AMC8 2019

First, we need to find the coordinates where the graphs intersect. We want the points x and y to be the same. Thus, we set $5=x+1,$ and get $x=4.$ Plugging this into the equation, $y=1-x,$ $y=5$, and $y=1+x$ intersect at $(4,5)$, we call this line x. Doing the same thing, we get $x=-4.$ Thus, $y=5$. Also, $y=5$ and $y=1-x$ intersect at $(-4,5)$, and we call this line y. It's apparent the only solution to $1-x=1+x$ is $0.$ Thus, $y=1.$ $y=1-x$ and $y=1+x$ intersect at $(0,1)$, we call this line z. Using the Shoelace Theorem we get: \[\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}\] $=$ So, our answer is $\boxed{\textbf{(E)}\ 16.}$ We might also see that the lines $y$ and $x$ are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by $-1$ to get the other. As the base is horizontal, this is an isosceles triangle with base 8, as the intersection points have a distance of 8. The height is $5-1=4,$ so $\frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}$ Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!