Set up the geometry

We have an equilateral triangle $EFG$ of side $a$, with masses $m$ at each vertex $E, F, G$.

The axis is the line $EX$:

• passes through $E$
• is perpendicular to $EG$
• lies in the plane of the triangle.

So the moment of inertia is $I = \sum m r_\perp^2$ where $r_\perp$ is the perpendicular distance of each mass from the line $EX$.

Choose coordinates

Let us place the triangle in the plane as follows: $E=(0,0), \qquad G=(a,0)$ Since the triangle is equilateral, $F=\left(\frac a2, \frac{\sqrt{3}a}{2}\right).$

Because $EG$ is along the $x$-axis, the line through $E$ perpendicular to $EG$ is the $y$-axis. Hence the axis $EX$ is simply the line $x=0.$

Perpendicular distances from the axis $EX$

• Mass at $E=(0,0)$ lies on the axis, so distance is $r_E=0.$

• Mass at $G=(a,0)$ has distance from $x=0$ equal to $r_G=a.$

• Mass at $F=\left(\frac a2, \frac{\sqrt{3}a}{2}\right)$ has distance from $x=0$ equal to $r_F=\frac a2.$

Compute moment of inertia

$I = m(0)^2 + m\left(\frac a2\right)^2 + m(a)^2$ $I = m\frac{a^2}{4} + ma^2$ $I = \frac{5}{4}ma^2.$

Match with given form

Given $I = \frac{N}{20}ma^2.$

So, $\frac{N}{20} = \frac{5}{4}$ $N = 20\cdot \frac{5}{4} = 25.$

Final answer

$\boxed{N=25}$