Question

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:

• A. 1.5
• B. 2.0
• C. 3.0
• D. 4.0

Hint:

Handy memory: $I_{\text{rectangle}}=\frac{bd^3}{12}$, $I_{\text{triangle,centroid}}=\frac{bd^3}{36}$.

Answer 1

The Correct Option is C

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

Step 2: Take 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: Conclusion.
Required ratio $=3.0$.