Distance from Seven Oaks, South Carolina to Lexington, South Carolina

Seven Oaks Lexington
Total Distance
11.1 mi
(17.9 km)
Travel Time
20 minutes

The driving distance from Seven Oaks, South Carolina to Lexington, South Carolina is 11.1 mi (17.9 km)

The straight-line (air) distance is approximately 4.3 mi (6.9 km)

Estimated travel time by road is 20 minutes

We found 2 routes between Seven Oaks and Lexington.

Available Routes

Shortest Fastest
I-20 W
10.5 mi
16.8 km
15 minutes
Bush River Rd/State Rd S-32-107 and SC-6 E/N Lake Dr
11.1 mi
17.9 km
20 minutes

Route Directions

Follow Banbury Rd to St Andrews Rd
0.3 mi
0 min
Take Bush River Rd/State Rd S-32-107 and N Lake Dr to E Main St in Lexington
10.8 mi
19 min

Route Directions

Get on I-20 W from Bush River Rd
2.4 mi
5 min
Follow I-20 W to US-1 in Lexington. Take exit 58 from I-20 W
5.6 mi
5 min
2.5 mi
5 min

Journey Details

Traveling from Seven Oaks, South Carolina to Lexington, South Carolina covers 10.5 miles (17 km) in a southwest direction. This mixed highways and local roads, short route typically takes 15 minutes under normal conditions.

Drivers have 2 route options between these locations (10-11 miles / 17-18 km). The primary route shown here saves 5 minutes compared to alternatives.

Journey Waypoints

Seven Oaks

Starting Point

Grenadier, South Carolina

10 mi to destination

1 mi
Twelvemile Creek, South Carolina

7 mi to destination

1 mi
Lexington

Destination

Destination
Lexington
Lexington
Address
Lexington · South Carolina · US
Coordinates
33.981537, -81.236211
Timezone
America/New_York
Local Time
9:12 PM - Wednesday, May 7
Origin
Address
Seven Oaks · South Carolina · US
Coordinates
34.048762, -81.146484
Timezone
America/New_York
Local Time
9:12 PM - Wednesday, May 7
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