{"id":12906,"date":"2017-10-31T00:06:33","date_gmt":"2017-10-31T00:06:33","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=12906"},"modified":"2022-01-16T00:35:38","modified_gmt":"2022-01-16T00:35:38","slug":"um-trapezio-retangulo","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=12906","title":{"rendered":"Um trap\u00e9zio ret\u00e2ngulo"},"content":{"rendered":"<p><ul id='GTTabs_ul_12906' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_12906' class='GTTabs_curr'><a  id=\"12906_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_12906' ><a  id=\"12906_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_12906'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"12907\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=12907\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio.png\" data-orig-size=\"338,308\" 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 ret\u00e2ngulo\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio.png\" class=\"alignright wp-image-12907\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio.png\" alt=\"\" width=\"240\" height=\"219\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio.png 338w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio-300x273.png 300w\" sizes=\"auto, (max-width: 240px) 100vw, 240px\" \/><\/a>O trap\u00e9zio ret\u00e2ngulo da figura tem 39 cm<sup>2<\/sup> de \u00e1rea.<\/p>\n<p>Determina, com erro inferior a 0,1, dois valores aproximados, um por defeito e outro por excesso, da medida do per\u00edmetro do trap\u00e9zio.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_12906' onClick='GTTabs_show(1,12906)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_12906'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"12907\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=12907\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio.png\" data-orig-size=\"338,308\" 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 ret\u00e2ngulo\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio.png\" class=\"alignright wp-image-12907\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio.png\" alt=\"\" width=\"240\" height=\"219\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio.png 338w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Trapezio-300x273.png 300w\" sizes=\"auto, (max-width: 240px) 100vw, 240px\" \/><\/a>Comecemos por determinar a altura do trap\u00e9zio, em cent\u00edmetros:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{{A_{Trap\u00e9zio}} = 39}&amp; \\Leftrightarrow &amp;{\\frac{{9 + 4}}{2} \\times h = 39}\\\\{}&amp; \\Leftrightarrow &amp;{h = 6}\\end{array}\\]<\/p>\n<p>Determinemos agora o comprimento, em cent\u00edmetros, do quarto lado do trap\u00e9zio, utilizando o Teorema de Pit\u00e1goras:<\/p>\n<p>\\[x = \\sqrt {{{\\left( {9 &#8211; 4} \\right)}^2} + {6^2}} = \\sqrt {25 + 36} = \\sqrt {61} \\]<\/p>\n<p>Enquadremos agora \\(\\sqrt {61} \\) com erro inferior a \\(0,1\\):<\/p>\n<p>\\[\\begin{array}{*{20}{c}}{{{78}^2} &lt; {{10}^2} \\times 61 &lt; {{79}^2}}\\\\{{{\\left( {\\frac{{78}}{{10}}} \\right)}^2} &lt; 61 &lt; {{\\left( {\\frac{{79}}{{10}}} \\right)}^2}}\\\\{7,8 &lt; \\sqrt {52} &lt; 7,9}\\end{array}\\]<\/p>\n<p>Portanto, o per\u00edmetro do trap\u00e9zio tem o seguinte enquadramento: \\(26,8\\;cm &lt; {P_{Trap\u00e9zio}} &lt; 26,9\\;cm\\).<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_12906' onClick='GTTabs_show(0,12906)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado O trap\u00e9zio ret\u00e2ngulo da figura tem 39 cm2 de \u00e1rea. Determina, com erro inferior a 0,1, dois valores aproximados, um por defeito e outro por excesso, da medida do per\u00edmetro do&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20323,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,258],"tags":[443,442],"series":[],"class_list":["post-12906","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-9--ano","category-aplicando","category-os-numeros-reais","tag-enquadramento","tag-valor-aproximado"],"views":2615,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/9V1Pag035-11_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/12906","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=12906"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/12906\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20323"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=12906"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=12906"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=12906"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=12906"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}