Question Number 151444 by liberty last updated on 21/Aug/21
Answered by mr W last updated on 21/Aug/21
$${say}\:{BE}={diameter}=\mathrm{2}×{AB} \\ $$$$\frac{{BC}}{{BO}}=\frac{{BO}}{{BE}} \\ $$$$\Rightarrow{BC}×{BE}={BO}^{\mathrm{2}} \\ $$$$\Rightarrow{BC}×\mathrm{2}×{AB}={BO}^{\mathrm{2}} \\ $$$$\Rightarrow\left[{ABCD}\right]={BC}×{AB}=\frac{{BO}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{10}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{50} \\ $$
Commented by mr W last updated on 21/Aug/21
Commented by Tawa11 last updated on 21/Aug/21
$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$
Answered by liberty last updated on 21/Aug/21