According to the Florida Building Code, what is the minimum clear width required for an accessible doorway?
Correct Answer
B) 32 inches
The Florida Building Code, following ADA guidelines, 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 stop.
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 accessibility standards. This measurement ensures wheelchair users and individuals with mobility devices can safely navigate through doorways. The 32-inch clear width is measured with the door in the open position at 90 degrees, from the face of the door to the opposite stop. This standard applies to all accessible routes and is a fundamental requirement for commercial and public buildings.
Why the Other Options Are Wrong
Option A: 30 inches
30 inches is insufficient width for wheelchair accessibility and does not meet ADA or Florida Building Code requirements for accessible doorways.
Option C: 34 inches
34 inches exceeds the minimum requirement, and while it would be acceptable, it is not the minimum standard specified in the code.
Option D: 36 inches
36 inches exceeds the minimum requirement and would be acceptable but is not the minimum clear width specified in the Florida Building Code for accessible doorways.
Memory Technique
Think '32 and through' - 32 inches gets wheelchairs through doorways safely. Also remember it's measured with door open 90 degrees, not the door width itself.
Reference Hint
Florida Building Code Chapter 11 - Accessibility, Section 1104 (Accessible Route) or ADA Standards Section 404.2.3
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