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Q)

In trapezium $ABCD$, if $AB = 9\; cm, DC = 6\; cm$ and $BD = 12\; cm$, then $BO$ is equal to

$\begin{array}{1 1} 7\;cm \\ 7.2 \;cm \\ 7.4 \;cm \\ 7.5 \;cm \end{array}$

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A)
Solution :
In $\Delta COD$ and $\Delta AOB$,
$\angle DOC = \angle AOB$ [vertically opposite ]
$\angle DOC = \angle AOB$ [Alternate angles ]
⇒$\Delta COD \approx \Delta AOB$ [AA similarity]
Let $OB = xx\; cm$
$\large\frac{AB}{CD} = \frac{OB}{OD}$
$\large\frac{9}{6} = \frac{x}{12-x}$
=> $108 -9x =6x$
=> $15 x =108$
=> $x = 7.2 \;cm$