Question

$\text{ABCD}$ is a trapezium such that $BC\parallel AD$ and $\text{AD}\text{=}\text{4cm}\text{.}$ If the diagonals $\text{AC}$ and $\text{BD}$ intersect at O such that $\frac{\text{AO}}{\text{OC}}\text{=}\frac{\text{DO}}{\text{OB}}\text{=}\frac{\text{1}}{\text{2}}\text{,}\text{then}\text{BC}\text{=}$

• A. $\text{7}\text{cm}$
• B. $8\text{cm}$
• C. $9\text{cm}$
• D. $6\text{cm}$

Answer 1

The correct option is B $8\text{cm}$

Given $BC\parallel AD$ and $AD=4$

Let $BC=m$

Let $\angle AOD=x\Rightarrow \angle COD=x$

Let $\angle OAD=y\Rightarrow \angle OCB=y$

Let $CO=2a\Rightarrow OA=a$

$\Rightarrow △OAD$ and $△OCB$ are similar.

$\Rightarrow \frac{OA}{AD}=\frac{OC}{CB}$

$\Rightarrow \frac{a}{4}=\frac{2a}{m}$

$\Rightarrow m=8$

Hence,$BC=m=8$