# From a square sheet of uniform density, one of the four portions within its diagonals is removed .. Find the of centre of mass of a remaining portion if the side of the square is a. The portion removed is right one

Lets find out the com of the triangle first.
Let the density be $\rho$
The center of the squre is taken as the origin
The x coordinate of the CoM is
${x}_{cm}=\frac{{\int }_{0}^{a/2}2\rho xx\mathrm{tan}\theta dx}{\rho \frac{{a}^{2}}{4}}=\frac{8}{{a}^{2}}\mathrm{tan}\left({45}^{0}\right){\int }_{0}^{a/2}{x}^{2}dx=\frac{8}{{a}^{2}}\frac{{\left(a}{2}\right)}^{3}}{3}=\frac{a}{3}$
Now using symmetry ${y}_{cm}=0$
Similarly, the coordinates of the other three triangles are (-a/3,0), (0,a/3),(0,-a/3)
So the CoM of the remaining square is
$\left({X}_{cm},{Y}_{cm}\right)=\left(\frac{-\frac{a}{3}+0+0}{3},\frac{0+\frac{a}{3}-\frac{a}{3}}{3}\right)\phantom{\rule{0ex}{0ex}}⇒\left({X}_{cm},{Y}_{cm}\right)=\left(-\frac{a}{9},0\right)$

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