`ABC` is a traiangular 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` repectively. Which one of the following relations is correct?

Options

A. `I_(CA)` is maximum
B. `I_(AB)gtI_(BC)`
C. `I_(BC)gtI_(AB)`
D. `I_(AB)+I_(BC)=I_(CA)`

Explanation

`M.I.` of a triangular plate about its base

`I_(base) =I_(1-1)=(mh^(2))/(6)`
For the given problem
`I_(AB)=(m(BC)^(2))/(6)=(m(3)^(2))/(6)=(3)/(2)m=1.5m unit`
`I_(BC)=(m(AB)^(2))/(6)=(m(4)^(2))/(6)=(8)/(3)m=2.67m unit`
For `AC`

`sintheta=(4)/(5)=(h)/(3)implies h=(12)/(5)=2.4`
`I_(AC)=(m(BD)^(2))/(6)=(m(2.4)^(2))/(6)=0.96 unit`
`I_(BC)gtI_(AB)gtI_(CA)`
OR
By observation, distribution of mass is the farthest from `BC`.