TE\u011eET – K\u0130R\u0130\u015e \u00d6ZELL\u0130KLER\u0130
\n1. Te\u011fet noktas\u0131ndan ve \u00e7emberin merkezinden ge\u00e7en do\u011fru, te\u011fet olan do\u011fruya diktir.AB do\u011frusu T noktas\u0131nda \u00e7embere te\u011fet
\nAB ^ OT Te\u011fet do\u011frusuna, te\u011fet noktas\u0131ndan \u00e7izilen dik do\u011fru \u00e7emberin merkezinden ge\u00e7er. <\/p>\n
2. \u00c7emberin d\u0131\u015f\u0131ndaki bir noktadan \u00e7embere \u00e7izilen te\u011fetlerin uzuluklar\u0131 birbirine
\ne\u015fittir.
\n[PA ve [PT
\n\u00e7embere te\u011fet<\/p>\n
|PA| = |PB| <\/p>\n
[PT ve [PS \u00e7embere te\u011fet ve O \u00e7emberin merkezi ise [PO, TPS a\u00e7\u0131s\u0131n\u0131n a\u00e7\u0131ortay\u0131d\u0131r.
\n|OT| = |OS| ve [PT] ^ [TO], [PS] ^ [SO] oldu\u011fundan PTOS d\u00f6rtgeni bir deltoid tir.
\n\u0130\u00e7ten ve d\u0131\u015ftan te\u011fet \u00e7emberlerde merkezleri birle\u015ftiren do\u011fru te\u011fet noktas\u0131ndan ge\u00e7er.
\nO1 ve O2 merkezli \u00e7emberler T noktas\u0131nda d\u0131\u015ftan te\u011fet ise, merkezleri birle\u015ftiren do\u011fru T noktas\u0131ndan ge\u00e7er.
\nAyn\u0131 \u00f6zellik i\u00e7ten te\u011fet \u00e7emberler i\u00e7in de ge\u00e7erlidir.O1 , O2 ve T noktalar\u0131 ayn\u0131 do\u011fru \u00fczerindedir.
\n3. Bir \u00e7emberin merkezinden kiri\u015fe indirilen dikme, kiri\u015fi ortalar.
\nBir \u00e7emberde, merkeze uzakl\u0131klar\u0131 e\u015fit olan kiri\u015flerin uzunluklar\u0131 da e\u015fittir.
\n|OF|=|OE|\u00db |AB|=|CD|
\nBir \u00e7emberde herhangi iki kiri\u015ften merkeze yak\u0131n olan\u0131 daha b\u00fcy\u00fckt\u00fcr.
\n|OH|<|ON|\u00db |AB|>|CD|
\n4. Bir \u00e7emberde e\u015fit uzunluktaki kiri\u015flerin g\u00f6rd\u00fc\u011f\u00fc yaylarda e\u015fittir.<\/p>\n
5. Bir \u00e7emberde paralel iki kiri\u015f aras\u0131nda kalan yaylar e\u015fittir.<\/p>\n
Bir \u00e7ember i\u00e7inde al\u0131nan herhangi bir P noktas\u0131ndan ge\u00e7en en k\u0131sa kiri\u015f, orta noktas\u0131 P olan kiri\u015ftir.
\n[AC] ^ [PO]
\nTE\u011eETLER D\u00d6RTGEN\u0130
\n1. Bir \u00e7embere te\u011fet d\u00f6rt do\u011fru par\u00e7as\u0131n\u0131n olu\u015fturdu\u011fu d\u00f6rtgene te\u011fetler d\u00f6rtgeni denir. ABCD d\u00f6rtgeninde K, L, M, N te\u011fetlerin de\u011fme noktas\u0131d\u0131r.<\/p>\n
2. Te\u011fetler d\u00f6rtgeninde kar\u015f\u0131l\u0131kl\u0131 kenarlar\u0131n uzunluklar\u0131 toplam\u0131 e\u015fittir.<\/p>\n
a+c=b+d
\n3. Te\u011fetler d\u00f6rtgeninin alan\u0131; i\u00e7te\u011fet \u00e7emberin yar\u0131\u00e7ap\u0131 ile \u00e7evresinin \u00e7arp\u0131m\u0131n\u0131n yar\u0131s\u0131d\u0131r. <\/p>\n
K\u0130R\u0130\u015eLER D\u00d6RTGEN\u0130
\nKiri\u015fler d\u00f6rtgeninde kar\u015f\u0131l\u0131kl\u0131 a\u00e7\u0131lar\u0131n toplam\u0131n\u0131n 180\u00b0 dir.
\nD\u00f6rtgeninin alan\u0131;<\/p>\n
A(ABCD)=\u00d6(u – a)(u – b)(u – c)(u – d)
\n KUVVET
\n1. \u00c7emberin D\u0131\u015f\u0131ndaki Bir Noktan\u0131n \u00c7embere G\u00f6re Kuvveti
\n[PT, T noktas\u0131nda \u00e7embere te\u011fet, [PB ve [PD \u00e7emberi
\nkesen \u0131\u015f\u0131nlar
\nKuvvet = |PT|2 = |PA| . |PB| = |PC| . |PD|
\n2. \u00c7emberin \u0130\u00e7indeki Bir Noktan\u0131n \u00c7embere G\u00f6re Kuvveti
\nBir \u00e7emberin i\u00e7indeki bir noktada kesi\u015fen iki kiri\u015f \u00fczerinde, kesim noktas\u0131n\u0131n ay\u0131rd\u0131\u011f\u0131 par\u00e7alar\u0131n uzunluklar\u0131 \u00e7arp\u0131m\u0131
\nsabittir.
