{"id":23553,"date":"2022-12-09T10:40:22","date_gmt":"2022-12-09T10:40:22","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=23553"},"modified":"2022-12-18T22:29:11","modified_gmt":"2022-12-18T22:29:11","slug":"bases-de-duas-pecas-metalicas","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=23553","title":{"rendered":"Bases de duas pe\u00e7as met\u00e1licas"},"content":{"rendered":"\n<p><ul id='GTTabs_ul_23553' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_23553' class='GTTabs_curr'><a  id=\"23553_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_23553' ><a  id=\"23553_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_23553'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Observa as medidas, em mil\u00edmetros, das bases de duas pe\u00e7as met\u00e1licas com a forma de trap\u00e9zio.<br \/>O segmento de reta a vermelho \u00e9 eixo de simetria de reflex\u00e3o da figura (B).<\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"23554\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=23554\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15.png\" data-orig-size=\"580,285\" 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_Pag067-15\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15.png\" class=\"aligncenter wp-image-23554\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15-300x147.png\" alt=\"\" width=\"380\" height=\"187\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15-300x147.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15.png 580w\" sizes=\"auto, (max-width: 380px) 100vw, 380px\" \/><\/a><\/p>\n<ol>\n<li>Determina <em>x<\/em> e <em>y<\/em>.<\/li>\n<li>Calcula o per\u00edmetro de cada uma das bases.<\/li>\n<li>Pretendendo forrar a papel as duas bases, quantos cm<sup>2<\/sup> de papel se v\u00e3o gastar em cada uma delas?<br \/>Apresenta o resultado arredondado \u00e0s cent\u00e9simas.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_23553' onClick='GTTabs_show(1,23553)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_23553'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p>Observa as medidas, em mil\u00edmetros, das bases de duas pe\u00e7as met\u00e1licas com a forma de trap\u00e9zio.<br \/>O segmento de reta a vermelho \u00e9 eixo de simetria de reflex\u00e3o da figura (B).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"23559\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=23559\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15_Letras.png\" data-orig-size=\"580,285\" 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_Pag067-15_Letras\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15_Letras.png\" class=\"aligncenter wp-image-23559\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15_Letras-300x147.png\" alt=\"\" width=\"380\" height=\"187\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15_Letras-300x147.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15_Letras.png 580w\" sizes=\"auto, (max-width: 380px) 100vw, 380px\" \/><\/p>\n<\/blockquote>\n<ol>\n<li>\n<blockquote>Determina <em>x<\/em> e <em>y<\/em>.<\/blockquote>\n<\/li>\n<li>\n<blockquote>Calcula o per\u00edmetro de cada uma das bases.<\/blockquote>\n<\/li>\n<li>\n<blockquote>Pretendendo forrar a papel as duas bases, quantos cm<sup>2<\/sup> de papel se v\u00e3o gastar em cada uma delas?<br \/>Apresenta o resultado arredondado \u00e0s cent\u00e9simas.<\/blockquote>\n<\/li>\n<\/ol>\n<p>\u00a0<\/p>\n<ol>\n<li>Consideremos o trap\u00e9zio (A) decomposto num ret\u00e2ngulo e num tri\u00e2ngulo ret\u00e2ngulo.<br \/>Aplicando o Teorema de Pit\u00e1goras neste tri\u00e2ngulo ret\u00e2ngulo, vem:<br \/>\\[x = \\sqrt {{{25}^2} + {{\\left( {30 &#8211; 20} \\right)}^2}} = \\sqrt {625 + 100} = \\sqrt {725} \\]<br \/>Portanto, \\(x = \\sqrt {725} \\) mm.<br \/>E \\(y = \\overline {AC} = 100\\) mm, pois o segmento de reta a vermelho \u00e9 eixo de simetria de reflex\u00e3o da figura (B).<br \/><br \/><\/li>\n<li>\\({P_{(A)}} = 20 + 25 + 30 + \\sqrt {750} = 75 + \\sqrt {750} \\)<br \/>\\({P_{(B)}} = 76 + 100 + 88 + 100 = 364\\)<br \/>Portanto, as bases t\u00eam os seguintes per\u00edmetros: \\({P_{(A)}} = 75 + \\sqrt {750} \\) mm e \\({P_{(B)}} = 364\\) mm.<br \/><br \/><\/li>\n<li>Comecemos por determinar a altura do trap\u00e9zio da figura (B), aplicando o Teorema de Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo [ABC]:<br \/>\\[\\overline {AB} = \\sqrt {{{\\overline {AC} }^2} &#8211; {{\\left( {\\frac{{88 &#8211; 76}}{2}} \\right)}^2}} = \\sqrt {{{100}^2} &#8211; {6^2}} = \\sqrt {10000 &#8211; 36} = \\sqrt {9964} \\]<br \/>Calculemos agora as \u00e1reas, em mm<sup>2<\/sup>, das bases de cada uma das pe\u00e7as:<br \/>\\[{A_{\\left( A \\right)}} = \\frac{{30 + 20}}{2} \\times 25 = 25 \\times 25 = 625\\]<br \/>\\[{A_{\\left( B \\right)}} = \\frac{{88 + 76}}{2} \\times \\sqrt {9964} = 82 \\times \\sqrt {9964} \\]<br \/>Para forrar as bases v\u00e3o-se gastar, respetivamente, \\({A_{\\left( A \\right)}} = 6,25\\) cm<sup>2<\/sup> e \\({A_{\\left( B \\right)}} \\approx 81,85\\) cm<sup>2<\/sup> de papel.<\/li>\n<\/ol>\n<p>\u00a0<\/p>\n<p>Para saber mais sobre a <a href=\"https:\/\/www.acasinhadamatematica.pt\/?p=5632\" target=\"_blank\" rel=\"noopener\">\u00c1rea do Trap\u00e9zio<\/a><\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_23553' onClick='GTTabs_show(0,23553)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Observa as medidas, em mil\u00edmetros, das bases de duas pe\u00e7as met\u00e1licas com a forma de trap\u00e9zio.O segmento de reta a vermelho \u00e9 eixo de simetria de reflex\u00e3o da figura (B). Determina&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":23555,"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-23553","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":246,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/12\/8_Pag067-15_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/23553","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=23553"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/23553\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/23555"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=23553"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=23553"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=23553"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=23553"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}