What is the area of the triangle formed by the lines $y=5$, $y=1+x$, and $y=1-x$?
$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$
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](https://artofproblemsolving.com/wiki/index.php/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 a isosceles triangle with base 8, as the intersection points have distance 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!