{"id":1836,"date":"2011-11-30T06:15:11","date_gmt":"2011-11-30T12:15:11","guid":{"rendered":"http:\/\/w0ep.us\/OM\/?p=1836"},"modified":"2011-11-30T06:24:17","modified_gmt":"2011-11-30T12:24:17","slug":"pythagorus-revisited","status":"publish","type":"post","link":"http:\/\/w0ep.us\/OM\/?p=1836","title":{"rendered":"Pythagoras revisited"},"content":{"rendered":"<p>[Today&#8217;s run: none yet]<\/p>\n<p>I fooled around some more with my large-triangle method.\u00a0\u00a0 I found three equations describing the large triangle, but they all simplified down to the same thing and I could not see how to go any farther.<\/p>\n<p>So I decided to try to find another method.<\/p>\n<p>The unique feature of a right triangle is that one of the angles is 90 degrees.\u00a0 All triangles have angles that add up to 180 degrees.\u00a0 So, the pointy parts of a right triangle have to add up to be the other 90 degrees. <\/p>\n<p><a rel=\"attachment wp-att-1837\" href=\"http:\/\/w0ep.us\/OM\/?attachment_id=1837\"><img loading=\"lazy\" decoding=\"async\" class=\"alignright size-full wp-image-1837\" title=\"pythag\" src=\"http:\/\/w0ep.us\/OM\/wp-content\/uploads\/2011\/11\/pythag.jpg\" alt=\"\" width=\"600\" height=\"235\" \/><\/a><\/p>\n<p>I saw that the big c-squared square needed to be part of the picture.  If I arrange copies of my right triangle around the big square, it makes a larger square.  The sides of the larger square are of length  (B+A).   There are two ways to describe the larger square.  I set those against each other and did some algebra.<\/p>\n<p>I don&#8217;t know exactly how I went wrong with my prior method.  It seems like it should work.  Maybe my three descriptions of the big triangle are not independent.  Or something.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>[Today&#8217;s run: none yet] I fooled around some more with my large-triangle method.\u00a0\u00a0 I found three equations describing the large triangle, but they all simplified down to the same thing and I could not see how to go any farther. So I decided to try to find another method. The unique feature of a right [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8],"tags":[],"class_list":["post-1836","post","type-post","status-publish","format-standard","hentry","category-other-thoughts"],"_links":{"self":[{"href":"http:\/\/w0ep.us\/OM\/index.php?rest_route=\/wp\/v2\/posts\/1836","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/w0ep.us\/OM\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/w0ep.us\/OM\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/w0ep.us\/OM\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/w0ep.us\/OM\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1836"}],"version-history":[{"count":13,"href":"http:\/\/w0ep.us\/OM\/index.php?rest_route=\/wp\/v2\/posts\/1836\/revisions"}],"predecessor-version":[{"id":1850,"href":"http:\/\/w0ep.us\/OM\/index.php?rest_route=\/wp\/v2\/posts\/1836\/revisions\/1850"}],"wp:attachment":[{"href":"http:\/\/w0ep.us\/OM\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1836"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/w0ep.us\/OM\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1836"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/w0ep.us\/OM\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1836"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}