D\u0130K PR\u0130ZMALARIN ALAN ve HAC\u0130MLER\u0130
\nAlt ve \u00fcst tabanlar\u0131 paralel e\u015f \u015fekillerden olu\u015fan cisimlere prizma denir. Yan y\u00fczeyleri taban d\u00fczlemine dik olan prizmalara dik prizma ad\u0131 verilir.<\/p>\n
Prizmalarda yan y\u00fczeyleri birle\u015ftiren ayr\u0131tlara yanal ayr\u0131t denir.
\n[AA’], [BB’], [CC’], [DD’]
\nyanal ayr\u0131tlard\u0131r.
\nDik prizmalarda yanal ayr\u0131t cismin y\u00fcksekli\u011fine e\u015fittir.
\nCismin y\u00fcksekli\u011fine h dersek
\nh = |AA’| = |BB’| = |CC’| = |DD’| olur.
\n Prizman\u0131n Hacmi
\nHacim=Taban Alan\u0131 x Y\u00fckseklik <\/p>\n
Dik prizman\u0131n taban bi\u00e7imi nas\u0131l olursa olsun, yanal y\u00fczeyi daima bir dikd\u00f6rtgen olur. Yanal y\u00fcz\u00fc olu\u015fturan dikd\u00f6rtgenin alt kenar\u0131 taban\u0131n \u00e7evresi kadard\u0131r. Di\u011fer kenar\u0131 ise h y\u00fcksekli\u011fi kadar olur.
\nYanal Alan = Taban \u00e7evresi x Y\u00fckseklik B\u00fct\u00fcn dik prizmalar\u0131n yanal alan\u0131 taban \u00e7evresi ile y\u00fcksekli\u011fin \u00e7arp\u0131m\u0131d\u0131r. B\u00fct\u00fcn Alan ise yanal alan ile iki taban alan\u0131n\u0131n toplam\u0131d\u0131r.
\nT\u00fcm Alan = Yanal Alan + 2. Taban Alan\u0131 1. Dikd\u00f6rtgenler Prizmas\u0131
\nDikd\u00f6rtgenler prizmas\u0131 yan y\u00fczeyleri kar\u015f\u0131l\u0131kl\u0131 iki\u015fer iki\u015fer e\u015f olan alt\u0131 adet dikd\u00f6rtgenden olu\u015fan prizmad\u0131r. Burada hacim, taban alan\u0131 olan (a.b) ile y\u00fckseklik olan (c) nin \u00e7arp\u0131m\u0131d\u0131r. Alan ise (a.b), (b.c) ve (a.c) y\u00fczey alanlar\u0131n\u0131n iki\u015fer katlar\u0131n\u0131n toplam\u0131d\u0131r. Dikd\u00f6rtgenler prizmas\u0131nda birbirine en uzak iki k\u00f6\u015feyi birle\u015ftiren do\u011fru par\u00e7as\u0131na cisim k\u00f6\u015fegeni denir.
\nCisim k\u00f6\u015fegeni daima prizman\u0131n i\u00e7inden ge\u00e7er. Y\u00fczeylerinden ge\u00e7mez. Sadece bir y\u00fczeyden ge\u00e7en k\u00f6\u015fegene o y\u00fcze ait y\u00fczey k\u00f6\u015fegeni denir. Burada k\u00f6\u015fegenlerin uzunluklar\u0131
\n|AC’| = |A’C| = |BD’| = |B’D| = e (cisim k\u00f6\u015fegeni)
\n|BD| = f (Y\u00fczey k\u00f6\u015fegeni) olsun. Bu durumda
\nHacim = a.b.c
\nAlan =2(ab+bc+ac)
\nAlan = 2 (ab + bc + ac)
\nCisim K\u00f6\u015fegeni: e =\u00d6a2 + b2 + c2
\nY\u00fczey K\u00f6\u015fegeni: f = \u00d6a2 + b2
\n2. Kare Prizma
\nTaban\u0131 kare olan prizmalara kare prizma denir. Yan y\u00fcz\u00fc d\u00f6rt adet e\u015f dikd\u00f6rtgenden olu\u015fur.<\/p>\n
Hacim = a2 . h Yanal Alan = 4 . a . h
\nAlan = 4.ah + 2.a2 Cisim k\u00f6\u015fegeni : e = \u00d6a2 + a2 + h2
\n3. K\u00fcp
\nB\u00fct\u00fcn ayr\u0131tlar\u0131 birbirine e\u015fit olan dik prizmaya k\u00fcp denir. T\u00fcm y\u00fczeyleri kare dir.<\/p>\n
Hacim = a3
\nAlan = 6a2
\nK\u00fcb\u00fcn y\u00fczey k\u00f6\u015fegenleri birbirine e\u015fittir.
