A triangular laminate has vertices $O = (0, 0), A = (0, a)$ and $B = (a, 0)$ and constant density $ρ_0$.
(i) Calculate the position of the centre of mass and show that the moment of inertia about the axis perpendicular to the plane of the laminate and passing through O is given by $\frac 13M a^2$, where $M$ is the mass of the laminate.
(ii) A pendulum consists of the triangular laminate which swings freely in a vertical plane about a horizontal axis through O . Show that the Lagrangian is given by $$L =\frac16Ma^2θ^2 +\frac M3\sqrt 2ga \cos θ$$ where M is the mass of the laminate and θ is an angle which should be defined.
(iii) Calculate the period of small oscillations about the downward vertical.
Can i merely state the centre of mass is given by $X=\frac a3 \ Y=\frac a3$? Would I need a double integral to find the moment of inertia?
You are right: the center of mass of a triangular lamina lies at the triangle centroid, i.e. it is the common point of its medians.
To see why, consider a thin stripe of the lamina, parallel to a side of the triangle: its center of mass is obviously at its midpoint, and it thus lies on the median corresponding to that side. The lamina can be seen as the sum of such thin stripes, whose centers of mass all lie on the same median: the center of mass of the whole lamina must therefore lie on that median too.
The same reasoning can be repeated for the other two medians, hence the center of mass of the lamina must be at the centroid of the triangle.
As for the moment of inertia, you can avoid the double integral by noticing that the moment of inertia at point $(a/2,a/2)$ is half that of a square at its center, and using then Steiner's theorem.
EDIT.
If you insist on computing the moment of inertia around $O$ with an integral, here it is: $$ I=\int_0^a\int_0^{a-x}\rho(x^2+y^2)\,dy\,dx, $$ where $\rho$ is the surface density and $M={1\over2}\rho a^2$.