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(C) Let the vertices of the equilateral triangle be $A$,$B$,and $C$. Let the axis of rotation be the side $AB$.The particles at $A$ and $B$ lie on the

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$\frac{3}{4} m a^2$

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(C) Let the vertices of the equilateral triangle be $A$,$B$,and $C$. Let the axis of rotation be the side $AB$.The particles at $A$ and $B$ lie on the axis of rotation,so their perpendicular distances from the axis are $0$. Thus,their contribution to the moment of inertia is $0$.The particle at $C$ is at a perpendicular distance $x$ from the side $AB$. In an equilateral triangle of side $a$,the altitude $x$ is given by $x = a \sin(60^\circ) = \frac{\sqrt{3}}{2} a$.The moment of inertia of the system about side $AB$ is $I = \sum m_i r_i^2 = m(0)^2 + m(0)^2 + m(x)^2$.$I = m x^2 = m \left( \frac{\sqrt{3}}{2} a \right)^2 = m \left( \frac{3}{4} a^2 \right) = \frac{3}{4} m a^2$.

Three particles of mass $m$ each are placed at the three vertices of an equilateral triangle. The length of each side of the triangle is $a$. The moment of inertia of this system about any side of the triangle is:

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