Seconsett Island is a census-designated place (CDP) in the town of Mashpee in Barnstable County, Massachusetts, United States.

[2] Seconsett Island is located in the southwestern part of the town of Mashpee at 41°34′0″N 70°30′43″W / 41.56667°N 70.51194°W / 41.56667; -70.51194 (41.566734, -70.511950).

[3] It is bounded by the town of Falmouth to the northwest, by Hamblin Pond to the northeast, by the Little River to the southeast (with the Monomoscoy Island CDP on the opposite bank), and by Waquoit Bay to the southwest.

The only road access is via Meadow Neck Road from Falmouth, which is built upon a natural land bridge with Hamblin Pond to one side and Waquoit Bay to the other; the presence of the land bridge means that Seconsett Island is technically not a true island.

According to the United States Census Bureau, the CDP has a total area of 0.11 square miles (0.28 km2), of which 0.004 square miles (0.01 km2), or 4.53%, is water.