Question
In a trapezium ABCD, it is given that AB || CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that $\text{ar}(\triangle\text{AOB})=84\text{cm}^2.$ Find $\text{ar}(\triangle\text{COD}).$
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Answer
The diagonals of a trapezium divide each other proportionally.$\angle\text{CDO}=\angle\text{OBA}$ ...(alternate angles)
$\angle\text{COD}=\angle\text{AOB}$ ...(vertically opposite angles)
$\Rightarrow\triangle\text{COD}=\triangle\text{AOB}$ ...(AA criterion for similarity)
$\Rightarrow\frac{\text{ar}(\triangle\text{COD})}{\text{ar}(\triangle\text{AOB})}=\frac{\text{CD}^2}{\text{AB}^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{COD})}{84}=\frac{1^2}{2^2}$
$\Rightarrow\text{ar}(\triangle\text{COD})=21\text{cm}^2$
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