jmc

The vertices of the triangle are the points where two of the lines intersect. The line $y=\frac{1}{2}x+3$ intersects $y=1$ when $$\frac{1}{2}x+3=1\Rightarrow x=-4.$$ The line $y=-2x+6$ intersects $y=1$ when $$-2x+6=1\Rightarrow x=\frac{5}{2}.$$ The line $y=\frac{1}{2}x+3$ intersects $y=-2x+6$ when $$\frac{1}{2}x+3=-2x+6\Rightarrow x=\frac{6}{5}.$$ and $$y=-2\left(\frac{6}{5}\right)+6=\frac{18}{5}$$

Thus the vertices of the triangle are $(-4,1)$, $\left(\frac{5}{2},1\right)$, and $\left(\frac{6}{5},\frac{18}{5}\right)$. We can let the base of the triangle lie along the line $y=1$. It will have length $$4+\frac{5}{2}=\frac{13}{2}.$$ The altitude from $\left(\frac{6}{5},\frac{18}{5}\right)$ to this line will have length $$\frac{18}{5}-1=\frac{13}{5}.$$ Thus the area of the triangle is $$\frac{1}{2}\ast \frac{13}{2}\ast \frac{13}{5}=\frac{169}{20}=\boxed{8.45}.$$