{"id":14432,"date":"2018-04-12T23:55:41","date_gmt":"2018-04-12T22:55:41","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=14432"},"modified":"2022-01-07T21:11:27","modified_gmt":"2022-01-07T21:11:27","slug":"um-trapezio-isosceles-2","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=14432","title":{"rendered":"Um trap\u00e9zio is\u00f3sceles"},"content":{"rendered":"<p><ul id='GTTabs_ul_14432' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_14432' class='GTTabs_curr'><a  id=\"14432_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_14432' ><a  id=\"14432_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_14432'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"14433\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=14433\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10.png\" data-orig-size=\"500,234\" 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=\"Trap\u00e9zio is\u00f3sceles\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10.png\" class=\"alignright wp-image-14433\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10-300x140.png\" alt=\"\" width=\"400\" height=\"187\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10-300x140.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10-520x245.png 520w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10.png 500w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/a>Num trap\u00e9zio is\u00f3sceles com 36 cm<sup>2<\/sup> de \u00e1rea, a base maior mede 10 cm e a base menor tem o dobro da altura.<br \/>\nQual \u00e9 o valor, arredondado \u00e0s cent\u00e9simas, do per\u00edmetro deste trap\u00e9zio? Explica a tua resposta.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_14432' onClick='GTTabs_show(1,14432)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_14432'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"14433\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=14433\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10.png\" data-orig-size=\"500,234\" 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=\"Trap\u00e9zio is\u00f3sceles\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10.png\" class=\"alignright wp-image-14433\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10-300x140.png\" alt=\"\" width=\"400\" height=\"187\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10-300x140.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10-520x245.png 520w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/2017-18-MF9P2-pag87-10.png 500w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/a>Sabe-se que:<\/p>\n<\/blockquote>\n<ul>\n<li>\n<blockquote><p>\\(\\overline {AB} = 10\\) cm<\/p><\/blockquote>\n<\/li>\n<li>\n<blockquote><p>\\(\\overline {CD} = 2 \\times \\overline {DE} \\)<\/p><\/blockquote>\n<\/li>\n<li>\n<blockquote><p>\\({A_{\\left[ {ABCD} \\right]}} = 36\\) cm<sup>2<\/sup><\/p><\/blockquote>\n<\/li>\n<\/ul>\n<p>\u00ad<\/p>\n<p>Comecemos por exprimir a \u00e1rea do trap\u00e9zio is\u00f3sceles em fun\u00e7\u00e3o de \\(\\overline {DE} \\):<\/p>\n<p>\\[{A_{\\left[ {ABCD} \\right]}} = \\frac{{\\overline {AB} + \\overline {CD} }}{2} \\times \\overline {DE} = \\frac{{10 + 2 \\times \\overline {DE} }}{2} \\times \\overline {DE} = \\left( {5 + \\overline {DE} } \\right) \\times \\overline {DE} = {\\overline {DE} ^2} + 5 \\times \\overline {DE} \\]<\/p>\n<p>Como o trap\u00e9zio tem 36 cm<sup>2<\/sup> de \u00e1rea, resulta:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{{A_{\\left[ {ABCD} \\right]}} = 36}&amp; \\Leftrightarrow &amp;{{{\\overline {DE} }^2} + 5 \\times \\overline {DE} = 36}\\end{array}\\]<\/p>\n<p>Determinemos, por isso, a solu\u00e7\u00e3o positiva da equa\u00e7\u00e3o \\({{x^2} + 5x = 36}\\):<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{{x^2} + 5x = 36}&amp; \\Leftrightarrow &amp;{{x^2} + 5x &#8211; 36 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 5 \\mp \\sqrt {25 &#8211; 4 \\times \\left( { &#8211; 36} \\right)} }}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 5 \\mp \\sqrt {169} }}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{ &#8211; 5 &#8211; 13}}{2}}&amp; \\vee &amp;{x = \\frac{{ &#8211; 5 + 13}}{2}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; 9}&amp; \\vee &amp;{x = 4}\\end{array}}\\end{array}\\]<\/p>\n<p>Assim, conclui-se que\u00a0\\(\\overline {DE} = 4\\) cm.<\/p>\n<p>Por aplica\u00e7\u00e3o do Teorema de Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo [<em>ADE<\/em>], vem:<\/p>\n<p>\\[\\overline {AD} = \\sqrt {{{\\overline {AE} }^2} + {{\\overline {DE} }^2}} = \\sqrt {{1^2} + {4^2}} = \\sqrt {17} \\]<\/p>\n<p>Resultando, desta forma, que o per\u00edmetro do trap\u00e9zio \u00e9, aproximadamente, 26,25 cm:<\/p>\n<p>\\[{P_{\\left[ {ABCD} \\right]}} = \\overline {AB} + \\overline {CD} + 2\\overline {AD} = 10 + 8 + 2\\sqrt {17} = 18 + 2\\sqrt {17} \\approx 26,25\\]<\/p><\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_14432' onClick='GTTabs_show(0,14432)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Num trap\u00e9zio is\u00f3sceles com 36 cm2 de \u00e1rea, a base maior mede 10 cm e a base menor tem o dobro da altura. Qual \u00e9 o valor, arredondado \u00e0s cent\u00e9simas, do&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":14434,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,303],"tags":[426,108,306,335],"series":[],"class_list":["post-14432","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","tag-perimetro"],"views":1850,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/04\/9V2Pag87-10a.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14432","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=14432"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14432\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14434"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=14432"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=14432"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=14432"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=14432"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}