\nY\u00fczey k\u00f6\u015fegeni: f = a\u00d62
\nCisim k\u00f6\u015fegeni: e = a\u00d63
\n4. \u00dc\u00e7gen Prizmalar
\nPrizmalar tabanlar\u0131n\u0131n \u015fekline g\u00f6re isim ald\u0131klar\u0131ndan taban\u0131 \u00fc\u00e7gen olan prizmalara \u00fc\u00e7gen prizma denir.
\n\u00dc\u00e7gen prizmalar taban\u0131n\u0131 olu\u015fturan \u00fc\u00e7gene g\u00f6re isimlenir.
\na. E\u015fkenar \u00dc\u00e7gen Prizma
\nE\u015fkenar \u00fc\u00e7gen prizman\u0131n tabanlar\u0131 e\u015fkenar \u00fc\u00e7gendir. Yan y\u00fczeyleri ise \u00fc\u00e7 tane e\u015f dikd\u00f6rtgenden olu\u015fur.Taban\u0131 e\u015fkenar \u00fc\u00e7gen oldu\u011fundan<\/p>\n
Taban\u0131 e\u015fkenar \u00fc\u00e7gen oldu\u011fundan
\nTaban alan\u0131 Hacim Taban \u00e7evresi 3a oldu\u011fundan, yanal alan 3a.h d\u0131r.
\nBuradan t\u00fcm alan\u0131
\nT\u00fcm alan b. Dik \u00dc\u00e7gen Prizma
\nDik \u00fc\u00e7gen prizman\u0131n taban\u0131 dik \u00fc\u00e7gendir. Yan y\u00fczeyleri ise \u00fc\u00e7 tane dikd\u00f6rtgenden olu\u015fur.<\/p>\n
Taban\u0131 dik \u00fc\u00e7gen oldu\u011fundan
\nTaban alan\u0131 = Hacim Taban \u00e7evresi a + b + c oldu\u011fundan,
\nYanal alan = (a + b + c) . h
\nT\u00fcm Alan = b . c + (a + b + c) . h
\n5. Silindir
\nTaban\u0131 daire olan prizmalara silindir denir. Silindirin yan y\u00fcz\u00fc dikd\u00f6rtgen bi\u00e7imindedir. Dikd\u00f6rtgenin bir kenar\u0131 y\u00fckseklik kadar, di\u011fer kenar\u0131 ise taban dairesinin \u00e7evresi kadard\u0131r.<\/p>\n
Taban alan\u0131= pr2
\nHacim= pr2h Taban \u00e7evresi 2pr oldu\u011fundan yanal alan 2prh olur.
\nT\u00fcm alan = 2prh+ 2pr Bir dikd\u00f6rtgen levha bir kenar\u0131 etraf\u0131nda d\u00f6nd\u00fcr\u00fcld\u00fc\u011f\u00fcnde silindir elde edilir.
\n 6. D\u00fczg\u00fcn \u00c7okgen Prizmalar
\nTaban\u0131 d\u00fczg\u00fcn \u00e7okgenlerden olu\u015fan prizmalara d\u00fczg\u00fcn \u00e7okgen prizmalar deriz. Taban ayr\u0131tlar\u0131 birbirine e\u015fittir. Di\u011fer dik prizmalarda oldu\u011fu gibi d\u00fczg\u00fcn \u00e7okgen prizmalarda da yanal ayr\u0131t ayn\u0131 zamanda y\u00fcksekliktir.
\nDik prizmalarda taban \u015fekli ne olursa olsun, hacmin taban alan\u0131 ile y\u00fcksekli\u011fin \u00e7arp\u0131m\u0131 ve yanal alan\u0131n ise taban \u00e7evresi ile y\u00fcksekli\u011fin \u00e7arp\u0131m\u0131 oldu\u011funu unutmayal\u0131m.
