# Difference between revisions of "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.

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 $(-4,5)$, 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 $z$ 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!

## 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

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