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The radius of the original lumen is:
r = d/2 = 5.0mm/2 = 2.5mm
The area of the original lumen is:
A = πr^2 = 3.14 x (2.5mm)^2 = 19.625mm²
The fatty deposit covers 65% of the original area, so the available area for blood flow is:
A_available = (100% - 65%) x A = 35% x 19.625mm² = 6.857mm²
Therefore, the area in mm² of the lumen that is available for blood flow is approximately 6.857mm².