The co-ordinates of center of mass of letter E which is cut from a uniform metal sheet are (Take origin at bottom left corner and width of letter 2 cm every where)

Density is same

${\mathrm{m}}_1=10 \times 2\times \mathrm{d}=20\mathrm{d}$

${\mathrm{m}}_2=6\times 2\times \mathrm{d}=12\mathrm{d}$

${\mathrm{m}}_3=2\times 2\times \mathrm{d}=4\mathrm{d}$

${\mathrm{m}}_4=6\times 2\times \mathrm{d}=12\mathrm{d}$

$\left({\mathrm{x}}_1, {\mathrm{y}}_1\right)=\left(1,5\right) \left({\mathrm{x}}_3, {\mathrm{y}}_3\right)=\left(3,5\right)$

$\left({\mathrm{x}}_2, {\mathrm{y}}_2\right)=\left(5,1\right) \left({\mathrm{x}}_4, {\mathrm{y}}_4\right)=\left(5,9\right)$

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2+{\mathrm{m}}_3{\mathrm{x}}_3+{\mathrm{m}}_4{\mathrm{x}}_4}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+{\mathrm{m}}_4}$

$=\frac{\mathrm{d}(20+60+12+60)}{\mathrm{d}\left(48\right)}=\frac{152}{48}=3.3\mathrm{cm}$

${\mathrm{y}}_{\mathrm{cm}}=\frac{\mathrm{d}\left(100+12+20+108\right)}{\mathrm{d}\left(20+12+4+12\right)}$

$=\frac{\mathrm{d}\left(240\right)}{\mathrm{d}\left(48\right)}=\frac{240}{48}=5$