# 2019 AMC 8 Problems/Problem 21

## Problem 21

What is the area of the triangle formed by the lines $y=5$, $y=1+x$, and $y=1-x$? $\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$

## Solution 1

First we need to find the coordinates where the graphs intersect. $y=5$, and $y=x+1$ intersect at $(4,5)$, $y=5$, and $y=1-x$ intersect at $(-4,5)$, $y=1-x$ and $y=1+x$ intersect at $(0,1)$.

Using the Shoelace Theorem we get: $$\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}$$ $=$ So our answer is $\boxed{\textbf{(E)}\ 16}$.

## Solution 2

Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get $\frac{4\cdot8}{2}$ which is equal to $\boxed{\textbf{(E)}\ 16}$.

## Video Solutions

https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 