{"id":2791,"date":"2023-03-05T11:54:22","date_gmt":"2023-03-05T10:54:22","guid":{"rendered":"http:\/\/matma.com.pl\/?p=2791"},"modified":"2023-05-06T10:04:31","modified_gmt":"2023-05-06T08:04:31","slug":"zadanie-32-1-0-2-zbior-rozszerzenie","status":"publish","type":"post","link":"http:\/\/matma.com.pl\/?p=2791","title":{"rendered":"zadanie 32.1 (0-2) zbi\u00f3r rozszerzenie"},"content":{"rendered":"<p>Zaznaczymy na obrazku k\u0105ty proste jakie tworzy promie\u0144 kuli z podstaw\u0105 sto\u017cka oraz jaki tworzy promie\u0144 kuli z tworz\u0105c\u0105 sto\u017cka. Zaznaczymy jednocze\u015bnie na naszym obrazku tr\u00f3jk\u0105ty prostok\u0105tne kt\u00f3re b\u0119d\u0105 do siebie podobne.<\/p>\n<table style=\"border-collapse: collapse; width: 100%; height: 24px;\">\n<tbody>\n<tr style=\"height: 24px;\">\n<td style=\"width: 50%; height: 24px;\">Oznaczymy przez <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?h\" alt=\"h\" align=\"absmiddle\" \/> wysoko\u015b\u0107 sto\u017cka oraz oznaczymy wierzcho\u0142ki tr\u00f3jk\u0105t\u00f3w prostok\u0105tnych podobnych.<\/p>\n<p>Mo\u017cemy zapisa\u0107 d\u0142ugo\u015bci odcink\u00f3w:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;|&amp;space;AB&amp;space;\\right&amp;space;|=r\" alt=\"\\left | AB \\right |=r\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;|&amp;space;AC&amp;space;\\right&amp;space;|=h\" alt=\"\\left | AC \\right |=h\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;|&amp;space;SC&amp;space;\\right&amp;space;|=h-1\" alt=\"\\left | SC \\right |=h-1\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" style=\"font-family: inherit; font-size: inherit;\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;|&amp;space;SA&amp;space;\\right&amp;space;|=\\left&amp;space;|&amp;space;SD&amp;space;\\right&amp;space;|=1\" alt=\"\\left | SA \\right |=\\left | SD \\right |=1\" align=\"absmiddle\" \/><\/p>\n<p>oraz oznaczmy <img decoding=\"async\" style=\"font-family: inherit; font-size: inherit;\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;|&amp;space;CD&amp;space;\\right&amp;space;|=x\" alt=\"\\left | CD \\right |=x\" align=\"absmiddle\" \/><\/p>\n<p>Z twierdzenia Pitagorasa z tr\u00f3jk\u0105ta <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?SDC\" alt=\"SDC\" align=\"absmiddle\" \/> mamy:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?1^2+x^2=\\left&amp;space;(&amp;space;h-1&amp;space;\\right&amp;space;)^2\" alt=\"1^2+x^2=\\left ( h-1 \\right )^2\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x^2=h^2-2h+1-1\" alt=\"x^2=h^2-2h+1-1\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x=\\sqrt{h^2-2h}\" alt=\"x=\\sqrt{h^2-2h}\" align=\"absmiddle\" \/><\/td>\n<td style=\"width: 50%; height: 24px;\"><img fetchpriority=\"high\" decoding=\"async\" class=\" wp-image-2794 aligncenter\" src=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-05-111204-220x300.jpg\" alt=\"\" width=\"250\" height=\"341\" srcset=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-05-111204-220x300.jpg 220w, http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-05-111204.jpg 332w\" sizes=\"(max-width: 250px) 100vw, 250px\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Korzystaj\u0105c z podobie\u0144stwa tr\u00f3jk\u0105t\u00f3w <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?ABC\" alt=\"ABC\" align=\"absmiddle\" \/> i <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?DSC\" alt=\"DSC\" align=\"absmiddle\" \/> mo\u017cemy zapisa\u0107:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\frac{\\left&amp;space;|&amp;space;AB&amp;space;\\right&amp;space;|}{\\left&amp;space;|&amp;space;AC&amp;space;\\right&amp;space;|}=\\frac{\\left&amp;space;|&amp;space;SD&amp;space;\\right&amp;space;|}{\\left&amp;space;|DC&amp;space;\\right&amp;space;|}\" alt=\"\\frac{\\left | AB \\right |}{\\left | AC \\right |}=\\frac{\\left | SD \\right |}{\\left |DC \\right |}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\frac{r}{h}=\\frac{1}{x}\" alt=\"\\frac{r}{h}=\\frac{1}{x}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\frac{r}{h}=\\frac{1}{\\sqrt{h^2-2h}}\" alt=\"\\frac{r}{h}=\\frac{1}{\\sqrt{h^2-2h}}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r\\cdot&amp;space;\\sqrt{h^2-2h}=h\\cdot&amp;space;1\\;&amp;space;\\;&amp;space;\/:\\sqrt{h^2-2h}\" alt=\"r\\cdot \\sqrt{h^2-2h}=h\\cdot 1\\; \\; \/:\\sqrt{h^2-2h}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r=\\frac{h}{\\sqrt{h^2-2h}}\" alt=\"r=\\frac{h}{\\sqrt{h^2-2h}}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r=\\sqrt{\\frac{h^2}{h\\left&amp;space;(&amp;space;h-2&amp;space;\\right&amp;space;)}}\" alt=\"r=\\sqrt{\\frac{h^2}{h\\left ( h-2 \\right )}}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r=\\sqrt{\\frac{h}{h-2}}\" alt=\"r=\\sqrt{\\frac{h}{h-2}}\" align=\"absmiddle\" \/><\/p>\n<p>Wyznaczmy funkcje opisuj\u0105ca obj\u0119to\u015b\u0107 sto\u017cka uzale\u017cnion\u0105 od d\u0142ugo\u015bci jego wysoko\u015bci<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V(h)=\\frac{1}{3}\\pi\\cdot&amp;space;\\left&amp;space;(&amp;space;\\sqrt{\\frac{h}{h-2}}&amp;space;\\right&amp;space;)^2\\cdot&amp;space;h\" alt=\"V(h)=\\frac{1}{3}\\pi\\cdot \\left ( \\sqrt{\\frac{h}{h-2}} \\right )^2\\cdot h\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V(h)=\\frac{1}{3}\\pi\\cdot\\frac{h}{h-2}&amp;space;\\cdot&amp;space;h\" alt=\"V(h)=\\frac{1}{3}\\pi\\cdot\\frac{h}{h-2} \\cdot h\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V(h)=\\frac{h^2\\pi&amp;space;}{3\\left&amp;space;(&amp;space;h-2&amp;space;\\right&amp;space;)}\" alt=\"V(h)=\\frac{h^2\\pi }{3\\left ( h-2 \\right )}\" align=\"absmiddle\" \/><\/p>\n<p style=\"text-align: right;\">Co nale\u017ca\u0142o pokaza\u0107.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Zaznaczymy na obrazku k\u0105ty proste jakie tworzy promie\u0144 kuli z podstaw\u0105 sto\u017cka oraz jaki tworzy promie\u0144 kuli z tworz\u0105c\u0105 sto\u017cka. Zaznaczymy jednocze\u015bnie na naszym obrazku tr\u00f3jk\u0105ty prostok\u0105tne kt\u00f3re b\u0119d\u0105 do siebie podobne. Oznaczymy przez wysoko\u015b\u0107 sto\u017cka oraz oznaczymy wierzcho\u0142ki tr\u00f3jk\u0105t\u00f3w prostok\u0105tnych podobnych. Mo\u017cemy zapisa\u0107 d\u0142ugo\u015bci odcink\u00f3w: oraz oznaczmy Z twierdzenia Pitagorasa z tr\u00f3jk\u0105ta mamy: Korzystaj\u0105c [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-2791","post","type-post","status-publish","format-standard","hentry","category-najnowsze-pliki"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.1 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>zadanie 32.1 (0-2) zbi\u00f3r rozszerzenie - Matma.com.pl<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/matma.com.pl\/?p=2791\" \/>\n<meta property=\"og:locale\" content=\"pl_PL\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"zadanie 32.1 (0-2) zbi\u00f3r rozszerzenie - Matma.com.pl\" \/>\n<meta property=\"og:description\" content=\"Zaznaczymy na obrazku k\u0105ty proste jakie tworzy promie\u0144 kuli z podstaw\u0105 sto\u017cka oraz jaki tworzy promie\u0144 kuli z tworz\u0105c\u0105 sto\u017cka. Zaznaczymy jednocze\u015bnie na naszym obrazku tr\u00f3jk\u0105ty prostok\u0105tne kt\u00f3re b\u0119d\u0105 do siebie podobne. Oznaczymy przez wysoko\u015b\u0107 sto\u017cka oraz oznaczymy wierzcho\u0142ki tr\u00f3jk\u0105t\u00f3w prostok\u0105tnych podobnych. 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