A_{\text{triangle}} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \cdot 16 = 4\sqrt{3} \text{ cm}^2

February 22, 2026

$A_{\text{triangle}} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \cdot 16 = 4\sqrt{3} \text{ cm}^2$

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$$Question: A virologist is studying the structure of a viral capsid modeled as a regular hexagonal prism with six congruent equilateral triangle faces. If each edge of the prism is 4 cm, what is the total surface area of the prism in square centimeters?
Solution: The surface area of a regular hexagonal prism consists of the area of the two hexagonal bases and the six triangular lateral faces. Each face is equilateral with side length $ s = 4 $ cm.
Area of one equilateral triangle:
Total area of six lateral triangles:
\cdot 4\sqrt{3} = 24\sqrt{3} \text{ cm}^2
Area of one regular hexagon (composed of 6 equilateral triangles):

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