D\u0130K \u00dc\u00c7GEN
\nBir a\u00e7\u0131s\u0131n\u0131n \u00f6l\u00e7\u00fcs\u00fc 90\u00b0 olan \u00fc\u00e7gene dik \u00fc\u00e7gen denir. Dik \u00fc\u00e7gende 90\u00b0 nin kar\u015f\u0131s\u0131ndaki kenara hipoten\u00fcs, di\u011fer kenarlara dik kenar ad\u0131 verilir. Hipoten\u00fcs \u00fc\u00e7genin daima en uzun kenar\u0131d\u0131r. \u015fekilde, m(A) = 90\u00b0
\n[BC] kenar\u0131 hipoten\u00fcs
\n[AB] ve [AC] kenarlar\u0131
\ndik kenarlard\u0131r.<\/p>\n
P\u0130SAGOR BA\u011eINTISI
\nDik \u00fc\u00e7gende dik kenarlar\u0131n uzunluklar\u0131n\u0131n kareleri toplam\u0131 hipoten\u00fcs\u00fcn uzunlu\u011funun karesine e\u015fittir. ABC \u00fc\u00e7geninde m(A) = 90\u00b0
\na2=b2+c2
\n\u00d6ZEL D\u0130K \u00dc\u00c7GENLER
\n1. (3 – 4 – 5) \u00dc\u00e7geni
\nKenar uzunluklar\u0131 (3 – 4 – 5) say\u0131lar\u0131 veya bunlar\u0131n kat\u0131 olan b\u00fct\u00fcn \u00fc\u00e7genler dik \u00fc\u00e7gendir. (6 – 8 – 10), (9 – 12 – 15), \u2026 gibi<\/p>\n
2. (5 – 12 – 13) \u00dc\u00e7geni
\nKenar uzunluklar\u0131 (5 – 12 – 13) say\u0131lar\u0131 ve bunlar\u0131n kat\u0131 olan b\u00fct\u00fcn \u00fc\u00e7genler dik \u00fc\u00e7genlerdir. (10 – 24 – 26), (15 – 36 – 39), \u2026 gibi.<\/p>\n
Kenar uzunluklar\u0131 8, 15, 17 say\u0131lar\u0131 ile orant\u0131l\u0131 olan \u00fc\u00e7genler dik \u00fc\u00e7genlerdir.
\n Kenar uzunluklar\u0131 7, 24, 25 say\u0131lar\u0131 ile orant\u0131l\u0131 olan \u00fc\u00e7genler dik \u00fc\u00e7genlerdir.
\n 3. \u0130kizkenar dik \u00fc\u00e7gen
\nABC dik \u00fc\u00e7gen |AB| = |BC| = a |AC| = a\u00d62
\nm(A) = m(C) = 45\u00b0 \u0130kizkenar dik \u00fc\u00e7gende
\nhipoten\u00fcs dik kenarlar\u0131n \u00d62 kat\u0131d\u0131r.
\n 4. (30\u00b0 \u2013 60\u00b0 \u2013 90\u00b0) \u00dc\u00e7geni
\nABC e\u015fkenar \u00fc\u00e7geni y\u00fckseklikle ikiye b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde
\nABH ve ACH (30\u00b0 – 60\u00b0 – 90\u00b0)
\n\u00fc\u00e7genleri elde edilir.
