$L_1 = 12 \, \text{ in.}$

$x_1 = 0$

$y_1 = 6$

$L_2 = \sqrt{6^2 + 6^2} = 6\sqrt{2} \, \text{ in.}$

$x_2 = \frac{1}{2}(6) = 3 \, \text{ in.}$

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

$L_3 = \sqrt{6^2 + 6^2} = 6\sqrt{2} \, \text{ in.}$

$x_3 = 6 + \frac{1}{2}(6) = 9 \, \text{ in.}$

$y_3 = \frac{1}{2}(6) = 3 \, \text{ in.}$

$L_4 = 6 \, \text{ in.}$

$x_4 = 12 \, \text{ in.}$

$y_4 = 3 \, \text{ in.}$

$L_5 = \sqrt{12^2 + 6^2} = 6\sqrt{5} \, \text{ in.}$

$x_5 = \frac{1}{2}(12) = 6 \, \text{ in.}$

$y_5 = 6 + \frac{1}{2}(6) = 9 \, \text{ in.}$

$L = L_1 + L_2 + L_3 + L_4 + L_5$

$L = 12 + 6\sqrt{2} + 6\sqrt{2} + 6 + 6\sqrt{5}$

$L = 48.387 \, \text{ in.}$

$L \, \bar{x} = \Sigma lx$

$48.387\bar{x} = 12(0)+ 6\sqrt{2}(3)+ 6\sqrt{2}(9) + 6(12)+ 6\sqrt{5}(6)$

$\bar{x} = 5.256 \, \text{ in.}$ *answer*

$L \, \bar{x} = \Sigma lx$

$48.387\bar{y} = 12(6)+ 6\sqrt{2}(3)+ 6\sqrt{2}(3) + 6(3)+ 6\sqrt{5}(9)$

$\bar{y} = 5.408 \, \text{ in.}$ *answer*

## Comments

## Thank you very much for the

Thank you very much for the example of centroids for closed straight lines.