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(C) The triangle $ABC$ has sides $AB=3 \,cm$,$AC=4 \,cm$,and $BC=5 \,cm$. Since $3^2 + 4^2 = 5^2$,it is a right-angled triangle with the right angle a

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$140 \,g$ at $B$

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(C) The triangle $ABC$ has sides $AB=3 \,cm$,$AC=4 \,cm$,and $BC=5 \,cm$. Since $3^2 + 4^2 = 5^2$,it is a right-angled triangle with the right angle at $A$.Let the mass of the plate be $M = 540 \,g$. The weight of the plate acts through its centroid $G$.To keep the side $BC$ horizontal,the torque about the pivot $A$ due to the weight of the plate must be balanced by the torque due to an additional mass $m_1$ placed at a vertex.Let $AE$ be the altitude from $A$ to $BC$. In $	riangle ABC$,$AE = (AB 	imes AC) / BC = (3 	imes 4) / 5 = 2.4 \,cm$.$BE = \sqrt{AB^2 - AE^2} = \sqrt{3^2 - 2.4^2} = \sqrt{9 - 5.76} = \sqrt{3.24} = 1.8 \,cm$.$EC = BC - BE = 5 - 1.8 = 3.2 \,cm$.The horizontal distance of the centroid $G$ from the vertical line passing through $A$ is the horizontal distance of $G$ from $AE$,which is $GH$. Since $G$ is the centroid,its distance from $BC$ is $1/3$ of the altitude $AE$,and its horizontal distance from $AE$ is $1/3$ of the distance of the centroid of the base $BC$ from $E$. The centroid $G$ divides the median $AD$ in ratio $2:1$. The horizontal distance $GH = (1/3) 	imes (EC - BE) / 2 = (1/3) 	imes (3.2 - 1.8) / 2 = 1.4 / 6 = 7/30 \,cm$.For equilibrium,$M 	imes g 	imes GH = m_1 	imes g 	imes BE$.$540 	imes (7/30) = m_1 	imes 1.8$.$18 	imes 7 = m_1 	imes 1.8$.$126 = 1.8 	imes m_1 \Rightarrow m_1 = 126 / 1.8 = 70 \,g$. Wait,re-evaluating the geometry: The horizontal distance of $G$ from $A$ is the projection of the vector $\vec{AG}$ onto the horizontal. The horizontal distance of $G$ from the vertical line through $A$ is $GH = (1/3) 	imes (BE + EC)/2$ is incorrect. The horizontal distance of $G$ from $A$ is $1/3$ of the horizontal distance of $B$ and $C$ from $A$. The horizontal distance of $B$ from $A$ is $BE = 1.8 \,cm$ and $C$ from $A$ is $EC = 3.2 \,cm$. The centroid $G$ is at a horizontal distance $(1.8 - 3.2)/3 = -1.4/3$ from $A$. To balance,we place $m_1$ at $B$. Torque balance: $M 	imes (1.4/3) = m_1 	imes 1.8 \Rightarrow 540 	imes 1.4 / 3 = m_1 	imes 1.8 \Rightarrow 180 	imes 1.4 = m_1 	imes 1.8 \Rightarrow 252 = 1.8 	imes m_1 \Rightarrow m_1 = 140 \,g$ at $B$.

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