When planning a telecommunication termination area, what is the estimated space needed to terminate 144 strands of optical fiber?

Question

Options

• A6350.01 mm2 (250 in2)
• B7315.21 mm2 (288 in2)
• C8890.01 mm2 (350 in2)
• D10972.82 mm2 (432 in2)
• E14376.43 mm2 (566 in2)

• A
  4% (1)
• B
  82% (23)
• D
  11% (3)
• E
  4% (1)

Explanation

Option B is correct because industry standards (BICSI TDMM) allocate 2 square inches per fiber strand for termination space planning. Multiplying 144 strands × 2 in² yields exactly 288 in² (7315.21 mm²), making B the precise match.

The distractors are wrong because they correspond to different per-strand ratios that don't match the standard: A (250 in²) falls short of the 288 in² requirement; C (350 in²) overshoots by roughly 62 in²; D (432 in²) would correspond to a 3 in² per strand allocation; and E (566 in²) is nearly double the required space, more appropriate for a much larger fiber count or additional equipment.

Memory tip: Think "2 square inches per fiber" - for 144 strands, pair each fiber with 2 in², then group them: 144 × 2 = 288. If you remember that a standard fiber patch panel row holds 12 fibers × 2 in² = 24 in² per row, 144 fibers = 12 rows × 24 in² = 288 in².

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