\nE\u011e\u0130K PR\u0130ZMALAR
\n1. E\u011fik Kare Prizma<\/p>\n
Taban\u0131, bir kenar\u0131 a olan kareden olu\u015fan prizma bir y\u00f6ne do\u011fru taban d\u00fczlemi ile a a\u00e7\u0131s\u0131 yapacak kadar e\u011filirse e\u011fik kare prizma elde edilir.
\nPrizman\u0131n yanal ayr\u0131tlar\u0131na l dersek,
\nPrizman\u0131n y\u00fcksekli\u011fi h =l .sin a olur.
\nE\u011fik prizman\u0131n yanal ayr\u0131tlar\u0131na dik olacak \u015fekilde olu\u015fan kesitine dik kesit denir. E\u011fik kare prizman\u0131n iki yan y\u00fczeyi dikd\u00f6rtgen, di\u011fer iki yan y\u00fczeyi ise paralelkenard\u0131r.
\nE\u011fik kare prizman\u0131n dik kesitinin bir kenar\u0131 taban kenar\u0131 a kadar, di\u011feri ise,
\na’=a.sin a kadard\u0131r.
\nBuradan;
\nDik Kesit Alan\u0131 = Taban Alan\u0131 x Sin a
\nDik kesit \u00e7evresi = 2a +2a.sin a E\u011fik prizmalar\u0131n yanal alanlar\u0131n\u0131n toplam\u0131
\nYanal alan= Dik kesit \u00e7evresi x Yanal Ayr\u0131t ba\u011f\u0131nt\u0131s\u0131 ile bulunur. Alt ve \u00fcst tabanlar ilave edildi\u011finde t\u00fcm alan bulunmu\u015f olur. B\u00fct\u00fcn prizmalarda oldu\u011fu gibi e\u011fik prizmalarda da hacim, taban alan\u0131 ile y\u00fcksekli\u011fin \u00e7arp\u0131m\u0131 ile bulunur.
\nHacim = Taban Alan\u0131 x Y\u00fckseklik Ayr\u0131ca dik kesit alan\u0131 ile yanal ayr\u0131t\u0131n \u00e7arp\u0131m\u0131 ile de hacim bulunabilir.
\nHacim = Dik Kesit Alan\u0131 x Yanal Ayr\u0131t
\n2. E\u011fik Silindir
\n|AA’| = |BB’| = l
\nYanal ayr\u0131t\u0131 l olan ve taban d\u00fczlemi ile a a\u00e7\u0131s\u0131 yapan e\u011fik silindirde y\u00fckseklik,
\nh=l.sin a
\nDik Kesit Alan\u0131=Taban Alan\u0131 x Sin a E\u011fik silindirin yan y\u00fcz alan\u0131, dik kesit \u00e7evresi ile yanal ayr\u0131t\u0131n\u0131n \u00e7arp\u0131m\u0131d\u0131r. B\u00fct\u00fcn e\u011fik prizmalarda oldu\u011fu gibi e\u011fik silindir de de hacim, dik kesit alan\u0131 ile yanal ayr\u0131t\u0131n \u00e7arp\u0131m\u0131na e\u015fittir.
\nHacim = Taban Alan\u0131 x Y\u00fckseklik
\nHacim = Dik Kesit Alan\u0131 x Yanal Ayr\u0131t
\nYanal Alan = Dik Kesit \u00c7evresi x Yanal Ayr\u0131t
\nD\u0130K PR\u0130ZMALARIN ALAN ve HAC\u0130MLER\u0130
\nAlt ve \u00fcst tabanlar\u0131 paralel e\u015f \u015fekillerden olu\u015fan cisimlere prizma denir. Yan y\u00fczeyleri taban d\u00fczlemine dik olan prizmalara dik prizma ad\u0131 verilir.<\/p>\n
Prizmalarda yan y\u00fczeyleri birle\u015ftiren ayr\u0131tlara yanal ayr\u0131t denir.
\n[AA’], [BB’], [CC’], [DD’]
\nyanal ayr\u0131tlard\u0131r.
\nDik prizmalarda yanal ayr\u0131t cismin y\u00fcksekli\u011fine e\u015fittir.
