Given, 1) X is 1 km northeast of Y.2) Y is 1 km southeast of Z.3) W is 1 km west of Z.4) P is 1 km south of W.5) Q is 1 km east of PWe can draw the following diagram according to the given conditions,From the figure,ZQ = 1 kmNow,?ZYX = 90° ? By Pythagoras theorem in triangle ZYX\({\rm{ZX}} = \sqrt {{{\left( {{\rm{ZY}}} \right)}^2} + {{\left( {{\rm{YX}}} \right)}^2}} \)\({\rm{ZX}} = {\rm{\;}}\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} \)\({\rm{ZX}} = \sqrt 2 \)In right-angle triangle ZQX,\({\rm{QX}} = \sqrt {{{\left( {{\rm{ZQ}}} \right)}^2} + {{\left( {{\rm{ZX}}} \right)}^2}} \)\({\rm{QX}} = {\rm{\;}}\sqrt {{{\left( 1 \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} \)\({\rm{ZX}} = \sqrt 3 \)Hence, "?3" is the correct answer.