Distance from Lexington Park to Dimitrovgrad
The shortest distance (air line) between Lexington Park and Dimitrovgrad is 4,885.68 mi (7,862.75 km).
How far is Lexington Park from Dimitrovgrad
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 04:45 (03.02.2025)
Dimitrovgrad is located in Pirotska oblast, Serbia within 43° 1' 0.12" N 22° 46' 59.88" E (43.0167, 22.7833) coordinates. The local time in Dimitrovgrad is 10:45 (03.02.2025)
The calculated flying distance from Dimitrovgrad to Dimitrovgrad is 4,885.68 miles which is equal to 7,862.75 km.
Lexington Park, Maryland, United States
Related Distances from Lexington Park
Dimitrovgrad, Pirotska oblast, Serbia