\nCismin y\u00fcksekli\u011fine h dersek
\nh = |AA’| = |BB’| = |CC’| = |DD’| olur.<\/p>\n
Prizman\u0131n Hacmi
\nHacim=Taban Alan\u0131 x Y\u00fckseklik <\/p>\n
Dik prizman\u0131n taban bi\u00e7imi nas\u0131l olursa olsun, yanal y\u00fczeyi daima bir dikd\u00f6rtgen olur. Yanal y\u00fcz\u00fc olu\u015fturan dikd\u00f6rtgenin alt kenar\u0131 taban\u0131n \u00e7evresi kadard\u0131r. Di\u011fer kenar\u0131 ise h y\u00fcksekli\u011fi kadar olur.
\nYanal Alan = Taban \u00e7evresi x Y\u00fckseklik B\u00fct\u00fcn dik prizmalar\u0131n yanal alan\u0131 taban \u00e7evresi ile y\u00fcksekli\u011fin \u00e7arp\u0131m\u0131d\u0131r. B\u00fct\u00fcn Alan ise yanal alan ile iki taban alan\u0131n\u0131n toplam\u0131d\u0131r.
\nT\u00fcm Alan = Yanal Alan + 2. Taban Alan\u0131 1. Dikd\u00f6rtgenler Prizmas\u0131
\nDikd\u00f6rtgenler prizmas\u0131 yan y\u00fczeyleri kar\u015f\u0131l\u0131kl\u0131 iki\u015fer iki\u015fer e\u015f olan alt\u0131 adet dikd\u00f6rtgenden olu\u015fan prizmad\u0131r. Burada hacim, taban alan\u0131 olan (a.b) ile y\u00fckseklik olan (c) nin \u00e7arp\u0131m\u0131d\u0131r. Alan ise (a.b), (b.c) ve (a.c) y\u00fczey alanlar\u0131n\u0131n iki\u015fer katlar\u0131n\u0131n toplam\u0131d\u0131r. Dikd\u00f6rtgenler prizmas\u0131nda birbirine en uzak iki k\u00f6\u015feyi birle\u015ftiren do\u011fru par\u00e7as\u0131na cisim k\u00f6\u015fegeni denir.
\nCisim k\u00f6\u015fegeni daima prizman\u0131n i\u00e7inden ge\u00e7er. Y\u00fczeylerinden ge\u00e7mez. Sadece bir y\u00fczeyden ge\u00e7en k\u00f6\u015fegene o y\u00fcze ait y\u00fczey k\u00f6\u015fegeni denir. Burada k\u00f6\u015fegenlerin uzunluklar\u0131
\n|AC’| = |A’C| = |BD’| = |B’D| = e (cisim k\u00f6\u015fegeni)
\n|BD| = f (Y\u00fczey k\u00f6\u015fegeni) olsun. Bu durumda
\nHacim = a.b.c
\nAlan =2(ab+bc+ac)
\nAlan = 2 (ab + bc + ac)
\nCisim K\u00f6\u015fegeni: e =\u00d6a2 + b2 + c2
\nY\u00fczey K\u00f6\u015fegeni: f = \u00d6a2 + b2
\n2. Kare Prizma
\nTaban\u0131 kare olan prizmalara kare prizma denir. Yan y\u00fcz\u00fc d\u00f6rt adet e\u015f dikd\u00f6rtgenden olu\u015fur.<\/p>\n
Hacim = a2 . h Yanal Alan = 4 . a . h
\nAlan = 4.ah + 2.a2 Cisim k\u00f6\u015fegeni : e = \u00d6a2 + a2 + h2
\n3. K\u00fcp
\nB\u00fct\u00fcn ayr\u0131tlar\u0131 birbirine e\u015fit olan dik prizmaya k\u00fcp denir. T\u00fcm y\u00fczeyleri kare dir.<\/p>\n
Hacim = a3
\nAlan = 6a2
\nK\u00fcb\u00fcn y\u00fczey k\u00f6\u015fegenleri birbirine e\u015fittir.
\nY\u00fczey k\u00f6\u015fegeni: f = a\u00d62
\nCisim k\u00f6\u015fegeni: e = a\u00d63
\n4. \u00dc\u00e7gen Prizmalar
\nPrizmalar tabanlar\u0131n\u0131n \u015fekline g\u00f6re isim ald\u0131klar\u0131ndan taban\u0131 \u00fc\u00e7gen olan prizmalara \u00fc\u00e7gen prizma denir.
