A uniform lamina ABCDE is made from a square ABDE and an equilaterial triangle BCD. Find the centre of mass of the lamina.

Explanation

Here, `A_(1)`= area of square ABDE =(a)(a) =`a^(2)`
`A_(2)` = area of triangle `=1/2(a)((sqrt(3)a)/2) = sqrt(3)/4 a^(2)`
`(x_(1),y_(1))` = co-ordiantes of centre of mass of square `=(a/2,0)`
`(x_(2),y_(2))` = co-ordinates of centre of mass of the triangle = `(a + a/(2sqrt(3)),0)`
Now, `x_(CM) =(A_(1)x_(1) + A_(2)x_(2))/(A_(1) + A_(2)) =((a^(2)).(a/2) + (sqrt(3)a^(2))/4 (a+a/(2sqrt(3))))/(a^(2) + sqrt(3)/4 a^(2)) = 0.74 a`
`Y_(CM) =0` as `y_(1)` and `y_(2)` both are zero.
Therefore co-ordinates of CM of the given lamina are (0.74a, 0).