$A_1 = 12(12) = 144 \, \text{ in.}^2$

$x_1 = \frac{1}{2}(12) = 6 \, \text{ in.}$

$y_1 = \frac{1}{2}(12) = 6 \, \text{ in.}$

$A_2 = \frac{1}{2}(12)(6) = 36 \, \text{ in.}^2$

$x_2 = \frac{1}{2}(12) = 6 \, \text{ in.}$

$y_2 = \frac{1}{3}(6) = 2 \, \text{ in.}$

$A_3 = \frac{1}{2}(12)(6) = 36 \, \text{ in.}^2$

$x_3 = \frac{2}{3}(12) = 8 \, \text{ in.}$

$y_3 = 6 + \frac{2}{3}(6) = 10 \, \text{ in.}$

$A = A_1 - A_2 - A_3$

$A = 144 - 36 - 36$

$A = 72 \, \text{ in.}^2$

$A \, \bar{x} = \Sigma ax$

$72\bar{x} = 144(6)- 36(6)- 36(8)$

$\bar{x} = 5 \, \text{ in.}$ *answer*

$A \, \bar{y} = \Sigma ay$

$72\bar{y} = 144(6)- 36(2)- 36(10)$

$y = 6 \, \text{ in.}$ *answer*

## Comments

## for x2, why is it (1/2)(12)

for x2, why is it (1/2)(12) and not (2/3)(12)?

## Based on the formula for

Based on the formula for centroids, solving for x in an equilateral triangle will only equate to zero if it is placed along the origin. Therefore, we are just going to figure out the distance of the triangle from the origin.