\nKuvvet = |PA| . |PB| = |PC| . |PD|
\n\u00c7emberin \u00fczerindeki bir noktan\u0131n \u00e7embere g\u00f6re kuvveti s\u0131f\u0131rd\u0131r
\n3. \u0130ki \u00c7emberin Kuvvet Ekseni
\nKuvvet ekseni \u00fczerindeki noktalar\u0131n her iki \u00e7embere g\u00f6re kuvvetleri e\u015fittir.
\na. D\u0131\u015ftan te\u011fet iki \u00e7emberin kuvvet ekseni te\u011fet noktas\u0131ndan ge\u00e7er. Kuvvet ekseni \u00e7emberin merkezlerini birle\u015ftiren do\u011fruya te\u011fet noktas\u0131nda diktir. |O1O2| = r1 + r2
\n b. \u0130\u00e7ten te\u011fet \u00e7emberlerin kuvvet ekseni te\u011fet noktas\u0131ndan ge\u00e7er. Kuvvet ekseni merkezlerden ge\u00e7en do\u011fruya te\u011fet noktas\u0131nda diktir. |O1O2| = r1 \u2013 r2<\/p>\n
c. Kesi\u015fen \u00e7emberlerde kuvvet ekseni \u00e7emberlerin kesi\u015fim noktalar\u0131ndan ge\u00e7er ve merkezleri birle\u015ftiren do\u011fruya diktir. |O1O2| < r1 + r2\n\n\u015fekildeki P noktas\u0131n\u0131n A noktas\u0131nda birbirine d\u0131\u015ftan te\u011fet olan O1 ve O2 merkezli \u00e7emberlere uygulam\u0131\u015f oldu\u011fu kuvvetler e\u015fittir.\n|PB|=|PA|=|PC| \u00db |BA]^[AC] \nYar\u0131\u00e7aplar\u0131 kesi\u015fim noktalar\u0131nda dik olan \u00e7emberlere dik kesi\u015fen \u00e7emberler denir.\nd. Kesi\u015fmeyen \u00e7emberlerin ortak noktas\u0131 yoktur. Kuvvet ekseni iki \u00e7emberin aras\u0131nda ve \u00e7emberlerin merkezlerini birle\u015ftiren do\u011fruya diktir. |O1O2| > r1 + r2<\/p>\n
4. Ortak Te\u011fet Par\u00e7as\u0131n\u0131n Uzunlu\u011fu<\/p>\n
Ortak te\u011fet uzunlu\u011funun bulunabilmesi i\u00e7in merkezlerden te\u011fetlere dikler \u00e7izilir.
\nO1O2C dik \u00fc\u00e7geninde |CO2| = |AB|<\/p>\n
|AB|2 =|O1O2|2 – |r1-r2|2
\n5. Bir Do\u011fru \u0130le Bir \u00c7emberin Durumlar\u0131<\/p>\n
Ayn\u0131 d\u00fczlemde bulunan O merkezli r yar\u0131\u00e7apl\u0131 bir \u00e7ember ile d do\u011frusu \u00fc\u00e7 farkl\u0131 durumda bulunur.
\na. |OH| > r ise
\ndo\u011fru \u00e7emberi kesmez ve do\u011fru \u00e7emberin d\u0131\u015f\u0131ndad\u0131r.
\n\u00c7ember \u00c7 d = \u00c6<\/p>\n
b. |OH| = r ise
\ndo\u011fru \u00e7emberi bir noktada keser. Yani do\u011fru \u00e7embere te\u011fettir.
\n\u00c7ember \u00c7 d = {H}<\/p>\n
c. |OH| < r ise \ndo\u011fru \u00e7emberi iki noktada keser.\n\u00c7ember \u00c7 d = {A, B}\n<\/p>\n","protected":false},"excerpt":{"rendered":"
TE\u011eET – K\u0130R\u0130\u015e \u00d6ZELL\u0130KLER\u0130 1. Te\u011fet noktas\u0131ndan ve \u00e7emberin merkezinden ge\u00e7en do\u011fru, te\u011fet olan do\u011fruya diktir.AB do\u011frusu T noktas\u0131nda \u00e7embere te\u011fet AB ^ OT Te\u011fet do\u011frusuna, te\u011fet noktas\u0131ndan \u00e7izilen dik do\u011fru \u00e7emberin merkezinden ge\u00e7er. 2. \u00c7emberin d\u0131\u015f\u0131ndaki bir noktadan \u00e7embere \u00e7izilen te\u011fetlerin uzuluklar\u0131 birbirine e\u015fittir. [PA ve [PT \u00e7embere te\u011fet |PA| = |PB| [PT ve …<\/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":[7317,7318],"class_list":["post-3111","post","type-post","status-publish","format-standard","hentry","category-matematik-odevleri","category-odevler","tag-cemberde-teget-kiris-ozellikleri","tag-tegetler-dortgeni"],"_links":{"self":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3111","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=3111"}],"version-history":[{"count":0,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3111\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/media?parent=3111"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/categories?post=3111"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/tags?post=3111"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}