The ABC is a triangular plate of uniform thickness. The sides are in the ratio shown in the figure. I_AB, I_BC and I_CA are the moments of inertia of the plate about AB, BC and CA respectively. Which one of the following relations is correct?

Question

Accepted Answer

(4) I_BC>I_AB. Given: a uniform triangular plate with AB=4, BC=3, CA=5, so the angle at B is a right angle. Idea: for a uniform triangular plate the moment of inertia about a side is I=(Mh²)/6, where h is the perpendicular distance from the opposite vertex to that side. About AB: the opposite vertex is C, and BC⊥ AB, so h=3 and I_AB=(9M)/6=1.5M. About BC: the opposite vertex is A, and AB⊥ BC, so h=4 and I_BC=(16M)/6=2.67M. About CA: h=(AB× BC)/(CA)=(4×3)/5=2.4, so I_CA=(5.76M)/6=0.96M. So I_BC>I_AB>I_CA, and the correct relation is I_BC>I_AB.

The ABC is a triangular plate of uniform thickness. The sides are in the ratio shown in the figure. I_AB, I_BC and I_CA are the moments of inertia of the plate about AB, BC and CA respectively. Which one of the following relations is correct?

Step-by-step solution

Why the other options are wrong