Region is bounded by : $y=x+4$, $y=5$, $y=-x-4$, $x=3$
Graph is here: https://www.desmos.com/calculator/cwmlwknywe
Can I just split the shape at the x-axis, find the individual centroids, and then add them or average them? I can't seem to find an example like this in my book.
On
This problem can be simplified by reminding that the centroid of any isosceles triangle divides the height in two parts with ratio $2:1$ (with the longest part towards the vertex). In this problem we have two right isosceles triangles. The height of the large one is included between $(-4,0)$ and $(3,0)$, so the centroid is in $(-\frac{1}{2},0)$. The height of the small one is included between $(3,5)$ and $(2,6)$, so the centroid is in $(\frac{7}{3},\frac{17}{3})$.
Now simply apply the standard method of averaging the two centroid, taking into account that the two areas are $49$ and $2$ and that the area of the small triangle has to be considered negative.
If $n$ masses $m_1,\;m_2,\ldots,\,m_n$ are placed in points $(x_1,y_1),\,(x_2,y_2),\ldots,(x_n,y_n)$, respectively, then the centroid of the systems is $(\overline{x},\overline{y})$ where $$\overline{x}=\dfrac{\sum_{i=1}^{n}{m_ix_i}}{\sum_{i=1}^{n}{m_i}}\qquad\text{and}\qquad \overline{y}=\dfrac{\sum_{i=1}^{n}{m_iy_i}}{\sum_{i=1}^{n}{m_i}}$$ In the present case, if $m_1$ is the area(mass) of the big triangle and $m_2$ is the mass(area) of the little one, we have $$\overline{x}=\dfrac{m_1x_1-m_2x_2}{m_1-m_2}\qquad\text{and}\qquad \overline{y}=\dfrac{m_1y_1-m_2y_2}{m_1-m_2}$$ where $(x_1,y_1)$ and $(x_2,y_2)$ are the centroid of the triangles big and little one, respectively.