Given:1) B is 40 m to the right of D,2) A is 60 m to the south of B. 3) C is 50 m to the west of D, and4) E is 70 m to the north of A.According to the given information, we get the following figure,Therefore, the distance between C and E is: Hypotenuse of triangle CEB= CE = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGcaaWdaeaapeGaaiikaiaaiMdacaaIWaWdamaaCaaaleqabaWd % biaaikdaaaGccqGHRaWkcaGGGcGaaGymaiaaicdapaWaaWbaaSqabe % aapeGaaGOmaaaakiaacMcaaSqabaaaaa!3EC3! \sqrt {({{90}^2} + \;{{10}^2})} \) = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGcaaWdaeaapeGaaiikaiaaiIdacaaIXaGaaGimaiaaicdacqGH % RaWkcaGGGcGaaGymaiaaicdacaaIWaGaaiykaaWcbeaaaaa!3ECD! \sqrt {(8100 +100)} \)) = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGcaaWdaeaapeGaaGioaiaaikdacaaIWaGaaGimaaWcbeaaaaa!3940! \sqrt {8200} \) = 90.55= Which is greater than CB(90 m) ,Hence, the correct answer is It must be definitely greater than 90 m.