\n|AB| = |AC| = a
\n|BH| = |HC| = pisagordan (30\u00b0 – 60\u00b0 – 90\u00b0) dik \u00fc\u00e7geninde; 30\u00b0’nin kar\u015f\u0131s\u0131ndaki kenar hipoten\u00fcs\u00fcn yar\u0131s\u0131na e\u015fittir. 60\u00b0 nin kar\u015f\u0131s\u0131ndaki kenar,
\n30\u00b0 nin kar\u015f\u0131s\u0131ndaki kenar\u0131n \u00d63 kat\u0131d\u0131r.<\/p>\n
5. (30\u00b0 – 30\u00b0 – 120\u00b0) \u00dc\u00e7geni (30\u00b0 – 30\u00b0 – 120\u00b0) \u00fc\u00e7geninde 30\u00b0 lik a\u00e7\u0131lar\u0131n kar\u015f\u0131lar\u0131ndaki kenarlara a dersek 120\u00b0 lik a\u00e7\u0131n\u0131n kar\u015f\u0131s\u0131ndaki kenar a\u00d63 olur.<\/p>\n
6. (15\u00b0 – 75\u00b0 – 90\u00b0) \u00dc\u00e7geni (15\u00b0 – 75\u00b0 – 90\u00b0) \u00fc\u00e7geninde
\nhipoten\u00fcse ait y\u00fckseklik |AH| = h dersek, hipoten\u00fcs
\n|BC| = 4h olur. Hipoten\u00fcs kendisine ait y\u00fcksekli\u011fin d\u00f6rt
\nkat\u0131d\u0131r.<\/p>\n
\u00d6KL\u0130T BA\u011eINTILARI
\nDik \u00fc\u00e7genlerde hipoten\u00fcse ait y\u00fcksekli\u011fin verildi\u011fi durumlarda benzerlikten kaynaklanan \u00f6klit ba\u011f\u0131nt\u0131lar\u0131 kullan\u0131l\u0131r.
\n 1. Y\u00fcksekli\u011fin hipoten\u00fcste ay\u0131rd\u0131\u011f\u0131 par\u00e7alar\u0131n \u00e7arp\u0131m\u0131 y\u00fcksekli\u011fin karesine e\u015fittir.
\nh2 = p.k 2. b2 = k.a c2 = p.a 3. ABC \u00fc\u00e7geninin alan\u0131n\u0131 iki farkl\u0131 \u015fekilde yaz\u0131p e\u015fitledi\u011fimizde<\/p>\n
a.h =b.c
\nYukar\u0131da anlat\u0131lan \u00f6klit ba\u011f\u0131nt\u0131lar\u0131 kullan\u0131larak elde edilir.<\/p>\n
Genellikle bu \u00f6klit ba\u011f\u0131nt\u0131s\u0131n\u0131 kullanmak yerine, yukar\u0131daki \u00f6klit ba\u011f\u0131nt\u0131lar\u0131 ve pisagor ba\u011f\u0131nt\u0131s\u0131n\u0131 kullanarak \u00e7\u00f6z\u00fcme gideriz.
\n\u0130K\u0130ZKENAR \u00dc\u00c7GEN
\n\u0130kizkenar \u00fc\u00e7genin tepe a\u00e7\u0131s\u0131ndan taban\u0131na \u00e7izilen y\u00fckseklik, hem a\u00e7\u0131ortay, hem de kenarortayd\u0131r.
\n 1. Bir \u00fc\u00e7gende, a\u00e7\u0131ortay ayn\u0131 zamanda y\u00fckseklik ise bu \u00fc\u00e7gen ikizkenar \u00fc\u00e7gendir. |AB| = |AC|
\n|BH| = |HC|
\nm(B) = m(C)<\/p>\n
2. Bir \u00fc\u00e7gende, a\u00e7\u0131ortay ayn\u0131 zamanda kenarortay ise bu \u00fc\u00e7gen ikizkenar \u00fc\u00e7gendir. |AB| = |AC|,
\n[AH] ^ [BC]
\nm(B) = m(C)<\/p>\n
3. Bir \u00fc\u00e7gende, y\u00fckseklik ayn\u0131 zamanda kenarortay ise bu \u00fc\u00e7gen ikizkenar \u00fc\u00e7gendir. |AB| = |AC|
\nm(BAH) = m(HAC)
\nm(B) = m(C)<\/p>\n
\u0130kizkenar \u00fc\u00e7gende a\u00e7\u0131ortay, kenarortay ve y\u00fcksekli\u011fin ayn\u0131 olmas\u0131 bir\u00e7ok yerde kar\u015f\u0131m\u0131za \u00e7\u0131kt\u0131\u011f\u0131ndan \u00e7ok iyi bilinmesi gereken bir \u00f6zelliktir.
\n4. \u0130kizkenar \u00fc\u00e7gende ikizkenara ait y\u00fckseklikler e\u015fittir. Bu durumda y\u00fcksekliklerin kesim noktas\u0131n\u0131n ay\u0131rd\u0131\u011f\u0131 par\u00e7alarda e\u015fit olur.
