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Fie ABCD un trapez,AB paralel cu CD,iar AC intersectat cu BD prin punctul O.
Daca AB=10 cm CD=25 cm,AC=35cm si BD=42cm,Aflati lungimile segmentelor AO,CO,BO,DO


Răspuns :

<BOA≡<DOC si AB || CD => ΔAOB~ΔCOD => [tex] \frac{AO}{CO}= \frac{BO}{DO}= \frac{AB}{CD}[/tex].

[tex] \frac{AB}{CD}= \frac{10}{25}= \frac{2}{5} => \frac{AO}{CO}= \frac{BO}{DO}= \frac{2}{5} .[/tex]

Din [tex] \frac{AO}{CO}= \frac{2}{5}=> AO=2k ~si~CO=5k. [/tex]
AO+CO=AC=35 <=> 2k+5k=35 <=> 7k=35 => k=5 => AO=2*5=10 (cm) si CO=5*5=25 (cm).

Din [tex] \frac{BO}{DO}= \frac{2}{5}=>BO=2p~si~DO=5p. [/tex]
BO+DO=BD=42 <=> 2p+5p=42 <=> 7p=42 => p=6 => BO=2*7=14 (cm) si CO=5*7=35 (cm).