The moment of inertia of a triangular section (height h, base b) about its base, is
\( \frac { bh^2 } { 12 }\)
\( \frac { b^2h} { 12 }\)
\( \frac { bh^3 } { 12 }\)
\( \frac { b^3h} { 12 }\)
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Answer : Option C
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