MCQ
In trapezium $\text{ABCD},$ if $\text{AB}\parallel\text{DC},\text{AB}\parallel\text{DC}, AB = 9\ cm, DC = 6\ cm$ and $BD = 12\ cm,$ then $BO$ is equal to :
- A$7.4\ cm.$
- B$7\ cm.$
- C$7.5\ cm.$
- ✓$7.2\ cm.$
✓
Answer
Correct option: D.
$7.2\ cm.$
In $\triangle\text{COD }$and $\triangle\text{AOB}$
$\angle\text{DOC}=\angle\text{AOB}\ [$vertically opposite$]$
And $\angle\text{DCO}=\angle\text{OAB} \ [$Alternate angles$]$
$\Rightarrow\triangle\text{COD}\sim\triangle\text{AOB}\ [$similarity$]$
Let $OB = x\ cm$
$\therefore\frac{\text{AB}}{\text{CD}}=\frac{\text{OB}}{\text{OD}}$
$\Rightarrow\frac{9}{6}=\frac{\text{x}}{12-\text{x}}$
$\Rightarrow108-9\text{x}=6\text{x}$
$\Rightarrow15\text{x}=108$
$\Rightarrow\text{x}=7.2\text{ cm}$
$\angle\text{DOC}=\angle\text{AOB}\ [$vertically opposite$]$
And $\angle\text{DCO}=\angle\text{OAB} \ [$Alternate angles$]$
$\Rightarrow\triangle\text{COD}\sim\triangle\text{AOB}\ [$similarity$]$
Let $OB = x\ cm$
$\therefore\frac{\text{AB}}{\text{CD}}=\frac{\text{OB}}{\text{OD}}$
$\Rightarrow\frac{9}{6}=\frac{\text{x}}{12-\text{x}}$
$\Rightarrow108-9\text{x}=6\text{x}$
$\Rightarrow15\text{x}=108$
$\Rightarrow\text{x}=7.2\text{ cm}$
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