Question Number 203282 by mr W last updated on 14/Jan/24
Commented by cortano12 last updated on 14/Jan/24
$$\mathrm{AD}\parallel\mathrm{BC}\:? \\ $$
Commented by mr W last updated on 15/Jan/24
$${square}\:{is}\:{inscribed}\:{in}\:{right}\:{angled} \\ $$$${trapezoid} \\ $$
Answered by behi834171 last updated on 15/Jan/24
$${SP}={x} \\ $$$$\frac{\frac{{x}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}}{\mathrm{2}}.\mathrm{8}=\frac{\sqrt{\mathrm{3}}{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \Rightarrow{x}=\mathrm{4} \\ $$$${SD}={DR}={y} \\ $$$${x}^{\mathrm{2}} =\mathrm{4}{y}^{\mathrm{2}} \Rightarrow{y}=\mathrm{2} \\ $$$${RC}={QR}=\mathrm{4}\Rightarrow\frac{{CR}}{{DR}}\overset{?} {=}\mathrm{2} \\ $$$${but}\bar {\:}{it}\:{seems}\:{that}\:{some}\:{thing}\:{went}\:{wrong}! \\ $$
Commented by mr W last updated on 16/Jan/24
$${indeed}\:{something}\:{went}\:{wrong}\:{sir}. \\ $$$${please}\:{recheck}. \\ $$
Answered by mr W last updated on 16/Jan/24
Commented by mr W last updated on 16/Jan/24
$${ABFE}\:{is}\:{also}\:{a}\:{square}. \\ $$$$\Rightarrow{AE}={BF}={AB}=\mathrm{8} \\ $$$${FC}=\mathrm{11}−\mathrm{8}=\mathrm{3} \\ $$$${DE}=\mathrm{8}−\mathrm{6}=\mathrm{2} \\ $$$$\frac{{CR}}{{DR}}=\frac{{FC}}{{DE}}=\frac{\mathrm{3}}{\mathrm{2}}\:\checkmark \\ $$