For an equilateral triangle, the distance \( r \) from the center of each edge to the centroid \( C \) is:
\[r = \frac{1}{\sqrt{3}}\]
The moment of inertia \( I \) about point \( C \) and perpendicular to the plane is given by:
\[I = r^2 \left[ 2 + 4 + 6 \right]\]
Substitute \( r = \frac{1}{\sqrt{3}} \):
\[I = \left( \frac{1}{\sqrt{3}} \right)^2 \times 12\]
\[I = \frac{1}{3} \times 12 = 4 \, \text{kg} \cdot \text{m}^2\]
For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through \( O \) (the center of mass) and \( O' \) (corner point) is:
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: