14448)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_14448'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-12.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"14451\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=14451\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-12.png\" data-orig-size=\"315,414\" 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=\"Jardim\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-12.png\" class=\"alignright wp-image-14451\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-12-228x300.png\" alt=\"\" width=\"300\" height=\"394\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-12-228x300.png 228w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-12.png 315w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a>Seja\u00a0\\(x &gt; 0\\), em metros, o valor acrescentado a cada lado do jardim.<br \/>\nAssim, a \u00e1rea do jardim aumentado, em fun\u00e7\u00e3o do aumento \\(x &gt; 0\\), em metros, pode ser expressa por:<br \/>\n\\[\\begin{array}{*{20}{l}}{{A_{JA}}}&amp; = &amp;{\\left( {x + 6 + x} \\right)\\left( {x + 4 + x} \\right)}\\\\{}&amp; = &amp;{\\left( {2x + 6} \\right)\\left( {2x + 4} \\right)}\\\\{}&amp; = &amp;{4{x^2} + 8x + 12x + 24}\\\\{}&amp; = &amp;{4{x^2} + 20x + 24}\\end{array}\\]<br \/>\nComo o jardim aumentado ficou com uma \u00e1rea de 143 m<sup>2<\/sup>, temos:<br \/>\n\\[\\begin{array}{*{20}{l}}{{A_{JA}} = 143}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{4{x^2} + 20x + 24 = 143}&amp; \\wedge &amp;{x &gt; 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{4{x^2} + 20x &#8211; 119 = 0}&amp; \\wedge &amp;{x &gt; 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{ &#8211; 20 \\mp \\sqrt {{{20}^2} &#8211; 4 \\times 4 \\times \\left( { &#8211; 119} \\right)} }}{{2 \\times 4}}}&amp; \\wedge &amp;{x &gt; 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{ &#8211; 20 \\mp \\sqrt {2304} }}{8}}&amp; \\wedge &amp;{x &gt; 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{ &#8211; 20 \\mp 48}}{8}}&amp; \\wedge &amp;{x &gt; 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{\\left( {\\begin{array}{*{20}{c}}{x = &#8211; 8,5}&amp; \\vee &amp;{x = 3,5}\\end{array}} \\right)}&amp; \\wedge &amp;{x &gt; 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{x = 3,5}\\end{array}\\]<br \/>\nPortanto, foram acrescentados 7 metros ao comprimento, quer \u00e0 largura do jardim.<br \/>\n\u00ad<\/li>\n<li>Calculemos, em metros, o per\u00edmetro do jardim aumentado:<br \/>\n\\[\\begin{array}{*{20}{l}}{{P_{JA}}}&amp; = &amp;{2 \\times \\left( {6 + 7} \\right) + 2 \\times \\left( {4 + 7} \\right)}\\\\{}&amp; = &amp;{26 + 22}\\\\{}&amp; = &amp;{48}\\end{array}\\]<br \/>\nN\u00e3o, um rolo com 45 m de rede n\u00e3o chega para vedar o novo jardim, pois o seu per\u00edmetro \u00e9 superior a 45 metros.<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_14448' onClick='GTTabs_show(0,14448)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Um jardim retangular tem 6 metros de comprimento e 4 metros de largura. Este jardim foi aumentado de modo a ficar com uma \u00e1rea de 143 m2. O acrescento a cada&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":14449,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,303],"tags":[426,108,306],"series":[],"class_list":["post-14448","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-area","tag-equacao-do-2-o-grau"],"views":2700,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-12a.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14448","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=14448"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14448\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14449"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=14448"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=14448"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=14448"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=14448"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}