In triunghiul isoscel ABC , AB = AC , se duc inaltimile AD perpendicular pe BC , D apartine pe BC si BE perpendicular pe AC , E apartine pe AC. Stiind ca AB = 25 , BC = 40 , aflati AD , BE ,AE , CE .
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2014-09-14T18:20:15+03:00
AD perpendicular pe BC=> AD e mediana=> D e mijlocul lui BC=>BD=BC/2=> BD=20 si DC=20.
In triunghiul ABD, dreptunghic in D aplicam Pitagora:
 AB^{2}= BD^{2} + AD^{2} =>  AD^{2}=625-400=225=> AD=15
In triunghiul BEC, dreptunghic in E aplicam Pitagora:
 BC^{2}= BE^{2} + EC^{2} =>  BE^{2}= 40^{2}-EC^{2} .
In triunghiul BEA, dreptunghic in E aplicam Pitagora:
 AB^{2}= BE^{2} + EA^{2} =>  BE^{2}= 25^{2}-EA^{2} .
Cumulate, cele doua relatii conduc catre   40^{2}-EC^{2}= 25^{2}-EA^{2} =>
1600- EC^{2}=625- EA^{2}  =>1600- EC^{2}=625- (AC-EC)^{2}=>
1600- EC^{2}=625- (25-EC)^{2}=>1600- EC^{2}=625-625+50EC- EC^{2} => 50EC=1600=> EC=32=> BE^{2}= 40^{2}-32^{2} => BE^{2}= 1600-1024=>BE=24=> 24^{2}= 25^{2}-EA^{2} =>EA=7.
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2014-09-14T18:43:35+03:00
AD perpendicular pe BC=> AD e mediana=> D e mijlocul lui BC=>BD=BC/2=> BD=20 si DC=20. In triunghiul ABD, dreptunghic in D aplicam Pitagora: AB^{2}= BD^{2} + AD^{2} => AD^{2}=625-400=225=> AD=15 In triunghiul BEC, dreptunghic in E aplicam Pitagora: BC^{2}= BE^{2} + EC^{2} => BE^{2}= 40^{2}-EC^{2} . In triunghiul BEA, dreptunghic in E aplicam Pitagora: AB^{2}= BE^{2} + EA^{2} => BE^{2}= 25^{2}-EA^{2} . Cumulate, cele doua relatii conduc catre 40^{2}-EC^{2}= 25^{2}-EA^{2} => 1600- EC^{2}=625- EA^{2} =>1600- EC^{2}=625- (AC-EC)^{2}=> 1600- EC^{2}=625- (25-EC)^{2}=>1600- EC^{2}=625-625+50EC- EC^{2} => 50EC=1600=> EC=32=> BE^{2}= 40^{2}-32^{2} => BE^{2}= 1600-1024=>BE=24=> 24^{2}= 25^{2}-EA^{2} =>EA=7
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