Distance from Lexington Park to Aleksandrow Kujawski
The shortest distance (air line) between Lexington Park and Aleksandrow Kujawski is 4,365.13 mi (7,025.00 km).
How far is Lexington Park from Aleksandrow Kujawski
Lexington Park is located in Maryland, United States within 38° 15' 10.08" N -77° 33' 27.36" W (38.2528, -76.4424) coordinates. The local time in Lexington Park is 13:22 (11.03.2025)
Aleksandrow Kujawski is located in Włocławski, Poland within 52° 52' 36.12" N 18° 41' 36.96" E (52.8767, 18.6936) coordinates. The local time in Aleksandrow Kujawski is 19:22 (11.03.2025)
The calculated flying distance from Aleksandrow Kujawski to Aleksandrow Kujawski is 4,365.13 miles which is equal to 7,025.00 km.
Lexington Park, Maryland, United States
Related Distances from Lexington Park
Aleksandrow Kujawski, Włocławski, Poland