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According to the Florida Building Code, what is the minimum door width required for an accessible entrance?

Correct Answer

B) 32 inches

The Florida Building Code requires a minimum clear width of 32 inches for accessible doorways. This measurement is taken with the door open 90 degrees, measured between the face of the door and the opposite stop.

Answer Options
A
30 inches
B
32 inches
C
34 inches
D
36 inches

Why This Is the Correct Answer

The Florida Building Code requires a minimum clear width of 32 inches for accessible doorways to comply with ADA standards. This measurement ensures wheelchair users and individuals with mobility devices can pass through comfortably. The 32-inch requirement is measured as clear width when the door is open 90 degrees, between the face of the door and the opposite stop. This standard is consistent with federal accessibility guidelines and is strictly enforced in Florida construction.

Why the Other Options Are Wrong

Option A: 30 inches

30 inches is insufficient for accessibility compliance and does not meet the minimum clear width requirements established by the Florida Building Code for accessible entrances.

Option C: 34 inches

34 inches exceeds the minimum requirement, and while it would be acceptable, it is not the minimum standard specified by the code.

Option D: 36 inches

36 inches exceeds the minimum requirement and while it would provide better accessibility, it is not the minimum standard required by the Florida Building Code.

Memory Technique

Remember '32 for ADA' - the number 32 is the key accessibility door width that appears in most building codes nationwide.

Reference Hint

Florida Building Code, Chapter 11 - Accessibility, Section 1133 (Doors and Doorways) or Florida Accessibility Code for Building Construction

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