Question:medium

A rectangular section has width $b$ and depth $d$ and a symmetrical triangular section has base $b$ and depth $d$. The ratio of moment of inertia of rectangular section to that of triangular section with respect to their respective centroidal axis will be:

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Handy memory: $I_{\text{rectangle}}=\frac{bd^3}{12}$, $I_{\text{triangle,centroid}}=\frac{bd^3}{36}$.
Updated On: Feb 18, 2026
  • 1.5
  • 2.0
  • 3.0
  • 4.0
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The Correct Option is C

Solution and Explanation

Step 1: Apply standard centroidal $I$-formulae.
For a rectangle about its centroidal axis parallel to the base: \[I_{\text{rect}}=\frac{b d^{3}}{12}.\]For a triangle (base $b$, depth $d$) about its centroidal axis parallel to the base:\[I_{\text{tri}}=\frac{b d^{3}}{36}.\]

Step 2: Calculate the ratio.
\[\frac{I_{\text{rect}}}{I_{\text{tri}}}=\frac{\frac{b d^3}{12}}{\frac{b d^3}{36}}=\frac{1/12}{1/36}=3.\]

Step 3: State the conclusion.
The required ratio is $3.0$.

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