Question Number 206227 by bett last updated on 09/Apr/24
$${OA}=\left(\overset{{x}} {\mathrm{4}}\right)\:{OB}=_{\mathrm{7}} ^{\mathrm{5}} \:{and}\:{AB}=\mathrm{5}\:{units} \\ $$
Answered by A5T last updated on 09/Apr/24
$$\sqrt{\left(\mathrm{5}−{x}\right)^{\mathrm{2}} +\left(\mathrm{7}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{5}\Rightarrow\mathrm{25}−\mathrm{16}=\mathrm{3}^{\mathrm{2}} =\left(\mathrm{5}−{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{5}−{x}=\underset{−} {+}\mathrm{3}\Rightarrow{x}=\mathrm{2}\:{or}\:\mathrm{8} \\ $$