Using Vectors With Moments

Taking moments can be tricky for compound shapes when the distance to the centre of some shape can be the sum of several distances, each the side of some triangle drawn on a diagram. There is an easier way. Resolve the weight of each area n some way to use the symmetry of the problem.
Example: Find the reaction of the wall on the compound shape below.

The weights are resolved parallel to the sides of the shape.

Take moments about O. Since the lamina is uniform density  
\[\rho\]
,  
\[w= \rho , \; W= 12 \rho\]
Clockwise moments are
\[1.5Wcos60=0.75 \rho\]

\[2.5wcos60=1.25 \rho\]

\[X \times 4sin 60=2X\]

Clockwise moments are
\[2Wcos30= 12 \rho \sqrt{3}\]

\[4.5wcos30=2.25 \rho \sqrt{3}\]

Equating clockwise and anticlockwise moments gives
\[0.75 \rho +1.25 \rho +2X=12 \rho \sqrt{3}+ 2.25 \rho \sqrt{3}\]

Solving for  
\[X\]
:  
\[X=(7.125 \sqrt{3} +1) \rho\]
.