Figure shows triangular lamina which can rotate about different axes. The moment of inertia is maximum about the axis
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To maximize the moment of inertia of any planar lamina, choose the axis that is located as far away from the center of mass as possible, or an edge that is directly opposite the most distant vertex. The further the mass stretches from the line of rotation, the larger the $r^2$ term becomes.
Step 1: Recall what moment of inertia depends on. The moment of inertia about an axis is $I = \sum m_i r_i^2$, where $r_i$ is the perpendicular distance of each mass element from that axis. The more mass that sits far from the axis, the larger $I$ becomes. Step 2: Set the deciding rule. Among the candidate axes, the one that keeps the bulk of the lamina's mass at the greatest perpendicular distance will give the maximum $I$. So we look for the axis from which the material reaches out the furthest. Step 3: Compare central axes with edge axes. An axis that runs through the interior of the triangle (such as a median or altitude like $QS$) has mass on both sides, much of it close to the line, so its $I$ stays small. Step 4: Examine the edge axes $PQ$ and $QR$. An axis lying along an edge keeps all the mass on one side. The whole triangle then leans away from that edge toward the opposite vertex, increasing the average distance. Step 5: Pick the best edge. For the given lamina, the edge $QR$ is the one for which the opposite vertex $P$ - and the mass spread toward it - lies at the largest perpendicular distance. That maximises $\sum m_i r_i^2$. Step 6: Conclude. Hence the lamina has its greatest moment of inertia about the axis $QR$. \[ \boxed{\text{Maximum about axis } QR} \]
