Uniform plate of constant areal mass distribution

There may be a more elegant way to do this, but here's how I did it. I chose to deal with small strips instead of treating the two large triangles as whole entities.

Steps:

1) Divide the plate into two triangles. Side lengths and triangle areas are easily found by inspection (see image).

2) Establish $(x,y)$ coordinates for the triangles as shown in the image.

3) Divide each triangle into infinitesimal strips of area $dA$ . Calculate the moment of inertia of each strip using the formula for a rod about its center $(\frac{ML^{2}}{12})$ and apply the Parallel Axis theorem to adjust for the distance from the rod's center of mass to point O. Integrate over all strips.

We will need to calculate some angles for use in the Triangle 2 calculations. From Law of Sines (solve by taking asin):

$\frac{sin{\theta_1}}{10} = \frac{1}{\sqrt{104}} \\ \frac{sin{\theta_2}}{\sqrt{52}} = \frac{1}{\sqrt{104}} \\ \theta_3 = 180 ^{\circ} - \theta_1 - \theta_2$

For Triangle 1

$y_1 = \sqrt{52} - x_1 \\ dA_1 = y_1 dx_1 = (\sqrt{52} - x_1)dx_1 \\ dm_1 = \frac{dA_1}{A_{total}}M_{total} = \frac{M_{total}}{A_{total}}(\sqrt{52} - x_1)dx_1$

$dI_1 = \frac{dm_1 \, y_1^2}{12} + dm_1(x_1^2 + (\frac{y_1}{2})^2) \\ = \frac{M_{total}}{A_{total}}(\sqrt{52} - x_1) \Big[\frac{(\sqrt{52} - x_1)^2}{12} +\big[ x_1^2 + (\frac{\sqrt{52} - x_1}{2})^2 \big] \Big] dx_1 \\ = \frac{M_{total}}{A_{total}} \Big[\frac{(\sqrt{52} - x_1)^3}{3} +\ x_1^2 (\sqrt{52} - x_1) \Big] dx_1$

I_1 = \frac{7}{36}\int_0^\sqrt{52} \Big[\frac{(\sqrt{52} - x_1)^3}{3} +\ x_1^2 (\sqrt{52} - x_1) \Big] dx_1 = \frac{2366}{27}

For Triangle 2

Calculate the horizontal and vertical distances from Point B to Point O.

$BO_y = 2 + \sqrt{52} cos \theta_3 = 6 \\ BO_x = \sqrt{52} sin \theta_3 = 6$

$y_2 = 2(1 - \frac{x_2}{10}) = 2 - \frac{x_2}{5}\\ dA_2 = y_2 dx_2 = (2 - \frac{x_2}{5})dx_2 \\ dm_2 = \frac{dA_2}{A_{total}}M_{total} = \frac{M_{total}}{A_{total}}(2 - \frac{x_2}{5})dx_2$

$dI_2 = \frac{dm_2 \, y_2^2}{12} + dm_2 \big[ (BO_x - x_2)^2 + (BO_y - \frac{y_2}{2})^2 \big] \\ = \frac{M_{total}}{A_{total}}(2 - \frac{x_2}{5}) \Big[\frac{(2 - \frac{x_2}{5})^2}{12} +\big[ (BO_x - x_2)^2 + \Big(BO_y - \frac{2 - \frac{x_2}{5}}{2}\Big)^2 \big] \Big] dx_2 \\ = \frac{M_{total}}{A_{total}}(2 - \frac{x_2}{5}) \Big[\frac{(2 - \frac{x_2}{5})^2}{12} + (6 - x_2)^2 + \Big(5 + \frac{x_2}{10}\Big)^2 \Big] dx_2$

$I_2 = \frac{7}{36} \int_0^{10} (2 - \frac{x_2}{5}) \Big[\frac{(2 - \frac{x_2}{5})^2}{12} + (6 - x_2)^2 + \Big(5 + \frac{x_2}{10}\Big)^2 \Big] dx_2 = \frac{2170}{27}$

$I_1 + I_2 = \frac{2366}{27} + \frac{2170}{27} = 168 = 84\lambda. \\ \boxed{\lambda = 2}$