Question Number 94508 by john santu last updated on 19/May/20
Answered by mr W last updated on 19/May/20
$${OD}={DB}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{2}}}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${OE}={ED}={OA}=\mathrm{6} \\ $$$$\mathrm{cos}\:\angle{ODE}=\frac{{OD}}{\mathrm{2}{ED}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${OC}^{\mathrm{2}} ={OD}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}{ODx}×\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\mathrm{36}=\mathrm{18}+{x}^{\mathrm{2}} +\mathrm{3}{x} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3} \\ $$$$ \\ $$$${or} \\ $$$${EC}=\mathrm{2}×{OE}×\mathrm{cos}\:\angle{OED} \\ $$$$=\mathrm{2}×\mathrm{6}×\mathrm{cos}\:\left(\pi−\mathrm{2}\angle{ODE}\right) \\ $$$$=\mathrm{12}×\left(\mathrm{1}−\mathrm{2}×\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\right)^{\mathrm{2}} \right)=\mathrm{9} \\ $$$${x}=\mathrm{9}−\mathrm{6}=\mathrm{3} \\ $$
Commented by PRITHWISH SEN 2 last updated on 19/May/20
$$\mathrm{sir}\:\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be} \\ $$$$\mathrm{OD}=\mathrm{DB}\:\left(\mathrm{1}^{\mathrm{st}} \:\mathrm{line}\right) \\ $$
Commented by mr W last updated on 19/May/20
$${yes},\:{thanks}! \\ $$
Commented by john santu last updated on 19/May/20
great ��������
Answered by john santu last updated on 22/May/20
Commented by john santu last updated on 22/May/20
$$\Rightarrow\left(\mathrm{6}−\mathrm{y}\right)\left(\mathrm{6}+\mathrm{y}\right)\:=\:\mathrm{6}{x} \\ $$$$\Rightarrow\mathrm{36}−\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{6}{x}\:\:\:\left({i}\right) \\ $$$$\Rightarrow\mathrm{2y}^{\mathrm{2}} =\:\mathrm{36}\:;\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{18}\:\:\:\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{then}\:\mathrm{we}\:\mathrm{get}\:\mathrm{36}−\mathrm{18}\:=\:\mathrm{6}{x} \\ $$$$\mathrm{6}{x}\:=\:\mathrm{18}\:;\:{x}\:=\:\mathrm{3}\: \\ $$
Commented by i jagooll last updated on 22/May/20
$$\mathrm{waw}….. \\ $$