Principal MI of 15 cm right triangle

Right angle at the origin O, legs of 15 cm along the X and Y axes (so $b = h = 15$ cm). The principal axes are required for the X–Y axes at O.

 Y
15|\
  | \
  |  \
  O---- 15  X

Moments and product of inertia about X–Y

$$\begin{aligned} I_x &= \frac{bh^3}{12} = \frac{15\times15^3}{12} = 4218.750\ \text{cm}^4\\ I_y &= \frac{hb^3}{12} = 4218.750\ \text{cm}^4\\ I_{xy} &= \frac{b^2h^2}{24} = \frac{15^2\times15^2}{24} = 2109.375\ \text{cm}^4 \end{aligned}$$

Principal moments of inertia

$$\begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} = 4218.750\pm\sqrt{0 + 2109.375^2}\\ I_{max} &= 6328.125\ \text{cm}^4,\qquad I_{min} = 2109.375\ \text{cm}^4 \end{aligned}$$

Principal axes

$$\tan2\theta_p = \frac{-2I_{xy}}{I_x-I_y} = \frac{-2(2109.375)}{0} = -\infty\;\Rightarrow\;2\theta_p = -90^\circ\ \text{or}\ 90^\circ$$

so $\theta_p = 45^\circ$ and $135^\circ$. Checking with $I(\theta) = \frac{I_x+I_y}{2} + \frac{I_x-I_y}{2}\cos2\theta - I_{xy}\sin2\theta$: at $\theta = 135^\circ$ it is $I = 4218.750 - 2109.375\sin270^\circ = 6328.125$ (maximum), and at $\theta = 45^\circ$ it is $2109.375$ (minimum).

Answer: $I_{max} = 6328.12\ \text{cm}^4$ about the axis through O at $135^\circ$ to X (parallel to the hypotenuse), and $I_{min} = 2109.38\ \text{cm}^4$ about the axis through O at $45^\circ$ to X (the line from O to the mid-point of the hypotenuse).