Q.436. Centre of area

Question 436.
Locate the centroid (x_c, y_c)
of the shaded area?
Given:  a= 1 in,   b= 6 in,   c= 3 in,   d= 3 in

Answer 436. Refer to the link: http://en.wikipedia.org/wiki/List_of_cen…

Area of triangle = (1/2) cd
Center of triangle = (b + c/3, d/3)

Area of rectangle = bd
Center of rectangle = (b/2, d/2)

Area of quarter circle = πd^2/4
Center of quarter circle = ( - 4d/3π, 4d/3π)

Area of semicircular hole = πa^2/2
Center of semicircular hole = (0, 4a/3π)

=> x_c
= [(1/2)cd * (d/3) + bd * d/2 + (πd^2/4) * (- 4d/3π) - (πa^2/2) * 0] / [(1/2)cd + bd + πd^2/4 - πa^2/2]
= [(9/2) * 7 + 18 * 3 + (7.0685775) * (- 1.27324) - (1.5708) *0] / [(9/2) + 18 + (7.0685775) - (1.5708)]
= (31.5 + 54 - 9 ) / (4.5 + 18 - 5.5)
= (76.5) / (28)
= 2.732.
and y_c
= [(1/2)cd * (b + c/3) + bd * b/2 - (πd^2/4) * (4d/3π) - (πa^2/2) * (4a/3π)] / [(1/2)cd + bd + πd^2/4 - πa^2/2]
= [(9/2) * 1 + 18 * 1.5 + (7.0685775) * (1.27324) - (1.5708) * (0.42441)] / [(9/2) + 18 + (7.0685775) - (1.5708)]
= (4.5 + 27 + 9 - 0.67) / (4.5 + 18 - 5.5)
= (39.83) / (28)
= 1.423.

Link to YA!