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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In $\triangle\text{AOB}$ and $\triangle\text{COD},$ we have:
$\angle\text{AOB}=\angle\text{COD}$ (Vertically opposite angles)
$\angle\text{OAB}=\angle\text{OCD}$ (Alternate angles as AB || CD)
Applying AA similiarity criterion, we get:
$\triangle\text{AOB}\sim\triangle\text{COD}$
$\therefore\frac{\text{ar}(\triangle\text{AOB})}{\text{ar}(\triangle\text{COD})}=\frac{\text{AB}^2}{\text{CD}^2}$
$\Rightarrow\frac{84}{\text{ar}(\triangle\text{COD})}=\Big(\frac{\text{AB}}{\text{CD}}\Big)^2$
$\Rightarrow\frac{84}{\text{ar}(\triangle\text{COD})}=\Big(\frac{2\text{CD}}{\text{CD}}\Big)^2$
$\Rightarrow\text{ar}(\triangle\text{COD})=\frac{84}{4}=21\text{cm}^2$
$\angle\text{AOB}=\angle\text{COD}$ (Vertically opposite angles)
$\angle\text{OAB}=\angle\text{OCD}$ (Alternate angles as AB || CD)
Applying AA similiarity criterion, we get:
$\triangle\text{AOB}\sim\triangle\text{COD}$
$\therefore\frac{\text{ar}(\triangle\text{AOB})}{\text{ar}(\triangle\text{COD})}=\frac{\text{AB}^2}{\text{CD}^2}$
$\Rightarrow\frac{84}{\text{ar}(\triangle\text{COD})}=\Big(\frac{\text{AB}}{\text{CD}}\Big)^2$
$\Rightarrow\frac{84}{\text{ar}(\triangle\text{COD})}=\Big(\frac{2\text{CD}}{\text{CD}}\Big)^2$
$\Rightarrow\text{ar}(\triangle\text{COD})=\frac{84}{4}=21\text{cm}^2$
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