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

Question

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

Answer 1

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}$

$\large\frac{9}{6} = \frac{x}{12-x}$

=> $108 -9x =6x$

=> $15 x =108$

=> $x = 7.2 \;cm$