\nEnunciado<\/b><\/span><\/p>\n
Sabendo que $\\hat{A}=\\hat{T}$, $\\hat{C}=\\hat{R}$ e tendo em conta as medidas indicadas na figura, determina x e y.<\/p>\n
\u00a0 $\"\"$ <\/a><\/p>\n
Resolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n
\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
$\"\"$ <\/a><\/p>\n
Os tri\u00e2ngulos s\u00e3o semelhantes, pois possuem dois \u00e2ngulos geometricamente iguais, cada um a cada um, de um para o outro dos tri\u00e2ngulos. Consequentemente, os lados correspondentes t\u00eam comprimentos diretamente proporcionais, isto \u00e9: \\[\\frac{\\overline{AB}}{\\overline{ST}}=\\frac{\\overline{BC}}{\\overline{RS}}=\\frac{\\overline{AC}}{\\overline{RT}}\\]<\/p>\n
Assim, temos: \\[\\frac{8}{10}=\\frac{6}{\\overline{RS}}\\Leftrightarrow \\overline{RS}=\\frac{10\\times 6}{8}\\Leftrightarrow \\overline{RS}=7,5\\]
\nLogo, $\\overline{RS}=7,5\\,cm$.<\/p>\n
Por outro lado, \\[\\frac{8}{10}=\\frac{9}{\\overline{RT}}\\Leftrightarrow \\overline{RT}=\\frac{10\\times 9}{8}\\Leftrightarrow \\overline{RT}=11,25\\]
\nLogo, $\\overline{RT}=11,25\\,cm$.<\/p><\/p>\n
<< Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"
Enunciado Resolu\u00e7\u00e3o Enunciado Sabendo que $\\hat{A}=\\hat{T}$, $\\hat{C}=\\hat{R}$ e tendo em conta as medidas indicadas na figura, determina x e y. \u00a0 Resolu\u00e7\u00e3o >> Resolu\u00e7\u00e3o << Enunciado<\/p>\n","protected":false},"author":1,"featured_media":20513,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,151],"tags":[149],"series":[],"class_list":["post-6686","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-semelhanca-de-triangulos-8--ano","tag-semelhanca-de-triangulos"],"views":2473,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8V1Pag132-5_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6686","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=6686"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6686\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20513"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6686"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6686"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6686"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6686"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}