\n\u00dc\u00e7gen prizmalar taban\u0131n\u0131 olu\u015fturan \u00fc\u00e7gene g\u00f6re isimlenir.
\na. E\u015fkenar \u00dc\u00e7gen Prizma
\nE\u015fkenar \u00fc\u00e7gen prizman\u0131n tabanlar\u0131 e\u015fkenar \u00fc\u00e7gendir. Yan y\u00fczeyleri ise \u00fc\u00e7 tane e\u015f dikd\u00f6rtgenden olu\u015fur.Taban\u0131 e\u015fkenar \u00fc\u00e7gen oldu\u011fundan<\/p>\n
Taban\u0131 e\u015fkenar \u00fc\u00e7gen oldu\u011fundan
\nTaban alan\u0131 Hacim Taban \u00e7evresi 3a oldu\u011fundan, yanal alan 3a.h d\u0131r.
\nBuradan t\u00fcm alan\u0131
\nT\u00fcm alan
\nb. Dik \u00dc\u00e7gen Prizma
\nDik \u00fc\u00e7gen prizman\u0131n taban\u0131 dik \u00fc\u00e7gendir. Yan y\u00fczeyleri ise \u00fc\u00e7 tane dikd\u00f6rtgenden olu\u015fur.<\/p>\n
Taban\u0131 dik \u00fc\u00e7gen oldu\u011fundan
\nTaban alan\u0131 = Hacim Taban \u00e7evresi a + b + c oldu\u011fundan,
\nYanal alan = (a + b + c) . h
\nT\u00fcm Alan = b . c + (a + b + c) . h
\n5. Silindir
\nTaban\u0131 daire olan prizmalara silindir denir. Silindirin yan y\u00fcz\u00fc dikd\u00f6rtgen bi\u00e7imindedir. Dikd\u00f6rtgenin bir kenar\u0131 y\u00fckseklik kadar, di\u011fer kenar\u0131 ise taban dairesinin \u00e7evresi kadard\u0131r.<\/p>\n
Taban alan\u0131= pr2
\nHacim= pr2h Taban \u00e7evresi 2pr oldu\u011fundan yanal alan 2prh olur.
\nT\u00fcm alan = 2prh+ 2pr Bir dikd\u00f6rtgen levha bir kenar\u0131 etraf\u0131nda d\u00f6nd\u00fcr\u00fcld\u00fc\u011f\u00fcnde silindir elde edilir.<\/p>\n
6. D\u00fczg\u00fcn \u00c7okgen Prizmalar
\nTaban\u0131 d\u00fczg\u00fcn \u00e7okgenlerden olu\u015fan prizmalara d\u00fczg\u00fcn \u00e7okgen prizmalar deriz. Taban ayr\u0131tlar\u0131 birbirine e\u015fittir. Di\u011fer dik prizmalarda oldu\u011fu gibi d\u00fczg\u00fcn \u00e7okgen prizmalarda da yanal ayr\u0131t ayn\u0131 zamanda y\u00fcksekliktir.
\nDik prizmalarda taban \u015fekli ne olursa olsun, hacmin taban alan\u0131 ile y\u00fcksekli\u011fin \u00e7arp\u0131m\u0131 ve yanal alan\u0131n ise taban \u00e7evresi ile y\u00fcksekli\u011fin \u00e7arp\u0131m\u0131 oldu\u011funu unutmayal\u0131m.
\nE\u011e\u0130K PR\u0130ZMALAR
\n1. E\u011fik Kare Prizma<\/p>\n
Taban\u0131, bir kenar\u0131 a olan kareden olu\u015fan prizma bir y\u00f6ne do\u011fru taban d\u00fczlemi ile a a\u00e7\u0131s\u0131 yapacak kadar e\u011filirse e\u011fik kare prizma elde edilir.
\nPrizman\u0131n yanal ayr\u0131tlar\u0131na l dersek,
\nPrizman\u0131n y\u00fcksekli\u011fi h =l .sin a olur.
