Question Number 44068 by peter frank last updated on 21/Sep/18
Answered by kunal1234523 last updated on 21/Sep/18
$${join}\:{OB}\:{and}\:{OD}\:={R} \\ $$$${l}^{\mathrm{2}} +{x}^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:\:…..\left(\mathrm{1}\right) \\ $$$${h}^{\mathrm{2}} +\left(\mathrm{2}{x}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:…….\left(\mathrm{2}\right) \\ $$$${from}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right) \\ $$$${l}^{\mathrm{2}} +{x}^{\mathrm{2}} ={h}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} ={l}^{\mathrm{2}} −{h}^{\mathrm{2}} \\ $$$${x}=\sqrt{\frac{\mathrm{1}}{\mathrm{3}}\left({l}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)}\:\:\:\:\:\:\:\:….\left(\mathrm{3}\right) \\ $$$${AB}=\mathrm{2}{x}\:\:\:\:\left({chord}\:{of}\:{the}\:{circle}\right) \\ $$$${AB}=\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{3}}\left({l}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)}\:\:\:\left({from}\:\left(\mathrm{3}\right)\right) \\ $$$${AB}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\sqrt{{l}^{\mathrm{2}} −{h}^{\mathrm{2}} } \\ $$$${AB}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\sqrt{{l}^{\mathrm{2}} −{h}^{\mathrm{2}} } \\ $$$${AB}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{3}\left({l}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)} \\ $$
Answered by kunal1234523 last updated on 21/Sep/18
