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Q2 Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

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Given: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.

 AB = 2 CD          ( Given )

In \triangle AOB\, and\, \triangle COD,

\angle COD=\angle AOB          (vertically opposite angles )

\angle OCD=\angle OAB           (Alternate angles)

\angle ODC=\angle OBA           (Alternate angles)

\therefore \triangle AOB\, \sim \, \triangle COD       (AAA similarity)

\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{AB^2}{CD^2}

\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{(2CD)^2}{CD^2}

\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{4.CD^2}{CD^2}

\Rightarrow \frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{4}{1}

\Rightarrow ar(\triangle AOB)=ar(\triangle COD)=4:1

 

 

 

 

 

 

 

 

 

 

Posted by

seema garhwal

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