#### Answer

(a) The coordinates of the center of gravity are (6 cm, 4 cm)
(b) $I = 0.0020~kg~m^2$
(c) $I = 0.0013~kg~m^2$

#### Work Step by Step

(a) We can find the x-coordinate of the center of gravity.
$x_{cog} = \frac{(100~g)(0)+(200~g)(6~cm)+(100~g)(12~cm)}{100~g+200~g+100~g}$
$x_{cog} = 6~cm$
We can find the y-coordinate of the center of gravity. Note that the y-coordinate of the 200-gram mass is 8 cm.
$y_{cog} = \frac{(100~g)(0)+(200~g)(8~cm)+(100~g)(0)}{100~g+200~g+100~g}$
$y_{cog} = 4~cm$
The coordinates of the center of gravity are (6 cm, 4 cm)
(b) We can find the moment of inertia about the axis that passes through mass A and is perpendicular to the page.
$I = \sum m_i~r_i^2$
$I = (0.10~kg)(0.10~m)^2+(0.20~kg)(0)^2+(0.10~kg)(0.10~m)^2$
$I = 0.0020~kg~m^2$
(c) We can find the moment of inertia about the axis that passes through masses B and C.
$I = \sum m_i~r_i^2$
$I = (0.10~kg)(0)^2+(0.20~kg)(0.080~m)^2+(0.10~kg)(0)^2$
$I = 0.0013~kg~m^2$