Centroid of half a pentagon - Number Crunching

We have the following coordinates for points $B, C, D$ :

$B = ( \dfrac{1}{2}, 0 )$

$C = ( \dfrac{1}{2} + \cos 72^{\circ} , \sin 72^{\circ} )$

$D = ( 0 , \sin 72^{\circ} + \sin 72^{\circ} + 180^{\circ} - 108^{\circ} ) = (0, \sin 72^{\circ} + \sin 36^{\circ})$

We'll segment the area of half the pentagon into two triangles, $\triangle OBC$ and $\triangle OCD$ ,

The centroid of $\triangle OBC$ is given by

$G_1 = (\dfrac{1}{3}) ( O + B + C )$

And, the centroid of $\triangle OCD$ is given by

$G_2 = (\dfrac{1}{3}) (O + C + D )$

The overall centroid is given by

$G = \dfrac{ [OBC] G_1 + [OCD] G_2 }{[OBC]+[OCD] }$

where, $[OBC]$ is the area of $\triangle OBC$ . $[OBC] = \dfrac{1}{2} B_x C_y$ , and $[OCD]$ is the area of $\triangle OCD$ . $[OCD] = \dfrac{1}{2} C_x D_y$

Crunching the numbers above, we get $G_x = 0.315737865, G_y = 0.68819096$ , so that $3 G_x + 4 G_y = 3.699977436 \approx 3.7$