\nE\u011fik prizman\u0131n yanal ayr\u0131tlar\u0131na dik olacak \u015fekilde olu\u015fan kesitine dik kesit denir. E\u011fik kare prizman\u0131n iki yan y\u00fczeyi dikd\u00f6rtgen, di\u011fer iki yan y\u00fczeyi ise paralelkenard\u0131r.
\nE\u011fik kare prizman\u0131n dik kesitinin bir kenar\u0131 taban kenar\u0131 a kadar, di\u011feri ise,
\na’=a.sin a kadard\u0131r.
\nBuradan;
\nDik Kesit Alan\u0131 = Taban Alan\u0131 x Sin a
\nDik kesit \u00e7evresi = 2a +2a.sin a E\u011fik prizmalar\u0131n yanal alanlar\u0131n\u0131n toplam\u0131
\nYanal alan= Dik kesit \u00e7evresi x Yanal Ayr\u0131t ba\u011f\u0131nt\u0131s\u0131 ile bulunur. Alt ve \u00fcst tabanlar ilave edildi\u011finde t\u00fcm alan bulunmu\u015f olur. B\u00fct\u00fcn prizmalarda oldu\u011fu gibi e\u011fik prizmalarda da hacim, taban alan\u0131 ile y\u00fcksekli\u011fin \u00e7arp\u0131m\u0131 ile bulunur.
\nHacim = Taban Alan\u0131 x Y\u00fckseklik Ayr\u0131ca dik kesit alan\u0131 ile yanal ayr\u0131t\u0131n \u00e7arp\u0131m\u0131 ile de hacim bulunabilir.
\nHacim = Dik Kesit Alan\u0131 x Yanal Ayr\u0131t
\n2. E\u011fik Silindir
\n|AA’| = |BB’| = l
\nYanal ayr\u0131t\u0131 l olan ve taban d\u00fczlemi ile a a\u00e7\u0131s\u0131 yapan e\u011fik silindirde y\u00fckseklik,
\nh=l.sin a
\nDik Kesit Alan\u0131=Taban Alan\u0131 x Sin a E\u011fik silindirin yan y\u00fcz alan\u0131, dik kesit \u00e7evresi ile yanal ayr\u0131t\u0131n\u0131n \u00e7arp\u0131m\u0131d\u0131r. B\u00fct\u00fcn e\u011fik prizmalarda oldu\u011fu gibi e\u011fik silindir de de hacim, dik kesit alan\u0131 ile yanal ayr\u0131t\u0131n \u00e7arp\u0131m\u0131na e\u015fittir.
\nHacim = Taban Alan\u0131 x Y\u00fckseklik
\nHacim = Dik Kesit Alan\u0131 x Yanal Ayr\u0131t
\nYanal Alan = Dik Kesit \u00c7evresi x Yanal Ayr\u0131t<\/p>\n","protected":false},"excerpt":{"rendered":"
D\u0130K PR\u0130ZMALARIN ALAN ve HAC\u0130MLER\u0130 Alt ve \u00fcst tabanlar\u0131 paralel e\u015f \u015fekillerden olu\u015fan cisimlere prizma denir. Yan y\u00fczeyleri taban d\u00fczlemine dik olan prizmalara dik prizma ad\u0131 verilir. Prizmalarda yan y\u00fczeyleri birle\u015ftiren ayr\u0131tlara yanal ayr\u0131t denir. [AA’], [BB’], [CC’], [DD’] yanal ayr\u0131tlard\u0131r. Dik prizmalarda yanal ayr\u0131t cismin y\u00fcksekli\u011fine e\u015fittir. Cismin y\u00fcksekli\u011fine h dersek h = |AA’| …<\/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":[7308,7306,7312,7295,7310,7220,7311,7305,7307,7309],"class_list":["post-3107","post","type-post","status-publish","format-standard","hentry","category-matematik-odevleri","category-odevler","tag-cisim-kosegeni","tag-dik-prizma","tag-dik-ucgen-prizma","tag-dikdortgenler-prizmasi","tag-egik-kare-prizma","tag-eskenar-ucgen","tag-kare-prizma","tag-prizmalar","tag-prizmanin-hacmi","tag-yuzey-kosegeni"],"_links":{"self":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3107","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=3107"}],"version-history":[{"count":0,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3107\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/media?parent=3107"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/categories?post=3107"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/tags?post=3107"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}