2019 AMC 10A Problem 7

The two lines through $(2,2)$ have equations $$\begin{aligned} y&=\frac12x+1,\\ y&=2x-2. \end{aligned}$$ Their intersections with $x+y=10$ are $(6,4)$ and $(4,6)$, respectively. Thus the vertices are $(2,2),(6,4),$ and $(4,6)$. The segment joining the last two points has length $2\sqrt2$, its midpoint is $(5,5)$, and the distance from $(2,2)$ to that midpoint is $3\sqrt2$. The area is therefore $$\frac12(2\sqrt2)(3\sqrt2)=6.$$ Thus, C is the correct answer.