{"id":23192,"date":"2022-11-19T17:47:54","date_gmt":"2022-11-19T17:47:54","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=23192"},"modified":"2022-11-24T19:04:20","modified_gmt":"2022-11-24T19:04:20","slug":"objetos-em-caixas-2","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=23192","title":{"rendered":"Objetos em caixas"},"content":{"rendered":"\n<p><ul id='GTTabs_ul_23192' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_23192' class='GTTabs_curr'><a  id=\"23192_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_23192' ><a  id=\"23192_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_23192'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"23193\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=23193\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag056-T8-2.png\" data-orig-size=\"200,185\" 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=\"8_Pag056-T8-2\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag056-T8-2.png\" class=\"size-full wp-image-23193 alignright\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag056-T8-2.png\" alt=\"\" width=\"200\" height=\"185\" \/>O Pedro tem uma caixa c\u00fabica.<br \/>A medida do comprimento da aresta da caixa \u00e9 6 cm.<\/p>\n<p>Ajuda-o a saber qual \u00e9 o comprimento m\u00e1ximo das palhinhas que cabem nessa caixa.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_23192' onClick='GTTabs_show(1,23192)'> Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_23192'>\n<span class='GTTabs_titles'><b> Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"23193\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=23193\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag056-T8-2.png\" data-orig-size=\"200,185\" 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=\"8_Pag056-T8-2\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag056-T8-2.png\" class=\"size-full wp-image-23193 alignright\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag056-T8-2.png\" alt=\"\" width=\"200\" height=\"185\" \/>O Pedro tem uma caixa c\u00fabica.<br \/>A medida do comprimento da aresta da caixa \u00e9 6 cm.<\/p>\n<p>Ajuda-o a saber qual \u00e9 o comprimento m\u00e1ximo das palhinhas que cabem nessa caixa.<\/p>\n<\/blockquote>\n<p>\u00a0<\/p>\n<p>Por aplica\u00e7\u00e3o do Teorema de Pit\u00e1goras, comecemos por determinar, em cent\u00edmetros, o comprimento da sombra da palhinha no fundo da caixa:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{{d_f}}&amp; = &amp;{\\sqrt {{6^2} + {6^2}} }\\\\{}&amp; = &amp;{\\sqrt {36 + 36} }\\\\{}&amp; = &amp;{\\sqrt {72} }\\end{array}\\]<\/p>\n<p>Ainda por aplica\u00e7\u00e3o do Teorema de Pit\u00e1goras, determinemos agora o comprimento, em cent\u00edmetros, da aresta espacial da caixa:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{{d_e}}&amp; = &amp;{\\sqrt {{6^2} + {{\\left( {\\sqrt {72} } \\right)}^2}} }\\\\{}&amp; = &amp;{\\sqrt {36 + 72} }\\\\{}&amp; = &amp;{\\sqrt {108} }\\\\{}&amp; \\simeq &amp;{10,39}\\end{array}\\]<\/p>\n<p>O comprimento m\u00e1ximo da palhinha a colocar na caixa c\u00fabica \u00e9, aproximadamente, 10,39 cm.<\/p>\n<p>\u00a0<\/p>\n<h6>Uma alternativa com menos c\u00e1lculos<\/h6>\n<p>Este m\u00e9todo de c\u00e1lculo \u00e9 conhecido por aplica\u00e7\u00e3o do Teorema de Pit\u00e1goras no espa\u00e7o:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{{d_e}}&amp; = &amp;{\\sqrt {{6^2} + {6^2} + {6^2}} }\\\\{}&amp; = &amp;{\\sqrt {36 + 36 + 36} }\\\\{}&amp; = &amp;{\\sqrt {108} }\\end{array}\\]<\/p>\n<p>A explica\u00e7\u00e3o encontra-se facilmente nos c\u00e1lculos acima. Vejamos:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{\\begin{array}{*{20}{l}}{{d_e}}&amp; = &amp;{\\sqrt {{6^2} + {6^2} + {6^2}} }\\\\{}&amp; = &amp;{\\sqrt {36 + 36 + 36} }\\\\{}&amp; = &amp;{\\sqrt {108} }\\end{array}}&amp;{}&amp;{\\begin{array}{*{20}{l}}{{d_e}}&amp; = &amp;{\\sqrt {{6^2} + {{\\left( {{d_f}} \\right)}^2}} }\\\\{}&amp; = &amp;{\\sqrt {{6^2} + {{\\left( {\\sqrt {{6^2} + {6^2}} } \\right)}^2}} }\\\\{}&amp; = &amp;{\\sqrt {{6^2} + {6^2} + {6^2}} }\\\\{}&amp; = &amp;{\\sqrt {36 + 36 + 36} }\\\\{}&amp; = &amp;{\\sqrt {108} }\\end{array}}&amp;{}&amp;{\\begin{array}{*{20}{l}}{{d_f}}&amp; = &amp;{\\sqrt {{6^2} + {6^2}} }\\\\{}&amp; = &amp;{\\sqrt {36 + 36} }\\\\{}&amp; = &amp;{\\sqrt {72} }\\end{array}}\\end{array}\\]<\/p>\n<p>\u00a0<\/p>\n<h6>Comprimento da diagonal da base e da diagonal espacial do paralelep\u00edpedo ret\u00e2ngulo<\/h6>\n<blockquote>\n<p><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"23201\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=23201\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Teorema-Pitagoras_Paralelepipedo-Diagonais.png\" data-orig-size=\"309,217\" 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=\"8_Teorema-Pitagoras_Paralelepipedo-Diagonais\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Teorema-Pitagoras_Paralelepipedo-Diagonais.png\" class=\"size-medium wp-image-23201 alignright\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Teorema-Pitagoras_Paralelepipedo-Diagonais-300x211.png\" alt=\"\" width=\"300\" height=\"211\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Teorema-Pitagoras_Paralelepipedo-Diagonais-300x211.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Teorema-Pitagoras_Paralelepipedo-Diagonais.png 309w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/>Comprimento da diagonal da base e da diagonal espacial do paralelep\u00edpedo ret\u00e2ngulo:<\/p>\n<\/blockquote>\n<ul>\n<li>\n<blockquote>\\(x = \\sqrt {{a^2} + {b^2}} \\)<\/blockquote>\n<\/li>\n<li>\n<blockquote>\\(d = \\sqrt {{a^2} + {b^2} + {c^2}} \\)<\/blockquote>\n<\/li>\n<\/ul>\n<p>\u00a0<\/p>\n<p>Demonstra\u00e7\u00e3o da express\u00e3o relativa \u00e0 diagonal espacial:<\/p>\n<p>Aplicando o Teorema de Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo [BDH], temos:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}d&amp; = &amp;{\\sqrt {{{\\overline {BD} }^2} + {{\\overline {DH} }^2}} }\\\\{}&amp; = &amp;{\\sqrt {{x^2} + {c^2}} }\\\\{}&amp; = &amp;{\\sqrt {{{\\left( {\\sqrt {{a^2} + {b^2}} } \\right)}^2} + {c^2}} }\\\\{}&amp; = &amp;{\\sqrt {{a^2} + {b^2} + {c^2}} }\\end{array}\\]<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_23192' onClick='GTTabs_show(0,23192)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado O Pedro tem uma caixa c\u00fabica.A medida do comprimento da aresta da caixa \u00e9 6 cm. Ajuda-o a saber qual \u00e9 o comprimento m\u00e1ximo das palhinhas que cabem nessa caixa. Resolu\u00e7\u00e3o&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":23195,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,682],"tags":[424,67,118],"series":[],"class_list":["post-23192","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-teorema-de-pitagoras","tag-8-o-ano","tag-geometria","tag-teorema-de-pitagoras"],"views":178,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag056-T8-2_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/23192","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=23192"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/23192\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/23195"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=23192"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=23192"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=23192"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=23192"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}