Area of Triangles

Rewriting the sides we get $y=\frac{-x}{5}+\frac{33}{5}; y= x-9; y=-2x+3$ ;

The Area is $\int_{-2}^{4}(\frac{-x}{5}+\frac{33}{5})-(-2x+3) dx +\int_{4}^{13} (\frac{-x}{5}+ \frac{33}{5})-(x-9)dx$ ;

$= \int_{-2}^{4}(\frac{9x}{5}+\frac{18}{5})dx + \int_{4}^{13} (\frac{-6x}{5}+\frac{78}{5})dx$ =

$(\frac{9x^2}{10}+\frac{18x}{5})\large|_{-2}^{4} + (\frac{-6x^2}{10}+\frac{78x}{5})\large|_{4}^{13}= \frac {405}{5}= \boxed{81}$

Another way to find the area is to calculate the distance between the vertices using pythagoras' theorem and using those lengths to calculate the area with Heron's formula.

Draw rectangle such that A is the upper left corner, right side includes B, and bottom includes C. Rectangle is 15 * 12 = 180. Three right triangles are formed along: AB .5* 15* 3 = 22.5, BC .5* 9* 9 = 40.5, and CA .5* 6* 12 = 36. Total is 99. 180 - 99 = 81.