\n 5. \u0130kizkenar \u00fc\u00e7gende ikizkenara ait kenarortaylar ve kenarortaylar\u0131n kesim noktas\u0131n\u0131n ay\u0131rd\u0131\u011f\u0131 par\u00e7alar da birbirine e\u015fittir.
\n6. \u0130kizkenar \u00fc\u00e7gende e\u015fit a\u00e7\u0131lara ait a\u00e7\u0131ortaylar da e\u015fittir. A\u00e7\u0131ortaylar birbirini ayn\u0131 oranda b\u00f6lerler.
\n 7. \u0130kizkenar \u00fc\u00e7gende ikiz olmayan kenar \u00fczerindeki herhangi bir noktadan ikiz kenarlara \u00e7izilen dikmelerin toplam\u0131, ikizkenarlara ait y\u00fcksekli\u011fi verir.
\n|AB| = |AC| Þ |LC| = |HP| + |KP|
\n8. \u0130kizkenar \u00fc\u00e7gende tabandan ikiz kenarlara \u00e7izilen paralellerin toplam\u0131, ikiz kenarlar\u0131n uzunlu\u011funa e\u015fittir. <\/p>\n
E\u015eKENAR \u00dc\u00c7GEN
\n1. E\u015fkenar \u00fc\u00e7gende b\u00fct\u00fcn a\u00e7\u0131ortay, kenarortay y\u00fckseklikler \u00e7ak\u0131\u015f\u0131k ve hepsinin uzunluklar\u0131 e\u015fittir. nA = nB = nC = Va = Vb = Vc = ha = hb = hc<\/p>\n
2. E\u015fkenar \u00fc\u00e7genin bir kenar\u0131na a dersek y\u00fck seklik Bu durumda e\u015fkenar \u00fc\u00e7genin alan\u0131
\n y\u00fckseklik cinsinden alan de\u011feri
\nAlan(ABC) =
\n3. E\u015fkenar \u00fc\u00e7genin i\u00e7indeki herhangi bir noktadan kenarlara \u00e7izilen dik uzunluklar\u0131n toplam\u0131, e\u015fkenar \u00fc\u00e7gene ait y\u00fcksekli\u011fi verir. Bir kenar\u0131 a olan e\u015fkenar \u00fc\u00e7gende;<\/p>\n
4. E\u015fkenar \u00fc\u00e7genin i\u00e7indeki herhangi bir noktadan kenarlara \u00e7izilen paralellerin toplam\u0131 bir kenar uzunlu\u011funa e\u015fittir.
\nBir kenar\u0131 a olan ABC e\u015fkenar \u00fc\u00e7geninde<\/p>\n","protected":false},"excerpt":{"rendered":"
D\u0130K \u00dc\u00c7GEN Bir a\u00e7\u0131s\u0131n\u0131n \u00f6l\u00e7\u00fcs\u00fc 90\u00b0 olan \u00fc\u00e7gene dik \u00fc\u00e7gen denir. Dik \u00fc\u00e7gende 90\u00b0 nin kar\u015f\u0131s\u0131ndaki kenara hipoten\u00fcs, di\u011fer kenarlara dik kenar ad\u0131 verilir. Hipoten\u00fcs \u00fc\u00e7genin daima en uzun kenar\u0131d\u0131r. \u015fekilde, m(A) = 90\u00b0 [BC] kenar\u0131 hipoten\u00fcs [AB] ve [AC] kenarlar\u0131 dik kenarlard\u0131r. P\u0130SAGOR BA\u011eINTISI Dik \u00fc\u00e7gende dik kenarlar\u0131n uzunluklar\u0131n\u0131n kareleri toplam\u0131 hipoten\u00fcs\u00fcn uzunlu\u011funun karesine …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1404,1403],"tags":[7224,7219,7220,7218,7225,7223,7222,7217,7221],"class_list":["post-3038","post","type-post","status-publish","format-standard","hentry","category-matematik-odevleri","category-odevler","tag-aciortay","tag-dik-ucgen","tag-eskenar-ucgen","tag-hipotenus","tag-ikizkenar-ucgen","tag-kenarortay","tag-oklit-bagintilari","tag-ozel-ucgenler","tag-pisagor"],"_links":{"self":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3038","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/comments?post=3038"}],"version-history":[{"count":0,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3038\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/media?parent=3038"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/categories?post=3038"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/tags?post=3038"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}