Question

For the given uniform square lamina $ABCD$, whose centre is $O$, which of the following is correct?

• A. $\sqrt{2}{I}_{AC}=I_{EF}$
• B. $I_{AD}=3I_{EF}$
• C. $I_{AC}=I_{EF}$
• D. $I_{AC}=\sqrt{2}{I}_{EF}$

Answer 1

The correct option is C $I_{AC}=I_{EF}$

Since it is a square lamina, line AC is perpendicular to line BD.
By perpendicular axis theorem,
$I_{square}=I_{AC}+I_{BD}$
Since $I_{AC}=I_{BD}$ by symmetry, $I_{square}=2I_{AC}=2I_{BD}$
$I_{square}=I_{EF}+I_{GH}$
Since $I_{EF}=I_{GH}$ by symmetry, $I_{square}=2I_{EF}=2I_{GH}$
$\therefore I_{square}=2I_{AC}=2I_{EF}$
$\Rightarrow I_{AC}=I_{EF}$