Question

I is the Moment of inertia of $△$ABC (equilateral) about and an axis passing through O (D,E,F are midpoints). And point O is the center of $△$ABC. $I_o$ is a moment of inertia of a remaining triangle when $△$DEF is cut from $△$ABC.

• A. $\frac{15}{16}{I}_o$
• B. $\frac{3}{4}{I}_o$
• C. $\frac{I_o}{4}$
• D. $\frac{16}{15}{I}_o$

Answer 1

The correct option is A $\frac{15}{16}{I}_o$

According to perpendicular Axis theorem-

$I_x=I_x+I_y$

Here, $I_0=m{R}^2$

Moment of inertia of cavity (DEF)

$I_{DEF}=\frac{m}{4}{\left(\frac{1}{2}\right)}^2=\frac{m{R}^2}{16}$

$I_{DEF}=\frac{I_0}{16}$

Therefore, remaining part$=I_{remain}=I_0-\frac{I_0}{16}=\frac{15}{16}{I}_0$