The centre of mass of a triangle shown in the figure has coordinates

What do you mean by mass element $dm$?

The masses and positions (in rectangular coordinates) of four particles are as follows: $1 \ kg$ at $(a, a)$,$2 \ kg$ at $(-a, a)$,$3 \ kg$ at $(-a, -a)$,and $4 \ kg$ at $(a, -a)$. The position vector of the centre of mass of the system of four particles is:

$A$ thin uniform wire is bent to form the two equal sides $AB$ and $AC$ of triangle $ABC$,where $AB = AC = 5\,cm$. The third side $BC$,of length $6\,cm$,is made from a uniform wire of twice the linear density of the first. The distance of the centre of mass from $A$ is

The distance of the centre of mass from end $A$ of a one-dimensional rod $(AB)$ having mass density $\rho = \rho_{0} \left(1 - \frac{x^{2}}{L^{2}}\right) \text{ kg/m}$ and length $L$ (in meters) is $\frac{3L}{\alpha} \text{ m}$. The value of $\alpha$ is $\ldots \ldots \ldots$ (where $x$ is the distance from end $A$).

$(n - 1)$ equal point masses each of mass $m$ are placed at the vertices of a regular $n$-polygon. The vacant vertex has a position vector $a$ with respect to the centre of the polygon. Find the position vector of the centre of mass.