For a triangular cross-section, what is the ratio of its second moment of area about the base to that about a centroidal axis parallel to the base?

Introduction / Context:
Second moments of area (area moments of inertia) are key for calculating bending stresses and deflections. For standard shapes, knowing ratios between base and centroidal axes helps quick checks and derivations.

Given Data / Assumptions:

• Right triangle with base b and altitude h.
• Axes considered: about the base, and through centroid parallel to the base.

Concept / Approach:
For a triangle, the formulae are I_base = b h^3 / 12 and I_centroidal = b h^3 / 36 for the axis parallel to the base through the centroid. The ratio I_base / I_centroidal = (1/12) / (1/36) = 3.0.

Step-by-Step Solution:
Recall I_base = b h^3 / 12.Recall I_centroidal = b h^3 / 36.Compute ratio ⇒ (b h^3 / 12) / (b h^3 / 36) = 36 / 12 = 3.0.

Verification / Alternative check:
Use the parallel-axis theorem: I_base = I_centroidal + A d^2 with d = h/3. Substituting A = b h / 2 gives I_base = b h^3 / 36 + (b h / 2) * (h/3)^2 = b h^3 / 36 + b h^3 / 18 = b h^3 / 12, confirming the ratio 3.0.

Why Other Options Are Wrong:

• 1.0, 1.5, 2.0, 2.5 do not satisfy the known formulae for triangular sections.

Common Pitfalls:

• Mixing up centroid location (h/3 from the base) leading to incorrect parallel-axis distances.

Final Answer:
3.0.