4

The moment of inertia (I) of an equilateral triangle about one of its sides is given by the formula:
$$ I = \frac{1}{3} m h^2 $$
where m is the mass and h is the height of the triangle.
The height (h) of an equilateral triangle with side length 'a' is given by:
$$ h = \frac{\sqrt{3}}{2} a $$
Therefore, substituting for h in the moment of inertia formula:
$$ I = \frac{1}{3} m \left(\frac{\sqrt{3}}{2} a\right)^2 = \frac{m \cdot 3a^2}{12} = \frac{ma^2}{4} $$
If we denote the side lengths of the first and second triangles as 'a' and '2a', their moments of inertia respectively would be:
$$ I_1 = \frac{m_1 a^2}{4} \text{ and } I_2 = \frac{m_2 (2a)^2}{4} = \frac{m_2 \cdot 4a^2}{4} = m_2 a^2 $$
The ratio of the moments of inertia is:
$$ \frac{I_2}{I_1} = \frac{m_2 a^2}{\frac{m_1 a^2}{4}} = \frac{4 m_2}{m_1} $$
Assuming that the triangles are uniform, their masses are proportional to the square of the side lengths (area).
Therefore,
$$ m_2 = 4m_1 $$
Thus,
$$ \frac{I_2}{I_1} = \frac{4 \cdot 4 m_1}{m_1} = 16 $$
Therefore, the ratio is 4.